Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
inequality $(a+c)(a+b+c)<0$, prove $(b-c)^2>4a(a+b+c)$
If $(a+c)(a+b+c)<0,$
prove $$(b-c)^2>4a(a+b+c)$$
I will use the constructor method that want to know can not directly prove it?
| Consider the quadratic
$$ f(x) = ax^2 - (b-c)x + (a+b+c) $$
$$f(1)f(0) = 2(a+c)(a+b+c) \lt 0$$
Thus if $a \neq 0$, then this has a real root in $(0,1)$ and so
$$(b-c)^2 \ge 4a(a+b+c)$$
If $(b-c)^2 = 4a(a+b+c)$, then we have a double root in $(0,1)$ in which case, $f(0)$ and $f(1)$ will have the same sign.
Thus $$(b-c)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$ I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f... | After doing some computations I found the following (lets hope I didn't make any mistakes). You need to solve the equations
$$f_x = 2x + y - \frac{1}{x^2} = 0 \quad f_y = 2y + x -\frac{1}{y^2} = 0$$
therefore after subtracting and adding them as in the hint we get
$$\begin{align}
f_x - f_y &= x - y - \frac{1}{x^2} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| You deal with the sum of functions, $f(x) = \frac{x}{2}$ and $g(x)= \frac{1}{4} \sin(2 x)$. So you would use linearity of the derivative:
$$
\frac{d}{d x} \left( f(x) + g(x) \right) = \frac{d f(x)}{d x} + \frac{d g(x)}{d x}
$$
To evaluate these derivatives, you would use $\frac{d}{d x}\left( c f(x) \right) = c \frac{... | {
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"timestamp": "2023-03-29T00:00:00",
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Is my Riemann Sum correct (Example # 2)?
Possible Duplicate:
Is my Riemann Sum correct?
This is my second attempt, the answer seems rather odd so I thought I would have it checked as well.
For the integral: $$\int_{-5}^{2} \left( x^{2} -4 \right) dx$$
My calculations:
$$\begin{align*}\Delta x &= \frac7n\\\\
x_i &= -... | The preliminary computations are fine. That means that the $n$th right hand Riemann sum will be:
$$\begin{align*}
\text{RHS} &= \sum_{i=1}^n f(x_i)\Delta x\\
&= \sum_{i=1}^n\left(21 - \frac{70i}{n} +\frac{49i^2}{n^2}\right)\frac{7}{n}\\
&= \frac{7(21)}{n}\sum_{i=1}^n1 - \frac{7(70)}{n^2}\sum_{i=1}^n i + \frac{7(49)}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/142015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How can one solve the equation $\sqrt{x\sqrt{x} - x} = 1-x$? $$\sqrt{x\sqrt{x} - x} = 1-x$$
I know the solution but have no idea how to solve it analytically.
| Just writing out Robert's manipulation:
$$\eqalign{
& \sqrt {x\sqrt x - x} = 1 - x \cr
& x\sqrt x - x = {\left( {1 - x} \right)^2} \cr
& x\left( {\sqrt x - 1} \right) = {\left( {1 - x} \right)^2} \cr
& \sqrt x - 1 = \frac{{{{\left( {1 - x} \right)}^2}}}{x} \cr
& \sqrt x = \frac{{{{\left( {1 - x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Alternative proof of the limitof the quotient of two sums. I found the following problem by Apostol: Let $a \in \Bbb R$ and $s_n(a)=\sum\limits_{k=1}^n k^a$. Find
$$\lim_{n\to +\infty} \frac{s_n(a+1)}{ns_n(a)}$$
After some struggling and helpless ideas I considered the following solution.
If $a > -1$, then
$$\int_0^1... | The argument below works for any real $a > -1$.
We are given that
$$s_n(a) = \sum_{k=1}^{n} k^a$$ Let $a_n = 1$ and $A(t) = \displaystyle \sum_{k \leq t} a_n = \left \lfloor t \right \rfloor$. Hence, $$s_n(a) = \int_{1^-}^{n^+} t^a dA(t)$$ The integral is to be interpreted as the Riemann Stieltjes integral. Now integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/150059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
An alternating series ... Find the limit of the following series:
$$ 1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{11} - \frac{1}{14} + \cdot \cdot \cdot $$
If i go the integration way all is fine for a while but then things become pretty ugly. I'm trying to find out if there is some easier way to follow.
| Let $S = 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots$. Then what you want is $\int_{0}^{1} S \ dx$. But we have
\begin{align*}
S &= 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots \\\ &= -(x^{3}+x^{8} + x^{13} + \cdots) + (1+x^{5} + x^{10} + \cdots) \\\ &= -\frac{x^{3}}{1-x^{5}} + \frac{1}{1-x^{5}}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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What is the Taylor series for $g(x) =\frac{ \sinh(-x^{1/2})}{(-x^{1/2})}$, for $x < 0$?
What is the Taylor series for $$g(x) = \frac{\sinh((-x)^{1/2})}{(-x)^{1/2}}$$, for $x < 0$?
Using the standard Taylor Series:
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}$$
I substituted in $x = x^{1/2}$, since... | As Arturo pointed out in a comment, It has to be $(-x)^{\frac{1}{2}}$ to be defined for $x<0$, then you have:
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}+\dots$$
Substituting $x$ with $(-x)^{\frac{1}{2}}$ we get:
$$\sinh (-x)^{\frac{1}{2}} = (-x)^{\frac{1}{2}} + \frac{({(-x)^{\frac{1}{2}}})^3}{3!} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given an alphabet with 6 non-distinct integers, how many distinct 4-digit integers are there?
How many distinct four-digit integers can one make from the digits
$1$, $3$, $3$, $7$, $7$ and $8$?
I can't really think how to get started with this, the only way I think might work would be to go through all the cases. F... | Distinct numbers with two $3$s and two $7$s: $\binom{4}{2}=6$.
Distinct numbers with two $3$s and one or fewer $7$s: $\binom{4}{2}3\cdot2=36$.
Distinct numbers with two $7$s and one or fewer $3$s: $\binom{4}{2}3\cdot2=36$.
Distinct numbers with one or fewer $7$s and one or fewer $3$s: $4\cdot3\cdot2\cdot1=24$.
Total: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum with binomial coefficients: $\sum_{k=1}^m \frac{1}{k}{m \choose k} $ I got this sum, in some work related to another question:
$$S_m=\sum_{k=1}^m \frac{1}{k}{m \choose k} $$
Are there any known results about this (bounds, asymptotics)?
| Consider a random task as follows. First, one chooses a nonempty subset $X$ of $\{1,2,\ldots,m\}$, each with equal probability. Then, one uniformly randomly selects an element $n$ of $X$. The event of interest is when $n=\max(X)$.
Fix $k\in\{1,2,\ldots,m\}$. The probability that $|X|=k$ is $\frac{1}{2^m-1}\,\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 1
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How to solve this quartic equation? For the quartic equation:
$$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$
I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions.
| $$x^4 - x^3 + 4x^2 + \underbrace{3x}_{4x-x} + \overbrace{5}^{4+1} = \\\color{red}{x^4-x^3}+4x^2+4x+4\color{red}{-x+1}\\={x^4-x^3}-x+1+4(x^2+x+1)\\={x^3(x-1)}-(x-1)+4(x^2+x+1)\\=(x-1)(x^3-1)+4(x^2+x+1)\\=(x-1)(x-1)(x^2+x+1)+4(x^2+x+1)\\=(x^2+x+1)(x-1)^2+4)=(x^2+x+1)(x^2-2x+5)$$
As for the roots, I assume you could solve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Simplify these expressions with radical sign 2 My question is
1) Rationalize the denominator:
$$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$
My answer is:
$$\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{18}$$
My question is
2) $$\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$$
My answer is: $$\frac... | $\begin{eqnarray*}
(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{12}+\sqrt{18}-\sqrt{30}) & = & (\sqrt{2}+\sqrt{3}+\sqrt{5})(2\sqrt{3}+3\sqrt{2}-\sqrt{2}\sqrt{3}\sqrt{5})\\& = & 12,
\end{eqnarray*}$
if you expand out the terms, so your first answer is incorrect. The denominator should be $12$.
$\begin{eqnarray*}
(\sqrt{2}+\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
| Here is a different approach. Let $$L = \lim_{x \to 0} \dfrac{\tan(x) - x}{x^3}$$
Replacing $x$ by $2y$, we get that
\begin{align}
L & = \lim_{y \to 0} \dfrac{\tan(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2y}{(2y)^3}\\
& = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recur... | Let us take a particular example that is large enough to illustrate the general situation. Concrete experience should precede the abstract.
Let $n=8$. We want to show that $2^0+2^1+2^2+\cdots +2^8=2^9-1$. We could add up on a calculator, and verify that the result holds for $n=8$. However, we would not learn much d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/158758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| I think you can do it this way.
\begin{align*}
\int \frac{1}{x^4 +1} \ dx & = \frac{1}{2} \cdot \int\frac{2}{1+x^{4}} \ dx \\\
&= \frac{1}{2} \cdot \int\frac{(1-x^{2}) + (1+x^{2})}{1+x^{4}} \ dx \\\ &=\frac{1}{2} \cdot \int \frac{1-x^2}{1+x^{4}} \ dx + \frac{1}{2} \int \frac{1+x^{2}}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 12
} |
Probably simple factoring problem I came across this in a friend's 12th grade math homework and couldn't solve it. I want to factor the following trinomial:
$$3x^2 -8x + 1.$$
How to solve this is far from immediately clear to me, but it is surely very easy. How is it done?
| A standard way of factorizing, when it is hard to guess the factors, is by completing the square.
\begin{align}
3x^2 - 8x + 1 & = 3 \left(x^2 - \dfrac83x + \dfrac13 \right)\\
& (\text{Pull out the coefficient of $x^2$})\\
& = 3 \left(x^2 - 2 \cdot \dfrac43 \cdot x + \dfrac13 \right)\\
& (\text{Multiply and divide by $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$ I try to solve the following sum:
$$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$
I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is v... | Here's another approach.
It depends primarily on the properties of telescoping series, partial fraction expansion, and the following identity for the $m$th harmonic number
$$\begin{eqnarray*}
\sum_{k=1}^\infty \frac{1}{k(k+m)}
&=& \frac{1}{m}\sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+m}\right) \\
&=& \frac{1}{m}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Limit of exponentials Why is $n^n (n+m)^{-{\left(n+m\over 2\right)}}(n-m)^{-{\left(n-m\over 2\right)}}$ asymptotically equal to $\exp\left(-{m^2\over 2n}\right)$ as $n,m\to \infty$?
| Let $m = n x$.
Take the logarithm:
$$
n \log(n) - n \frac{1+x}{2} \left(\log n + \log\left(1+x\right) \right) - n \frac{1-x}{2} \left( \log n + \log\left(1-x\right) \right)
$$
Notice that all the terms with $\log(n)$ cancel out, so we are left with
$$
-\frac{n}{2} \left( (1+x) \log(1+x) - (1-x) \log(1-x) \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$p=4n+3$ never has a Decomposition into $2$ Squares, right? Primes of the form $p=4k+1\;$ have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma.
Is it correct to say that, primes of the form $p=4n+3$, never have a decomposition into $2$ squares, because sum of the quadratic resi... | Yes, as it has been pointed out, if $a^2+b^2$ is odd, one of $a$ and $b$ must be even and the other odd. Then
$$
(2m)^2+(2n+1)^2=(4m^2)+(4n^2+4n+1)=4(m^2+n^2+n)+1\equiv1\pmod{4}
$$
Thus, it is impossible to have $a^2+b^2\equiv3\pmod{4}$.
In fact, suppose that a prime $p\equiv3\pmod{4}$ divides $a^2+b^2$. Since $p$ cann... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/163519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following? ${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ?
A. ${21 \choose 5}$
B. ${20 \choose 5}-{11 \choose 4}$
C. ${21 \choose 5}-{10 \choose 5}$
D. ${20 \choose 4}$
Plea... | What is the problem in this solution?
If $S=\binom{10}{4} + \binom{11}{4} + \cdots + \binom{20}{4}$ we have
\begin{eqnarray}
S&=& \left \{ \binom{4}{4} + \binom{5}{4} \cdots + \binom{20}{4}\right \} - \left \{\binom{4}{4} + \binom{5}{4} \cdots + \binom{9}{4}\right \} &=& \binom{21}{5} - \binom{10}{5}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$... | I have a Cauchy-Schwarz proof of it,hope you enjoy.:D
first,mutiply $2$ to each side,your inequality can be rewrite into
$$ \sum_{cyc}{\frac{2ab}{a^2+3b^2}}\leq \frac{3}{2}$$
Or
$$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a^2+3b^2}}\geq \frac{3}{2}$$
Now,Using Cauchy-Schwarz inequality,we have
$$ \sum_{cyc}{\frac{(a-b)^2+2b^2}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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A problem dealing with even perfect numbers. Question: Show that all even perfect numbers end in 6 or 8.
This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.
What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $... | $p$ is prime so it is 1 or 3$\mod 4$.
So, the ending digit of $2^p$ is (respectively) 2 or 8
(The ending digits of powers of 2 are $2,4,8,6,2,4,8,6,2,4,8,6...$
So, the ending digit of $2^{p-1}$ is (respectively) 6 or 4; and
the ending digit of $2^p-1$ is (respectively) 1 or 7.
Hence the ending digit of $2^{p-1}(2^p-1)$... | {
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"url": "https://math.stackexchange.com/questions/168504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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When is $(6a + b)(a + 6b)$ a power of two?
Find all positive integers $a$ and $b$ for which the product $(6a + b)(a + 6b)$ is a power of $2$.
I havnt been able to get this one yet, found it online, not homework!
any help is appreciated thanks!
| Assume, (6a + b)(a + 6b) = 2^c
where, c is an integer ge 0
Assume, (6a + b) = 2^r
where, r is an integer ge 0
Assume, (a + 6b) = 2^s
where, s is an integer ge 0
Now, (2^r)(2^s) = 2^c
i.e. r + s = c
Now, (6a + b) + (a + 6b) = 2^r + 2^s
i.e. 7(a + b) = 2^r + 2^s
i.e. a + b = (2^r)/7 + (2^s)/7
Now, (6a + b) - ( a + 6b) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorize $f$ as product of irreducible factors in $\mathbb Z_5$ Let $f = 3x^3+2x^2+2x+3$, factorize $f$ as product of irreducible factors in $\mathbb Z_5$.
First thing I've used the polynomial reminder theorem so to make the first factorization:
$$\begin{aligned} f = 3x^3+2x^2+2x+3 = (3x^2-x+3)(x+1)\end{aligned}$$
Ob... | If $f(X) = aX^2 + bX + c$ is a quadratic polynomial with roots $x_1$ and $x_2$ then $f(X) = a(X-x_1)(X-x_2)$ (the factor $a$ is necessary to get the right leading coefficient). You found that $3x^2-x+3$ has a double root at $x_1 = x_2 = 1$, so $3x^2-x+3 = 3(x-1)^2$. Your mistakes were
*
*You forgot to multiply by th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x... | \begin{eqnarray}
\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)&=&2\cos x\\
\sin x + 1 &=& 2 \cos^2 x \\
\sin x + 1 &=& 2(1-\sin^2 x)\\
\end{eqnarray}
Then $\sin x = -1$ or $\sin x = 1/2$ and the solutions are $-\pi + 2k \pi, \pi/6 + 2 k\pi$ and $5\pi/6 + 2k \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
... | Another approach:
Lets, start by arranging the expression:
$$\tan(20°) \tan(30°) \tan(40°) = \tan(30°) \tan(40°) \tan(20°)$$$$=\tan(30°) \tan(30°+10°) \tan(30° - 10°)$$
Now, we will express $\tan(30° + 10°) $ and $\tan(30° - 10°)$ as the ratio of Prosthaphaeresis Formulas, giving us:
$$\tan(30°) \left( \frac{\tan(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Prove $ (r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$ How would I verify the following trig identity?
$$(r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$$
My work thus far is
$$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$$
But how would I continue? My math skills fail me.
| Just use the distributive property and $\sin^2(x)+\cos^2(x)=1$:
$$
\begin{align}
&(r\sin(A)\cos(A))^2+(r\sin(A)\sin(A))^2+(r\cos(A))^2\\
&=r^2\sin^2(A)(\cos^2(A)+\sin^2(A))+r^2\cos^2(A)\\
&=r^2\sin^2(A)+r^2\cos^2(A)\\
&=r^2(\sin^2(A)+\cos^2(A))\\
&=r^2\tag{1}
\end{align}
$$
This can be generalized to
$$
\begin{align}
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$
I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox.
If the series continues infin... | Solution as per David Mitra's hint in a comment.
Write the given series as a telescoping series and evaluate its sum:
$$\begin{eqnarray*}
S &=&\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots \\
&=&\sum_{n=1}^{\infty }\frac{1}{\left( 2n-1\right) \left( 2n+1\right) } \\
&=&\sum_{n=1}^{\infty }\left( \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/177373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Trigonometry proof involving sum difference and product formula How would I solve the following trig problem.
$$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$
I am not sure what to really I know it involves the sum and difference identity but I know not what to do.
| $$\require{cancel}
\frac1{16} [ 5(\cos 3x+\cos x)+\cos 5x+5\cos x ]\\
=\frac1{16}[10\cos x \cos 2x+ \cos 5x +5 \cos x]\\
=\frac1{16} [5\cos x(2\cos 2x+1)+\cos 5x]\\
=\frac1{16} [5\cos x(2(2\cos^2 x-1)+1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the domain of $f(x)=\frac{3x+1}{\sqrt{x^2+x-2}}$
Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$
This is my work so far:
$$\dfrac{3x+1}{\sqrt{x^2+x-2}}\cdot \sqrt{\dfrac{x^2+x-2}{x^2+x-2}}$$
$$\dfrac{(3x+1)(\sqrt{x^2+x-2})}{x^2+x-2}$$
$(3x+1)(\sqrt{x^2+x-2})$ = $\alpha$ (Just because it's too much to type... | Note that $x^2+x-2=(x-1)(x+2)$. There is a problem only if $(x-1)(x+2)$ is $0$ or negative. (If it is $0$, we have a division by $0$ issue, and if it is negative we have a square root of a negative issue.)
Can you find where $(x-1)(x+2)$ is $0$? Can you find where it is negative? Together, these are the numbers which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solving linear system of equations when one variable cancels I have the following linear system of equations with two unknown variables $x$ and $y$. There are two equations and two unknowns. However, when the second equation is solved for $y$ and substituted into the first equation, the $x$ cancels. Is there a way o... | $$\begin{equation*}
\left\{
\begin{array}{c}
2.6513=\frac{3}{2}y+\frac{x}{2} \\
1.7675=y+\frac{x}{3}
\end{array}
\right.
\end{equation*}$$
If we multiply the first equation by $2$ and the second by $3$ we get
$$\begin{equation*}
\left\{
\begin{array}{c}
5.3026=3y+x \\
5.3025=3y+x
\end{array}
\right.
\end{equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
About the sequence satisfying $a_n=a_{n-1}a_{n+1}-1$ "Consider sequences of positive real numbers of the form x,2000,y,..., in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence?
(A) 1 (B) 2 (... | (This is basically EuYu's answer with the details of periodicity added; took a while to type up.)
Suppose that $a_0 , a_1 , \ldots$ is a generalised sequence of the type described, so that $a_i = a_{i-1} a_{i+1} - 1$ for all $i > 0$. Note that this condition is equivalent to demanding that $$a_{i+1} = \frac{ a_i + 1 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Can you construct a field with 6 elements?
Possible Duplicate:
Is there anything like GF(6)?
Could someone tell me if you can build a field with 6 elements.
| If such a field $F$ exists, then the multiplicative group $F^\times$ is cyclic of order 5. So let $a$ be a generator for this group and write
$F = \{ 0, 1, a, a^2, a^3, a^4\}$.
From $a(1 + a + a^2 + a^3 + a^4) = 1 + a + a^2 + a^3 + a^4$, it immediately follows that
$1 + a + a^2 + a^3 + a^4 = 0$. Let's call this (*).
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 2
} |
Rigorous proof of the Taylor expansions of sin $x$ and cos $x$ We learn trigonometric functions in high school, but their treatment is not rigorous.
Then we learn that they can be defined by power series in a college.
I think there is a gap between the two.
I'd like to fill in the gap in the following way.
Consider the... | Since $x^2 + y^2 = 1$, $y = \sqrt{1 - x^2}$,
$y' = \frac{-x}{\sqrt{1 - x^2}}$
By the arc length formula,
$\theta = \int_{0}^{x} \sqrt{1 + y'^2} dx = \int_{0}^{x} \frac{1}{\sqrt{1 - x^2}} dx$
We consider this integral on the interval [$-1, 1$] instead of [$0, 1$].
Then $\theta$ is a monotone strictly increasing functio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What is the best way to solve a problem given remainders and divisors? $x$ is a positive integer less than $100$. When $x$ is divided by $5$, its remainder is $4$. When $x$ is divided by $23$, its remainder is $7$. What is $x$ ?
| So, $x=5y+4$ and $x=23z+7$ for some integers $y,z$
So, $5y+4=23z+7=>5y+20=23z+23$ adding 16 to either sides,
$5(y+4)=23(z+1)=>y+4=\frac{23(z+1)}{5}$, so,$5|(z+1)$ as $y+4$ is integer and $(5,23)=1$
$=>z+1=5w$ for some integer $w$.
$x=23(5w-1)+7=115w-16$
As $0<x<100$ so,$x=99$ putting $w=1$
Alternatively, We have $5y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expressio... | The claim is that
$\sum_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
We will verify this by induction.
Clearly $n = 1$ holds.
Suppose the formula holds for $n$. Lets verify it holds for $n + 1$.
$$\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2
\\
= \frac{n(n + 1)(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ ... | Another proof:
Denote $r=b/a$. We know that the sum of a geometric progression of the type $1+r+r^2+\ldots+r^{n-1}$ is equal to $\frac{1-r^n}{1-r}$. Thus, we have
\begin{align}
1-r^n&=(1-r)(1+r+r^2+\ldots+r^{n-1}),\quad\text{substituting $r=b/a$ gives:}\\
a^n-b^n &= (a-b)\color{red}{(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 8,
"answer_id": 2
} |
Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can... | Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.
You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$
Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4... | It would be much easier to just compute the coefficient at $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$. You can begin by computing:
$$ (2-x)^6
= 64 - 6 \cdot 2^5 x + 15 \cdot 2^4 x^2 + x^3 \cdot (...)
= 64 - 192 x + 240 x^2 + x^3 \cdot (...) $$
Now, multiply this by $(1+2x+ax^2)$. Again, you're only interested in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int\frac{dx}{\sin(x+a)\sin(x+b)}$ Please help me evaluate:
$$
\int\frac{dx}{\sin(x+a)\sin(x+b)}
$$
| The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$
The given integral can write:
$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$
We substition
$$\frac{\sin(x+a)}{\sin(x+b)}=t$$
By the substition of the above have:
$$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Representations of integers by a binary quadratic form Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$.
Let $f(x, y) = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$.
We write $f^\alpha(x, y) ... | Lemma 1
Let $f = ax^2 + bxy + cy^2 \in \mathfrak{F}$.
Let $\alpha = \left( \begin{array}{ccc}
p & q \\
r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$.
Then $f^\alpha(x, y) = f(px + qy, rx + sy) = kx^2 + lxy + my^2$,
where
$k = ap^2 + bpr + cr^2$
$l = 2apq + b(ps + qr) + 2crs$
$m = aq^2 + bqs + cs^2$.
Pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
| Square both sides, and you get
$$x - 4 + 20\sqrt{x - 4} + 100 = x + 4$$
This simplifies to
$$20\sqrt{x - 4} = -92$$
or just
$$\sqrt{x - 4} = -\frac{92}{20}$$
Since square roots of numbers are always nonnegative, this cannot have a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Sum of the sequence What is the sum of the following sequence
$$\begin{align*}
(2^1 - 1) &+ \Big((2^1 - 1) + (2^2 - 1)\Big)\\
&+ \Big((2^1 - 1) + (2^2 - 1) + (2^3 - 1) \Big)+\ldots\\
&+\Big( (2^1 - 1)+(2^2 - 1)+(2^3 - 1)+\ldots+(2^n - 1)\Big)
\end{align*}$$
I tried to solve this. I reduced the equation into the followi... | Let's note that $$(2^1 - 1) + (2^2 - 1) + \cdots + (2^k - 1) = 2^{k+1} - 2-k$$ where we have used the geometric series. Thus, the desired sum is actually $$\sum_{k=1}^n{2^{k+1}-2-k}$$. As this is a finite sum, we can evaluate each of the terms separately. We get the sum is $$2\left(\frac{2^{n+1}-1}{2-1}-1\right) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| Inequality can be written as:
$$\left(a+b+c\right) \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 .$$
And now we apply the $AM-GM$ inequality for both of parenthesis. So:
$\displaystyle \frac{a+b+c}{3} \geq \sqrt[3]{abc} \tag{1}$ and $\displaystyle \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
The integral relation between Perimeter of ellipse and Quarter of Perimeter Ellipse Equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
$x=a\cos t$ ,$y=b\sin t$
$$L(\alpha)=\int_0^{\alpha}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$
$$L(\alpha)=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$
... | I proved the relation via using analytic way. I would like to share the solution with you.
$$\int_0^{\pi/2}\sqrt{b^2+(a^2-b^2)\sin^2 4u}\,du=K$$
$u=\pi/4-z$
$$K=\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^2-b^2)\sin^2 (\pi-4z)}\,dz$$
$\sin (\pi-4z)=\sin \pi \cos 4z-\cos \pi \sin 4z= \sin 4z$
$$\int_{-\pi/4}^{\pi/4}\sqrt{b^2+(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Inequality. $\sum{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$ Let $a,b,c$ be the side-lengths of a triangle. Prove that:
I.
$$\sum_{cyc}{(a+b)(b+c)\sqrt{a-b+c}} \geq 4(a+b+c)\sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
What I have tried:
\begin{eqnarray}
a-b+c&=&x\\
b-c+a&=&y\\
c-a+b&=&z.
\end{eqnarray}
... | Notice that the inequality proposed is proved once we estabilish
$$\frac{(a+b)(b+c)}{\sqrt{a+b-c}\sqrt{-a+b+c}}+\frac{(b+c)(c+a)}{\sqrt{a-b+c}\sqrt{-a+b+c}}+\frac{(c+a)(a+b)}{\sqrt{a-b+c}\sqrt{a+b-c}}\geq 4(a+b+c).$$
Using AM-GM on the denominators in the LHS, we estabilish that
*
*$\sqrt{a+b-c}\sqrt{-a+b+c}\leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
| Consider, $z_1= \frac{1+2i}{\sqrt{5}}$, $z_2= \frac{1+3i}{\sqrt{10} }$, and $z_3= \frac{1+i}{\sqrt{2} }$, then:
$$ z_1 z_2 z_3 = \frac{1}{10} (1+2i)(1+3i)(1+i)=-1 $$
Take arg of both sides and use property that $\arg(z_1 z_2 z_3) = \arg(z_1) + \arg(z_2) + \arg(z_3)$:
$$ \arg(z_1) + \arg(z_2) + \arg(z_3) = -1$$
The L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 6,
"answer_id": 5
} |
Question: Find all values of real number a such that $ \lim_{x\to1}\frac{ax^2+a^2x-2}{x^3-3x+2} $ exists. Thanks in advance for looking at my question.
I was tackling this limits problem using this method, but I can't seem to find any error with my work.
Question:
Find all values of real number a such that
$$
\li... | Since the denominator's limit is 0, the numerator cannot have a nonzero limit if the limit of the quotient is to be defined. The only hope is that the numerator's limit is also 0, and that after analyzing the indeterminate form, it does have a limit.
So, it must be the case that $\lim_{x\to1} ax^2+a^2x-2=0$, and conseq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to take the absolute value of both sides of an equation? I have a problem that says: Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$.
Since the quadratic is greater than 0, I assume that there are no real solutions since
$y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$
since $y>0$ ... | What you've written is an inequality, not an equation. If you have an equation, say $a=b$, you can conclude that $|a|=|b|$.
But notice that $3>-5$, although $|3|\not>|-5|$.
If $3x^2+bx+7>0$ for every value of $x$, then the quadratic equation $3x^2+bx+7=0$ has no solutions that are real numbers. THat implies that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Stuck with solving a polynomial I am doing a problem for homework that says:
Suppose $s(x)=3x^3-2$. Write the expression $\frac{s(2+x)-s(2)}{x}$ as a sum of terms, each of which is a constant times power of $x$.
I was able to do the following work for this problem:
$\frac{3(2+x)^3-3(2)^3-2}{x}$
$\frac{3(x^3+6x^2+12x+8)... | You went astray at the first step, when you got $$\frac{3(2+x)^3-3(2)^3-2}{x}\;;$$ in fact
$$s(2+x)-s(2)=\Big(3(2+x)^3-2\Big)-\Big(3\cdot2^3-2\Big)=3(2+x)^3-3\cdot2^3\;.$$
Can you straighten the rest out from there? When you do it correctly, there will be no constant term in the numerator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this equation have different number of answers? I have a simple equation:
$$\frac{x}{x-3} - \frac{2}{x-1} = \frac{4}{x^2-4x+3}$$
By looking at it, one can easily see that $x \not= 1$ because that would cause $\frac{2}{x-1} $ to become $\frac{2}{0}$, which is illegal.
However, if you do some magic with it. Firs... | If $\dfrac AB = 0$ then $A=0\cdot B$. But you can't say that if $A=0\cdot B$ then $\dfrac AB=0$ unless you know that $B\ne 0$. So if $A$ and $B$ are complicated expressions that can be solved for $x$, there may be values of $x$ that make $B$ equal to $0$, and if they also make $A$ equal to $0$, then they are solution... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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For integers $a$ and $b \gt 0$, and $n^2$ a sum of two square integers, does this strategy find the largest integer $x | x^2 \lt n^2(a^2 + b^2)$? Here is some background information on the problem I am trying to solve. I start with the following equation:
$n^2(a^2 + b^2) = x^2 + y^2$, where $n, a, b, x, y \in \mathbb ... | Let $N = n^2(a^2+b^2)$ and consider its prime factorization.
Let $p$ be a prime divisor of $N$.
If $p\equiv 3\pmod4$, then necessarily $p^2|N$, $p|x$ and $p|y$ (and the factor comes from $p|n$).
If $p\equiv 1\pmod 4$, find $u,v$ such that $u^2+v^2=p$.
Let $k$ be the exponet of $p$ in $N$, i.e. $p^k||N$.
Then $x+iy$ mus... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction $\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$ for $n\ge1$ Prove the following statement $S(n)$ for $n\ge1$:
$$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$
To prove the basis, I substitute $1$ for $n$ in $S(n)$:
$$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$
Great. For the inductive step, I assume $S(n)$ to be true and ... | Here's yet another way:
You have $\frac{n^2(n+1)^2}{4}+(n+1)^3$ and you want $\frac{(n+1)^2(n+2)^2}{4}$
So manipulate it to get there;
$\frac{n^2(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n^2 + 4n + 4)(n+1)^2 - (4n + 4)(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}- \frac{ (4n + 4)(n+1)^2}{4}+(n+1)^3 =$
$\frac{(n+2)^2(n+1)^2}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation wi... | Riffing on @Yves' "little known" solutions ...
The above trigonograph shows a scenario with $a^2 + b^2 = c^2 + d^2$, for $d \geq 0$, and we see that
$$\theta = \operatorname{atan}\frac{a}{b} + \operatorname{atan}\frac{d}{c} \tag{1}$$
(If the "$a$" triangle were taller than the "$b$" triangle, the "$+$" would become "... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 2
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The intersection of a line with a circle Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$
What I did:
$y^2=10-x^2$
$y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$
$ x+ 2 = y=\sqrt{10-x^2}$
If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$
But then, and here is where the problems... | Let the intersection be $(a,b)$, so it must satisfy both the given eqaution.
So, $a=b+2$ also $a^2+b^2=10$
Putting $b=a+2$ in the given circle $a^2+(a+2)^2=10$
$2a^2+4a+4=10\implies a=1$ or $-3$
If $a=1,b=a+2=3$
If $a=-3,b=-3+2=-1$
So, the intersections are $(-3,-1)$ and $(1,3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the limit without l'Hôpital's rule Find the limit
$$\lim_{x\to 1}\frac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)}.$$
I'm a little rusty with limits, can somebody please give me some pointers on how to solve this one? Also, l'Hôpital's rule isn't allowed in case you were thinking of using it. Thanks in advance.
| $$\dfrac{\sin(3x-3)}{\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)}{x^2(x-1)^2}$$
Hence, $$\dfrac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)(x^2-1)}{x^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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About the homework of differentiation This problem is not solved.
$$
\begin{align}
f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr
\frac{df}{dx}&=\mathord?
\end{align}
$$
| The answer is $$\frac{2\sqrt{2}}{1+x^4}$$
Hints: $$\frac{d\left(\log(1+x^2\pm\sqrt{2}x\right)}{dx}=\frac{2x\pm\sqrt{2}}{1+x^2\pm\sqrt{2}x}$$ and $$\left(1+x^2+\sqrt{2}x\right)\left(1+x^2-\sqrt{2}x\right)=\left(1+x^2\right)^2-\left(\sqrt{2}x\right)^2$$
Similarly,
$$\frac{d\left(\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\ri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.
The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
| $$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2-2$$
Let $x+\frac{1}{x}=z$.
Then we get, $$f(z)=z^2-2.$$
Hence we put x on the place of z. And we get
$f(x)=x^2-2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Do these $\delta-\epsilon$ proofs work? I'm new to $\delta, \epsilon$ proofs and not sure if I've got the hand of them quite yet.
$$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$
$|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|(2x + 4 - 3)(x + 2)| < \epsilon$
$|(2(x+2)^2 -3(x + 2)| \leq 2|x+2|^2 +3|x + 2| < \epsil... | $$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$
Finding $\delta$: $|2x^2 + 5x + 3 - 1| < \epsilon$
$|(2x + 1)(x + 2)| < \epsilon$
$|x - (-2)| < \delta $, pick $\delta = 3$
$|x+2| < 3 \Rightarrow -5 < x < 1 \Rightarrow -9 < 2x + 1 < 3$
This implies $|2x + 1||x + 2| < 3 \cdot |x + 2| < \epsilon \Rightarrow |x+2| < \frac{\epsilon}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/221259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y=x$ and $y = x^4$. Find the volume of the solid under the plane $x + 2y - z = 0$ and above the region bounded by $y = x$ and $y = x^4$.
$$
\int_0^1\int_{x^4}^x{x+2ydydx}\\
\int_0^1{x^2-x^8dx}\\
\frac{1}{3}-\frac{1}{9} = \fra... | I think it should be calculated as
\begin{eqnarray*}
V&=&\int_0^1\int_{x^4}^x\int_0^{x+2y}dzdydx\\
&=&\int_0^1\int_{x^4}^x(x+2y)dydx\\
&=&\int_0^1\left.\left(xy+y^2\right)\right|_{x^4}^xdx\\
&=&\int_0^1(2x^2-x^5-x^8)dx\\
&=&\left.(\frac{2}{3}x^3-\frac{1}{6}x^6-\frac{1}{9}x^9)\right|_0^1\\
&=&\frac{7}{18}
\end{eqnarray*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find all Laurent series of the form... Find all Laurent series of the form $\sum_{-\infty} ^{\infty} a_n $ for the function
$f(z)= \frac{z^2}{(1-z)^2(1+z)}$
There are a lot of problems similar to this. What are all the forms? I need to see this example to understand the idea.
| Here is related problem. First, convert the $f(z)$ to the form
$$f(z) = \frac{1}{4}\, \frac{1}{\left( 1+z \right)}+\frac{3}{4}\, \frac{1}{\left( -1+z \right) }+\frac{1}{2}\,\frac{1}{\left( -1+z \right)}$$
using partial fraction. Factoring out $z$ gives
$$ f(z)= \frac{1}{4z}\frac{1}{(1+\frac{1}{z})}- \frac{3}{4z}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sines and cosines law Can we apply the sines and cosines law on the external angles of triangle ?
| This answer assumes a triangle with angles $A, B, C$ with sides $a,b,c$.
Law of sines states that $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$
Knowing the external angle is $\pi - \gamma$ if the angle is $\gamma$, $\sin(\pi-\gamma) = \sin \gamma$ because in the unit circle, you are merely reflecting the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Squeeze Theorem Problem I'm busy studying for my Calculus A exam tomorrow and I've come across quite a tough question. I know I shouldn't post such localized questions, so if you don't want to answer, you can just push me in the right direction.
I had to use the squeeze theorem to determine:
$$\lim_{x\to\infty} \dfrac{... | I assume you meant $$\lim_{x \to \infty} \dfrac{2x^3 + \sin(x^2)}{1+x^3}$$ Note that $-1 \leq \sin(\theta) \leq 1$. Hence, we have that $$\dfrac{2x^3 - 1}{1+x^3} \leq \dfrac{2x^3 + \sin(x^2)}{1+x^3} \leq \dfrac{2x^3 + 1}{1+x^3}$$
Note that
$$\dfrac{2x^3 - 1}{1+x^3} = \dfrac{2x^3 +2 -3}{1+x^3} = 2 - \dfrac3{1+x^3}$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/225374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Derivative of a split function We have the function:
$$f(x) = \frac{x^2\sqrt[4]{x^3}}{x^3+2}.$$
I rewrote it as
$$f(x) = \frac{x^2{x^{3/4}}}{x^3+2}.$$
After a while of differentiating I get the final answer:
$$f(x)= \frac{- {\sqrt[4]{\left(\frac{1}{4}\right)^{19}} + \sqrt[4]{5.5^7}}}{(x^3+2)^2}$$(The minus isn't behind... | Let $y=\frac{x^2\cdot x^{3/4}}{x^3+2}$ so $y=\frac{x^{11/4}}{x^3+2}$ and therefore $y=x^{11/4}\times(x^3+2)^{-1}$. Now use the product rule of two functions: $$(f\cdot g)'=f'\cdot g+f\cdot g'$$ Here $f(x)=x^{11/4}$ and $g(x)=(x^3+2)^{-1}$. So $f'(x)=\frac{11}{4}x^{7/4}$ and $g'(x)=(-1)(3x^2)(x^3+2)^{-2}$. But thinking... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/225438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?
| Consider the recurrence given by
$$x_{n+1} = 7x_n - 3 x_{n-1}$$ where $x_0 = 2$, $x_1 = 7$.
Note that $$x_{n+1} \equiv (7x_n - 3 x_{n-1}) \pmod{3} \equiv (x_n + 3(2x_n - x_{n-1})) \pmod{3} \equiv x_n \pmod{3}$$
Since $x_1 \equiv 1 \pmod{3}$, we have that $x_n \equiv 1 \pmod{n}$. Ths solution to this recurrence is give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to the equation
$$2^a + 3^b = 5^c$$
Of course $(1,\space 1, \space 1)$ is a solution.
| There are three solutions which can all be found by elementary means.
If $b$ is odd
$$2^a+3\equiv 1 \bmod 4$$
Therefore $a=1$ and $b$ is odd.
If $b>1$, then $2\equiv 5^c \bmod 9$ and $c\equiv 5 \bmod 6$
Therefore $2+3^b\equiv 5^c\equiv3 \bmod 7$ and $b\equiv 0 \bmod 6$, a contradiction.
The only solution is $(a,b,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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For which values of $\alpha \in \mathbb R$ is the following system of linear equations solvable? The problem I was given:
Calculate the value of the following determinant:
$\left|
\begin{array}{ccc}
\alpha & 1 & \alpha^2 & -\alpha\\
1 & \alpha & 1 & 1\\
1 & \alpha^2 & 2\alpha & 2\alpha\\
1 & 1 & \alpha^2 & -\alpha
\en... | Let me first illustrate an alternate approach. You're looking at $$\left[\begin{array}{ccc}
\alpha & 1 & \alpha^2\\
1 & \alpha & 1\\
1 & \alpha^2 & 2\alpha\\
1 & 1 & \alpha^2
\end{array}\right]\left[\begin{array}{c} x_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c} -\alpha\\ 1\\ 2\alpha\\ -\alpha\end{array}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Prove $\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$ Prove inequality $$\frac{1}{a^3} + \frac{1}{b^3} +\frac{1}{c^3} ≥ 3$$ where $a+b+c=3abc$ and $a,b,c>0$
| If $a, b, c >0$ then $a+b+c=3abc \ \Rightarrow \ \cfrac 1{ab} + \cfrac 1{bc}+ \cfrac 1{ca} = 3$
See that $2\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ \cfrac 1{c^3}\right) +3 =\left(\cfrac 1{a^3} +\cfrac 1{b^3}+ 1\right)+\left(\cfrac 1{b^3} +\cfrac 1{c^3}+ 1\right)+\left(\cfrac 1{c^3} +\cfrac 1{a^3}+ 1\right) $
Use $AM-GM$ in... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $3^{2n+1} + 2^{n+2}$ is divisible by $7$ for all $n\ge0$ Expanding the equation out gives
$(3^{2n}\times3)+(2^n\times2^2) \equiv 0\pmod{7}$
Is this correct? I'm a little hazy on my index laws.
Not sure if this is what I need to do? Am I on the right track?
| Note that
$$3^{2n+1} = 3^{2n} \cdot 3^1 = 3 \cdot 9^n$$ and $$2^{n+2} = 4 \cdot 2^n$$
Note that $9^{3k} \equiv 1 \pmod{7}$ and $2^{3k} \equiv 1 \pmod{7}$.
If $n \equiv 0 \pmod{3}$, then $$3 \cdot 9^n + 4 \cdot 2^n \equiv (3+4) \pmod{7} \equiv 0 \pmod{7}$$
If $n \equiv 1 \pmod{3}$, then $$3 \cdot 9^n + 4 \cdot 2^n \equi... | {
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"timestamp": "2023-03-29T00:00:00",
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A limit $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$ maybe relate to riemann sum Find
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$$
I think this maybe relate to Riemann sum. but I can't deal with $k$ before $\sin$
| If there is no typo, then the answer is $\infty$. Indeed, let $m$ be any fixed positive integer and consider the final $m$ consecutive terms:
$$ \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}
= \sum_{k=1}^{m} \frac{(n-k) \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}. $$
As $n \to \inft... | {
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"source": "stackexchange",
"question_score": "2",
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Are there five complex numbers satisfying the following equalities? Can anyone help on the following question?
Are there five complex numbers $z_{1}$, $z_{2}$ , $z_{3}$ , $z_{4}$
and $z_{5}$ with $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|+\left|z_{5}\right|=1$
such that the smallest a... | Suppose you have solutions and express $z_i$ as $r_i e^{\theta_i}$.
(I use $s_i = \sin( \theta_i )$ and $c_i = \sin( \theta_i )$ to make notations shorter)
Then$$\begin{align*}
|z_i| + |z_j| - |z_i + z_j| & = |r_i e^{\theta_i}| + |r_j e^{\theta_j}| - |r_i e^{\theta_i} + r_j e^{\theta_j}| \\
& = r_i + r_j - |r_i(c_i... | {
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement hold... | Due to a recent comment on my other answer, I took a second look at this question and tried to apply a double generating function.
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=-n}^\infty\binom{2n+k}{n}x^ny^k\\
&=\sum_{n=0}^\infty\sum_{k=n}^\infty\binom{k}{n}\frac{x^n}{y^{2n}}y^k\\
&=\sum_{n=0}^\infty\frac{x^n}{y^{2n}}\fr... | {
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numbers' pattern It is known that
$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\
9+10+11+12 &=& 13+14+15 \\\
16+17+18+19+20 &=& 21+22+23+24 \\\
25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots
\end{array}$$
There is something similar for square numbers:
$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13... | Let's start by proving the basic sequences and then where and why trying to step it up to cubes fails. I don't prove anything just reduce the problem to a two variable quartic Diophantine equation.
Lemma $1 + 2 + 3 + 4 + \ldots + n = T_1(n) = \frac{n(n+1)}{2}$.
Corollary $(k+1) + (k+2) + \ldots + (k+n) = -T_1(k) + T_1... | {
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Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$
Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.
The answer is $Θ (n \log \log n)$.
Can anyone arrive at the solution.
| Let $n = m^{2^k}$. We then get that
$$T(m^{2^k}) = m^{2^{k-1}} T (m^{2^{k-1}}) + m^{2^{k}}$$
\begin{align}
f_m(k) & = m^{2^{k-1}} f_m(k-1) + m^{2^k} = m^{2^{k-1}}(m^{2^{k-2}} f_m(k-2) + m^{2^{k-1}}) + m^{2^k}\\
& = 2 m^{2^k} + m^{3 \cdot 2^{k-2}} f_m(k-2)
\end{align}
$$m^{3 \cdot 2^{k-2}} f_m(k-2) = m^{3 \cdot 2^{k-2}}... | {
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$p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{(\frac{p-1}{2})}*(p-1)! \equiv 1 \pmod p$ $p$ an odd prime, $p \equiv 3 \pmod 8$. Show that $2^{\left(\frac{p-1}{2}\right)}\cdot(p-1)! \equiv 1 \pmod p$
From Wilson's thm: $(p-1)!= -1 \pmod p$.
hence, need to show that $2^{\left(\frac{p-1}{2}\right)} \equiv -1 \pmo... | All the equalities below are in the ring $\mathbb{Z}/p\mathbb{Z}$.
Note that $-1 = (p-1)! = \prod_{i=1}^{p-1}i = \prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} 2i = 2^{\frac{p-1}{2}} \prod_{i=1}^{\frac{p-1}{2}}(2i-1)\prod_{i=1}^{\frac{p-1}{2}} i$
Now let $S_1, S_2$ be the set of respectively all odd and... | {
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Need someone to show me the solution Need someone to show me the solution. and tell me how !
$$(P÷N) × (N×(N+1)÷2) + N×(1-P) = N×(1-(P÷2)) + (P÷2)$$
| \begin{align}
\dfrac{P}{N} \times \dfrac{N(N+1)}2 + N\times (1-P) & = \underbrace{P \times \dfrac{N+1}{2} + N \times (1-P)}_{\text{Cancelling out the $N$ from the first term}}\\
& = \underbrace{\dfrac{PN + P}2 + N - NP}_{\text{$P \times (N+1) = PN + P$ and $N \times (1-P) = N - NP$}}\\
& = \underbrace{\dfrac{PN + P +2N... | {
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Example 2, Chpt 4 Advanced Mathematics (I)
$$\int \frac{x+2}{2x^3+3x^2+3x+1}\, \mathrm{d}x$$
I can get it down to this:
$$\int \frac{2}{2x+1} - \frac{x}{x^2+x+1}\, \mathrm{d}x $$
I can solve the first part but I don't exactly follow the method in the book.
$$ = \ln \vert 2x+1 \vert - \frac{1}{2}\int \frac{\left(2x... | The post indicates some difficulty with finding $\int \frac{dx}{x^2+x+1}$. We solve a more general problem. But I would suggest for your particular problem, you follow the steps used, instead of using the final result.
Suppose that we want to integrate $\dfrac{1}{ax^2+bx+c}$, where $ax^2+bx+c$ is always positive, or a... | {
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Two questions with mathematical induction First hello all, we have a lecture. It has 10 questions but I'm stuck with these two about 3 hours and I can't solve them.
Any help would be appreciated.
Question 1
Given that $T(1)=1$, and $T(n)=2T(\frac{n}{2})+1$, for $n$ a power of $2$, and
greater than $1$. Using mathem... | 1)
Prove by induction on $k$:
If $0\le k\le m$, then $T(2^m)=2^kT(2^{m-k})+2^k-1$.
The case $k=0$ is trivial.
If we already know that $T(2^m)=2^{k-1}T(2^{m-(k-1)})+2^{k-1}-1$ for all $m\ge k-1$, then for $m\ge 2^k$ we have
$$\begin{align}T(2^m)&=2^{k-1}T(2^{m-(k-1)})+2^{k-1}-1\\
&=2^{k-1}\left(2\cdot T\left(\tfrac{2^... | {
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Partial fractions integration Re-express $\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}$ in terms of partial fractions and compute the indefinite integral $\dfrac{1}5{}\int f(x)dx $ using the result from the first part of the question.
| Hint
Use $$\dfrac{6x^5 + x^2 + x + 2}{(x^2 + 2x + 1)(2x^2 - x + 4)(x+1)}=\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F$$
and solve for $A,B,C,D,E$ and $F$.
Explanation
Note that this partial fraction decomposition is a case where the degree of the numerator and denominator are the same. (Jus... | {
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Showing $\sqrt{2}\sqrt{3} $ is greater or less than $ \sqrt{2} + \sqrt{3} $ algebraically How can we establish algebraically if $\sqrt{2}\sqrt{3}$ is greater than or less than $\sqrt{2} + \sqrt{3}$?
I know I can plug the values into any calculator and compare the digits, but that is not very satisfying. I've tried to s... | Method 1: $\sqrt{2}+\sqrt{3}>\sqrt{2}+\sqrt{2}=2\sqrt{2}>\sqrt{3}\sqrt{2}$.
Method 2: $(\sqrt{2}\sqrt{3})^2=6<5+2<5+2\sqrt{6}=2+3+2\sqrt{2}\sqrt{3}=(\sqrt{2}+\sqrt{3})^2$, so $\sqrt{2}\sqrt{3}<\sqrt{2}+\sqrt{3}$.
Method 3: $\frac{196}{100}<2<\frac{225}{100}$ and $\frac{289}{100}<3<\frac{324}{100}$, so $\sqrt{2}\sqrt{3}... | {
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Where is the mistake in the calculation of $y'$ if $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $? Plase take a look here.
If $ y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} $
\begin{eqnarray}
y'&=& \dfrac{1}{4} \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \left \{ \dfrac{2x(x^2-1) - 2x(x^2+1) }{(x^2-1)^2} \right \}\\... | I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms).
Edited to explain further:
In your calculation, you write
\begin{align} \frac{dy}{dx} &= y\frac14 \left\{ \frac{2x}{(x^2+1)} - \frac{2x}{(x^2-1)} \right\} \\
&= \frac14 \frac{x^2+1}{x^2-1} \cdot 2x\frac{(x^2-1)-(x^2+1)}{(x... | {
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How to simplify polynomials I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
I tried combining like terms
$$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
$$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$
$$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$
It says the answer is $$3x^4-7x^3+x^2+8x+17$... | You cannot combine terms like that, you have to split your terms by powers of $x$.
So for example $$5x^2+5x+2x^2 = (5+2)x^2+5x = 7x^2+5x$$ and not $5x^3+2x^2$. Using this, you should end up with your answer.
| {
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Advanced integration, how to integrate 1/polynomial ? Thanks I have been trying to integrate a function with a polynomial as the denominator.
i.e, how would I go about integrating
$$\frac{1}{ax^2+bx+c}.$$
Any help at all with this would be much appreciated, thanks a lot :)
ps The polynomial has NO real roots
$${}{}$$... | If $a=0$, then
$$I = \int \dfrac{dx}{bx+c} = \dfrac1b \log (\vert bx+c \vert) + \text{constant}$$
$$I = \int \dfrac{dx}{ax^2 + bx + c} = \dfrac1a \int\dfrac{dx}{\left(x + \dfrac{b}{2a}\right)^2 + \left(\dfrac{c}a- \dfrac{b^2}{4a^2} \right)}$$
If $b^2 < 4ac$, then recall that
$$\int \dfrac{dt}{t^2 + a^2} = \dfrac1a\arct... | {
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Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this su... | Apparently I'm a little late to the party, but my answer has a punchline!
We have
$$
\frac{1}{z} \int_0^z \sum_{k=0}^{n} \binom{n}{k} s^k\,ds = \sum_{k=0}^{n} \binom{n}{k} \frac{z^k}{k+1},
$$
so that
$$
- \int_0^z \frac{1}{t} \int_0^t \sum_{k=0}^{n} \binom{n}{k} s^k\,ds\,dt = - \sum_{k=0}^{n} \binom{n}{k} \frac{z^{k+1}... | {
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greatest common divisor is 7 and the least common multiple is 16940 How many such number-pairs are there for which the greatest common divisor is 7 and the least common multiple is 16940?
| Let the two numbers be $7a$ and $7b$.
Note that $16940=7\cdot 2^2\cdot 5\cdot 11^2$.
We make a pair $(a,b)$ with gcd $1$ and lcm $2^2\cdot 5\cdot 11^2$ as follows. We "give" $2^2$ to one of $a$ and $b$, and $2^0$ to the other. We give $5^1$ to one of $a$ and $b$, and $5^0$ to the other. Finally, we give $11^2$ to one ... | {
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How to prove the given series is divergent? Given series
$$
\sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right],
$$
try to show that it is divergent!
The criterion will show that it is the case of limits $1$, so maybe need some other methods? any suggestions?
| Let $x > 1$. Then the inequality $$\frac{1}{t} \leq \frac{1}{x}(1-t) + 1$$ holds for all $t \in [1,x]$ (the right hand side is a straight line between $(1,1)$ and $(x, \tfrac{1}{x})$ in $t$) and in particular $$\log(x) = \int_1^x \frac{dt}{t} \leq \frac{1}{2} \left(x - \frac{1}{x} \right)$$ for all $x > 1$. Substitute... | {
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Number of Divisor which perfect cubes and multiples of a number n = $2^{14}$$3^9$$5^8$$7^{10}$$11^3$$13^5$$37^{10}$
How many positive divisors that are perfect cubes and multiples of $2^{10}$$3^9$$5^2$$7^{5}$$11^2$$13^2$$37^{2}$.
I'm able to solve number of perfect square and number of of perfect cubes.
But the extra ... | The numbers you are looking for must be perfect cubes. If you split them into powers of primes, they can have a factor $2^0$, $2^3$, $2^6$, $2^9$ and so on but not $2^1, 2^2, 2^4$ etc. because these are not cubes. The same goes for powers of $3, 5, 7$ and any other primes.
The numbers must also be multiples of $2^{10}... | {
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$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$ Suppose $a, b, c$ are the lengths of three triangular edges. Prove that:
$$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$$
| As the hint give in the comment says (I denote by $S$ the area of $ABC$ and by $R$ the radius of its circumcircle), if you multiply your inequality by $\sqrt{a+b+c}$ you'll get
$$4S \leq \frac{3\sqrt{3}abc}{a+b+c}$$
which is eqivalent to
$$a+b+c \leq 3\sqrt{3}\frac{abc}{4S}=3\sqrt{3}R.$$
This inequality is quite known... | {
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)... | Hint: $ \sum \frac{a^2 - b^2}{a+b} = \sum (a-b) = 0$.
(How is this used?)
Hint: $\sum \frac{a^2 + b^2}{a+b} \geq \sum \frac{a+b}{2} = a+b+c$ by AM-GM.
Hence, $\sum \frac{ a^2}{ a+b} \geq \frac{1}{2}(a+b+c)$.
| {
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Proving a Geometric Progression Formula, Related to Geometric Distribution I am trying to prove a geometric progression formula that is related to the formula for the second moment of the geometric distribution. Specifically, I am wondering where I am going wrong, so I can perhaps learn a new technique.
It is known, an... | You have $$\sum_{k=0}^{\infty} ka^k = \dfrac{a}{(1-a)^2}$$
Differentiating with respect to $a$ gives us
$$\sum_{k=0}^{\infty} k^2 a^{k-1} = \dfrac{(1-a)^2 - a \times 2 \times (a-1)}{(1-a)^4} = \dfrac{1-a + 2a}{(1-a)^3} = \dfrac{a-1+2}{(1-a)^3}\\ = \dfrac2{(1-a)^3} - \dfrac1{(1-a)^2}$$
| {
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"question_score": "3",
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A non-linear maximisation We know that $x+y=3$ where x and y are positive real numbers. How can one find the maximum value of $x^2y$? Is it $4,3\sqrt{2}, 9/4$ or $2$?
| By AM-GM
$$\sqrt[3]{2x^2y} \leq \frac{x+x+2y}{3}=2 $$
with equality if and only if $x=x=2y$.
Second solution
This one is more complicated, and artificial (since I needed to know the max)$.
$$x^2y=3x^2-x^3=-4+3x^2-x^3+4=4- (x-2)^2(x+1)\leq 4$$
since $(x-2)^2(x+1) \geq 0$.
| {
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how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
| You seek numbers which, when divided by $k$ (for $k=2,3,4,5,6$) gives a remainder of $k-1$. Thus the numbers you seek are precisely those which are one less than a multiple of $k$ for each of these values of $k$. To find all such numbers, consider the lowest common multiple of $2$, $3$, $4$, $5$ and $6$, and count how ... | {
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$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem
I tried this approach:
$$
\frac{1}{\sqrt{n^3+1}}\le\frac... | Hint: use $\frac{i}{\sqrt{n^3+n}} \le \frac{i}{\sqrt{n^3+i}} \le \frac{i}{\sqrt{n^3+1}}$. I even think the Squeeze theorem can be avoided.
| {
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If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.
Things should be known:
*
*Roots of a Quadratic Equations can be identified by:
The roots can be figured out by:
$$\frac{-b \pm \sqrt{d}}{2a},$$
whe... | As the two roots are equal the discriminant must be equal to $0$.
$$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$
Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/270344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find intersection of two 3D lines I have two lines $(5,5,4) (10,10,6)$ and $(5,5,5) (10,10,3)$ with same $x$, $y$ and difference in $z$ values.
Please some body tell me how can I find the intersection of these lines.
EDIT: By using the answer given by coffemath I would able to find the intersection point for the above ... | The direction numbers $(a,b,c)$ for a line in space may be obtained from two points on the line by subtracting corresponding coordinates. Note that $(a,b,c)$ may be rescaled by multiplying through by any nonzero constant.
The first line has direction numbers $(5,5,2)$ while the second line has direction numbers $(5,5,-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/270767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Calculate value of expression $(\sin^6 x+\cos^6 x)/(\sin^4 x+\cos^4 x)$
Calculate the value of expresion:
$$
E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}
$$
for $\tan(x) = 2$.
Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this:
Can you explain to me why t... | You can factor the term $\cos^6(x)$ from $\sin^6(x)+\cos^6(x)$ in the numerator to find: $$\cos^6(x)\left(\frac{\sin^6(x)}{\cos^6(x)}+1\right)=\cos^6(x)\left(\tan^6(x)+1\right)$$ and factor $\cos^4(x)$ from the denominator to find: $$\cos^4(x)\left(\frac{\sin^4(x)}{\cos^4(x)}+1\right)=\cos^4(x)\left(\tan^4(x)+1\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/271458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$ Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$
| (Presumably with tools from Calc I...) Using the conjugate identities
$$
\sqrt{a}-1=\frac{a-1}{1+\sqrt{a}},\qquad1-\sqrt[3]{b}=\frac{1-b}{1+\sqrt[3]{b}+\sqrt[3]{b^2}},
$$
one gets
$$
\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}=x_n+y_n,
$$
with
$$
x_n=\sqrt{1+a_n}-1=\frac{a_n}{1+\sqrt{1+a_n}},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| As $x,y,z$ are +ve real, we can set $\frac 1x=\sin^2A,\frac1y+\frac1z=\cos^2A$
again, $\frac1{y\cos^2A}+\frac1{z\cos^2A}=1$
we can put $\frac1{y\cos^2A}=\cos^2B, \frac1{z\cos^2A}=\sin^2B\implies y=\cos^{-2}A\cos^{-2}B,z=\cos^{-2}A\sin^{-2}B$
So, $$x^2+8y^2+27z^2=\sin^{-4}A+\cos^{-4}A(8\cos^{-4}B+27\sin^{-4}B)$$
We nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\
(|x-5|)^2=(|2x+6|-1)^2\\
...\\
9x^4+204x^3+1188x^2+720x=0?$
| Consider different cases:
Case 1: $x>5$
In this case, both $x-5$ and $2x+6$ are positive, and you can resolve the absolute values positively. hence
$$
x-5=2x+6-1 \Rightarrow x = -10,
$$
which is not compatible with the assumption that $x>5$, hence no solution so far.
Case 2: $-3<x\leq5$
In this case, $x-5$ is negative,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
$x^4 + y^4 = z^2$ $x, y, z \in \mathbb{N}$,
$\gcd(x, y) = 1$
prove that $x^4 + y^4 = z^2$ has no solutions.
It is true even without $\gcd(x, y) = 1$, but it is easy to see that $\gcd(x, y)$ must be $1$
| This has been completely revised to match the intended question. The proof is by showing that there is no minimal positive solution, i.e., by infinite descent. It’s from some old notes; I’ve no idea where I cribbed it from in the first place.
Suppose that $x^4+y^4=z^2$, where $z$ is the smallest positive integer for wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
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.