Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate $\sum\limits_{1 \le j\le k\le n} jk $ I am trying to solve an exercise from Concrete Mathematics but I seem to be stuck on the sum $$\sum_{1 \le j\le k\le n} jk $$ How to proceed? I have tried using Iverson's bracket condition like
$$ [1\le j\le k\le n] = [1\le j \le n][j\le k \le n] $$
but I am not sure how ... | An alternative is to set $j-1=a, k-j=b, n-k=c$. Then we are looking for
$$ S(n)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(a+1)(a+b+1)=\sum_{\substack{a+b+c=n-1\\a,b,c\geq 0}}(1+2a+b+a^2+ab) $$
where
$$ \sum_{d\geq 0}x^d = \frac{1}{1-x},\qquad \sum_{d\geq 0}dx^d = \frac{x}{(1-x)^2},\qquad \sum_{d\geq 0}d^2 x^d = \frac{x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Multiplication of matrices --$P^{-1}Q^{n}P^{-1}$ If $P$ be a square matrix - $\begin{pmatrix} √3/2 &1/2 \\ 1/2&√3/2 \end{pmatrix}$
and $A$ be another square matrix -$\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}$. Let $Q$ be the matrix given by $Q=PAP^{-1}$ then the question is to find the value of $P^{-1}(Q^{n})P^{-1}$ fo... | Hint: $Q^n = P A^n P^{-1}$ so the problem reduces to calculating $A^n$. Write:
$$
A = \begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}=I+A_1
$$
Note that $I,A_1$ commute, $I^k=I$ and $A_1^2=0$ so by binomial expansion:
$$
\require{cancel}
A^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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By knowing, that $a_{n+1}=(n+3)a_n$, how can I find $a_n$? $$a_1=1$$
$$a_{n+1}=(n+3)a_n$$
How can I get to the answer of this, which is:
$$a_n=\frac{(n+2)!}{6}$$
| write it as follows
$$\frac { { a }_{ n+1 } }{ { a }_{ n } } =n+3\\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } =4,\frac { { a }_{ 3 } }{ { { a } }_{ 2 } } =5,\frac { { a }_{ 4 } }{ { a }_{ 3 } } =6,...\frac { { a }_{ n } }{ { a }_{ n-1 } } =n+2\\ \\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } \cdot \frac { { a }_{ 3 } }{ { { a } }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the limit of $x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$ as $x\to +\infty$.
Calculate $\lim_{x\to +\infty} x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$.
First thing came to my mind is to simplify this to something easier. So multiply the numerator and the denominator by something like we used to do when two square roots involves. ... | Hint :
Use that
$$a-b=\frac{a^6-b^6}{(a^2+ab+b^2)(a^3+b^3)}$$
where $a=\sqrt{x^2+1}$ and $b=\sqrt[3]{x^3+1}$ to have
$$\begin{align}\lim_{x\to\infty}x(a-b)&=\lim_{x\to\infty}\frac{(a^6-b^6)x}{(a^2+ab+b^2)(a^3+b^3)}\\\\&=\lim_{x\to\infty}\frac{((x^2+1)^3-(x^3+1)^2)x}{(x^2+1+\sqrt{x^2+1}\sqrt[3]{x^3+1}+(x^3+1)^{2/3})((x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Find all $f:\mathbb{N} \to \mathbb{N}$ such that $f(m)^2 + f(n) \mid (m^2+n)^2;\; \forall n,m \in \mathbb{N}$
Find all $f:\mathbb{N} \to \mathbb{N}$ such that $f(m)^2 + f(n) \mid (m^2+n)^2;\; \forall n,m \in \mathbb{N}$
I tried to write $k(f(m)^2 + f(n)) = (m^2 + n) ^2$ and tried to find if there is any fixed $k$ bu... | (Note the obvious fact that the identity does satisfy the requirements.)
This is a partial answer, whose result suffices to show that $f(n) = n$ for all $n \leq 100000$.
Letting $n = m^2$, we obtain $f(m)^2 + f(m^2) \mid 4m^4$, and in particular $f(1)^2 + f(1) \mid 4$. It's easy to check that $f(1) = 1$ is the only pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?
I want to find limit of this problem by a very specific method . I am uploading the pic... | Alternate solution:
$$x_n < \frac{n}{n^3+1} + \frac{2n}{n^3+1} + \cdots + \frac{n\cdot n}{n^3+1} = \frac{n(1+2+\cdots + n)}{n^3+1} = \frac{n\cdot n(n+1)/2}{n^3+1}.$$
The limit of the last expression is $1/2.$ From below we have
$$\frac{n}{n^3+n} + \frac{2n}{n^3+n} + \cdots + \frac{n\cdot n}{n^3+n} = \frac{n\cdot n(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How many positive integers are divisors of at least one of $24^5, 20^6$, and $45^7$? I have a vague idea of prime factorizing 24, 20, and 45, then distributing the exponent, find how many numbers are divisible by them (factors), and then subtract the overcounts. How do I do this?
| $24=2^3\times 3\\
\implies 24^5=2^{15}\times 3^5$
Positive divisors of $24^5=16 \times 6=96$
$20=2^2\times 5\\
\implies 20^6=2^{12}\times 5^6$
Positive divisors of $20^6=13 \times 7=91$
$45=3^2\times 5\\
\implies 45^7=3^{14}\times 5^7$
Positive divisors of $45^7=15 \times 8=120$
Positive divisors of both $24^5$ and $20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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I have done this induction proof for the Fibonacci-sequence Would you please let me know how do I improve this proof?
The numbers $$F_{0},F_{1},F_{2},...$$ are defined as follows (this is a definition
by mathematical induction, by the way): $$F_{0} = 0, F_{1} = 1, F_{n+2} =
F_{n+1} + F_{n} \ for \ n=0,1,2,...$$
Prove ... | Base cases are fine.
At the inductive hypothesis you must assume that $P(k)$ and $P(k-1)$ are true. You have only said to assume $P(k)$
You could use "Strong induction" and assume that for all $i\le k, P(i)$ is true.
And then you seem to spin a while, to get to the point.
Show that $P(k+1)$ is true based on the assump... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If n is a positive multiple of 6, show that ${n\choose1}-3{n\choose3}+3^2{n\choose5}-\cdots=0$ If n is a positive multiple of 6, show that
$${n\choose1}-3{n\choose3}+3^2{n\choose5}-\cdots=0$$
$${n\choose1}-\frac{1}{3}{n\choose3}+\frac{1}{3^2}{n\choose5}-\cdots=0$$
I know that $\frac{1}{2}[(1+x)^n-(1-x)^n]$ will isolate... | As you already stated,
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$
and
\begin{align*}
\frac{1}{2}\left((1+x)^n - (1-x)^n\right) &= \sum_{k \geq 0} \binom{n}{2k+1} x^{2k+1}\\
&= \binom{n}{1}x + \binom{n}{3}x^3 + \binom{n}{5}x^5 + \ldots
\end{align*}
Now, to get alternate powers of odd terms, what rings a bell, is that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{... | The inequality is iff
\begin{aligned}
20-10\sqrt{3}<2\sqrt{3}\iff 20<12\sqrt{3}\iff 400<144\times3=432
\end{aligned}
which is clearly true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$
In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way.
My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}... | $$r=\frac{2S}{a+b+c}=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{a^2+2ab+b^2-c^2}=\frac{a+b-c}{2}$$
$$r_c=\frac{2S}{a+b-c}=\frac{ab}{a+b-c}=\frac{ab(a+b+c)}{a^2+2ab+b^2-c^2}=\frac{a+b+c}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$
Consider
$$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
I want to show that $I=3\pi$ and that $I=J$.
F... | Note that the integrands are even, so $$I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{4}}dx
$$ and $$J=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\left(2x\right)^{4}}{\left(1-x^{2}+x^{4}\right)^{3}}dx.
$$ Now let us consider the semicircular contour of radius $R$ centred in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Difficult trigonometry problem to find the minimum value Find the minimum value of $5\cos A + 12\sin A + 12$.
I don't know how to approach this problem. I need help.
I'll show you how much I got...
$$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$
| As we know that,
$−\sqrt{a^2+b^2} \le a \cosθ + b \sinθ \le \sqrt{a^2+b^2}$
So for $5 \cos θ+12 \sin θ$ we have,
$−\sqrt{5^2+12^2} \le 5cosθ+12sinθ \le \sqrt{5^2 +12^2}$
= $−13 ≤ 5cosθ +12sinθ ≤ 13$
Now adding 12 we get,
= $−13 + 12 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 13 + 12$
= $− 1 ≤ 5 \cosθ + 12 \sinθ + 12 ≤ 25$
So minimu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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How to i solve this Exponential equation How to solve this exponential equation?
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
| $$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
$$21 \cdot 3^{x} -25\cdot 5^{x}= 81\cdot3^{x}- 125\cdot5^{x}$$
$$100\cdot 5^{x}= 60\cdot3^{x}$$
$$\left(\frac{5}{3}\right)^x=\frac{3}{5}=\left(\frac{5}{3}\right)^{-1}$$
$$x=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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The digits 0-7 are used to form 4-digit numbers... Restrictions for all questions: No repetition and the thousands place cannot be 0 (ex. 0123, 0555, etc.)
A. How many 4-digit numbers are possible?
B. How many of these 4-digit numbers are odd?
C. How many of these 4-digit numbers are even?
D. How many of these 4-digit ... |
How many of these four-digit numbers can be formed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ if no two digits are the same?
Since $0$ cannot be placed in the thousands place, we have seven choices for the thousands digit, which leaves seven choices for the hundreds digit (since we can now use $0$ but cannot select th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding maxima and minima of $f(x,y)= x^4 + y^4 - 2x^2 - 2y^2 + 4xy$ For $f(x,y)=x^4+y^4-2x^2-2y^2+4xy$, I need to find maxima or minima. There are three critical points: $(0,0),(\sqrt2, -\sqrt2),(-\sqrt2,\sqrt2)$
So at $(\sqrt2, -\sqrt2)$, $f$ has minimum value, $-8$ and at $(-\sqrt2,\sqrt2),$ it has same minimum valu... | It's obvious that a maximal value does not exist.
We'll prove that $-8$ is a minimal value.
Let $x=\sqrt2a$ and $y=-\sqrt2b$.
Hence, we need to prove that
$$4a^4+4b^4-4a^2-4b^2-8ab+8\geq0$$ or
$$a^4+b^4+2\geq(a+b)^2,$$
which is AM-GM and C-S:
$$a^4+b^4+2\geq2\sqrt{(a^4+1)(1+b^4)}\geq2(a^2+b^2)\geq(a+b)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
System of equations - conics
What would be the best way to solve the following the following system of equations? $$x^2+y^2+6y+5=0$$ $$x^2+y^2-2x-8=0$$
This is how I did it, but I am hoping there is a simpler way.
Subtracting the two equations, I get $$6y+2x+13=0$$which simplifies to $$x=-3y-\dfrac{13}2$$I substitut... | After the substitution, we have
$$10y^2 + 45 y + \dfrac{189}{4} \implies y_{1, 2} = -\dfrac{9}{4} \pm \dfrac{3~ \sqrt{\dfrac{3}{5}}}{4}$$
This leads to
$$ x_{1, 2} = \dfrac{1}{4} \mp \dfrac{9~ \sqrt{\dfrac{3}{5}}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of given indefinite integral Evaluate the given integral
$$\int e^x \bigg[\frac{2-x^2}{(1-x)\sqrt{1-x^2}} \bigg]dx$$
I was trying to convert it to $\int e^x (f(x)+f'(x))dx=e^x \cdot f(x)+C$ but did not succeed in algebraic manipulations. Could someone hint me to something so that I could proceed?
| $$\begin{aligned}
\frac{2-x^2}{(1-x)\sqrt{1-x^2}}
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{1-x^2}{(1-x)\sqrt{1-x^2}} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1-x^2}}{(1-x)} \\
&=\frac{1}{(1-x)\sqrt{1-x^2}}+\frac{\sqrt{1+x}}{\sqrt{1-x}} \\
\end{aligned}$$
The only thing left is to notice that,I'll leave that to you.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$ Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$
| Assume without loss of generality that $a$ is the maximum of $a,b,c$. This gives us that $$ 3a \ge a+b+c=3 \iff 2 \ge a \ge 1 \tag{1}$$
From the condition $0 \le b,c \le a \le 2$. Note that we have that $$a^3+b^3+c^3 \le a^3+(b+c)^3=a^3+(3-a)^3=9\left(a-\frac{3}{2} \right)^2+\frac{27}{4} \tag{2}$$
Which follows from t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmat... | If either $AB$ or $B^TA^T$ is singular, there is no chance to get a solution.
Therefore the $3 \times 3$ matrices, have to be nonsingular (and they are, I just checked).
So, to get $X$ you need to invert both of them and get to $X=B^{-1}A^{-1}A^{-T}B^{-T} = B^{-1}(A^TA)^{-1}B^{-T}$
Usually you would decompose $AB$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $\sin\frac{\pi}n·\sin\frac{2\pi}n···\sin\frac{(n-1)\pi}n=\frac{n}{2^{n-1}}$ How to prove that
$$
\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}
$$
using the roots of $(z+1)^n-1=0$?
My rough idea is to solve $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roo... | Clearly $z=0$ is a root of $(z+1)^n-1$, so we need to exclude it to get something nontrivial for the product. Dividing and taking the limit,
$$ \lim_{z \to 0} \frac{(z+1)^n-1}{z} = n, $$
so the product of the remaining roots is $(-1)^{n-1} n$. Now we have to find expressions for the roots. We have roots
$$ z_k+1 = e^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| There's a sort of completion of the square that goes like this:
\begin{align}
x^4+1 & = \underbrace{(x^4+2x^2 + 1)}_\text{This is a square.} - \underbrace{(2x^2)}_\text{So is this.} \\[10pt]
& = \left( x^2+ 1 \right)^2 - (\sqrt 2\ x)^2 \\[10pt]
& = (x^2 + 1 - \sqrt 2\ x)(x^2 + 1 + \sqrt2\ x).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 1
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Proof in Arithmetic Progression My maths teacher at school asked a question which I am finding difficult to crack down.
We are given that $a^2 , b^2$ and $c^2$ are in AP. We need to prove that $\frac{a}{b+c} , \frac{b}{a+c}$ and $\frac{c}{a+b}$ are in AP.
This is what I tried.
Let the common difference of the AP be... | $AP1$: $$a^2\\ b^2\\ c^2$$
$$\text{Common Difference, }\quad d=b^2-a^2=c^2-b^2\qquad \qquad $$
$XP2$:
$$P=\frac a{b+c}\color{lightgrey}{\cdot\frac{(c-b)}{(c-b)}}=\frac {a(c-b)}{c^2-b^2}=\frac {a(c-b)}d\\
Q=\frac b{c+a}\color{lightgrey}{\cdot\frac{(c-a)}{(c-a)}}=\frac {b(c-a)}{c^2-a^2}=\frac {b(c-a)}{2d}\\
R=\frac c{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Can a factor of $x^2+4$ for odd $x$ be $\equiv 3\mod 4$? When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$?
Examples:
$$7^2+4=53$$
$$11^2+4= 5^3$$
$$31^2+4= 5 \cdot 193$$
$$47^2+4=2213$$
$$89^2+4=5^2 \cdot 317$$
All of the prime factors above are $\equiv 1 \mod 4$.
Is there no... | If $p$ is an odd prime divisor of $x^2+4$, then we have $-4 \equiv x^2 \pmod p$
So $1=\left(\frac{-4}{p}\right)=\left(\frac{4}{p}\right) \left(\frac{-1}{p}\right) = \left(\frac{-1}{p}\right)$ which means that $p \equiv 1 \pmod 4$ by the first supplement to quadratic reciprocity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Whats is the limit of $\lim_{(x,y)\to(a,a)}\frac{x^3-y^3}{x^2-y^2}$ for $a\ne0$? What is $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}$ for $a\ne 0$? I have found that the limit coould be $\frac{3a}{2}$ to proof it I have come to the point where $\left|\frac{x^3-y^3 }{x^2-y^2}-\frac{3a}{2}\right| = \left|... | $$\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{(x-y)(x^2+xy+y^2) }{(x-y)(x+y)}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^2+xy+y^2}{x+y}=\frac{3a}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x_n$ be a sequence in a complete metric space $(X,d)$ such that $d(x_n,x_{n+1}) \le \frac {1}{n^2}$. Is $x_n$ convergent? I have a proof that $x_n$ is convergent. Here it is:
Let $\epsilon \gt 0$ be given and let $m \gt n$ where $n,m \in \Bbb N$.
Then $d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+...+d(x_{m-1},... | How about
\begin{align*}
d(x_n,x_m) \le \frac 1{n^2} +\frac 1{(n+1)^2}+...\frac 1{(m-1)^2}&\le \frac 1{n(n-1)} +\frac 1{(n+1)n}+...\frac 1{(m-1)(m-2)}\\
&\le \frac 1{n-1} -\frac 1{m-1}\le\frac 1{n-1},
\end{align*}
but this is in fact one of the approaches to show $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit? Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit?
$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$
I dont know how to simplify $\frac{2^n }{ n}$.
I cannot just do $\frac{... | $\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{5} \dots \frac{2}{n} < \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{4} \dots \frac{2}{4} = \frac{8}{6}\cdot(\frac{1}{2})^{n-3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Least Positive Residue How would I find the least positive residue of
say
$6! \bmod 7$
or
$12! \bmod 13$
I just learned modular arithmetic and my book doesn't explain what least positive residues are so I'm a bit lost.
| Do you already know this?: $a\equiv b \bmod m \implies ac\equiv bc \bmod m$.
You can use this to keep the numbers small as you investigate these results:
$4! = 1\cdot 2\cdot 3\cdot 4 \equiv 24\equiv 3 \bmod 7$
$5! = 4!\cdot 5 \equiv 3\cdot 5 \equiv 15 \equiv 1 \bmod 7$
$6! = 5!\cdot 6 \equiv 1\cdot 6 \equiv 6 \bmod 7$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove using induction the inequality.
$\forall$:n∈${N}$
$\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$
I tried with no any success...
| $$n=1 :\binom{2}{1}=2 \geq \dfrac{4}{3}
\\n=k \to \binom{2k}{k}\geq \dfrac{4^k}{2k+1} \\
n=k+1 \to \binom{2(k+1)}{k+1}\geq \dfrac{4^{k+1}}{2(k+1)+1}$$
$$\binom{2(k+1)}{k+1}=\dfrac{(2k+2)!}{(k+1)!(k+1)!}=\dfrac{(2k+2)(2k+1)(2k)!}{(k+1)^2k!}=\\
\dfrac{(2k+2)(2k+1)}{(k+1)^2} \binom{2k}{k}\geq \dfrac{4.4^k}{2k+3}=4.\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that a limit goes to infinity Looking for some help in proving why the following infinity goes to $\infty$ as n approaches $\infty$
$$\frac{(1-\frac{1}{n})^n}{1-(1-\frac{1}{n})}$$
| $$\dfrac{(1-\dfrac{1}{n})^n}{1-(1-\dfrac{1}{n})}=\\
\dfrac{\binom{n}{0}1^{n}{(-\dfrac{1}{n})}^{0}+\binom{n}{1}1^{n-1}{(-\dfrac{1}{n})}^{1}+\binom{n}{2}1^{n}{(-\dfrac{1}{n-2})}^{2}+...+\binom{n}{n}1^{0}{(-\dfrac{1}{n})}^{n}
}{1-(1-\dfrac{1}{n})}=\\
\dfrac{1+n{(-\dfrac{1}{n})}+\dfrac{n(n-1)}{2}{(-\dfrac{1}{n})}^{2}+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $|z - w| \geq \big||z|-|w|\big|$
Show that $|z - w| \geq \big||z|-|w|\big|$ given $z = x + iy$ and $w = u + iv$.
So far I have,
$$|z - w| \geq \big||z|-|w|\big|$$
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
At this point I get a little 'undone'.
I tried a few different route... | Other answers address your issue in a better way but I think it will be nice if a proof by your line of thought is given.
$$\sqrt{(x - u)^2 + (y - v)^2} \geq |\sqrt{x^2 + y^2} -\sqrt{u^2 + v^2}|$$
Since both sides are $>0$,
$$(x - u)^2 + (y - v)^2 \geq x^2 + y^2 - u^2 + v^2 -2\sqrt{x^2 + y^2}\sqrt{u^2 + v^2}$$
$$(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Shifted Legendre Polynomial We know that the Legendre Polynomial is $P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}[(x^2-1)^n]$. The Shifted Legendre Polynomial $\tilde P_n(x)$ is defined as $P_n(2x-1)$. Can you please tell me how to get $\tilde P_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}[(x^2-x)^n]$? Thanks.
| $$P_n(X) = \frac{1}{2^nn!}\frac{d^n}{dX^n}[(X^2-1)^n]$$
With $X=(2x-1)$
$$\tilde P_n(x)=P_n(2x-1)=P_n(X)=\frac{1}{2^nn!}\frac{d^n}{d(2x-1)^n}[\left((2x-1)^2-1\right)^n]$$
$d(2x-1)=2dx \quad\to\quad d(2x-1)^n=2^n dx^n$
$\left((2x-1)^2-1\right)^n=(4x^2-4x)^n=4^n(x^2-x)^n$
$$\tilde P_n(x)=\frac{1}{2^n n!}\frac{d^n}{2^n d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Validating the inequality. $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}$, $x>0$
Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.
We have done the Lagrange's mean value theorem and intermediate value th... | defining $$f(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ln(1+x)$$ for $x=0$ we get $$f(0)=0$$ and $$f'(x)=1-x+x^2-\frac{1}{x+1}=\frac{(1-x+x^2)(1+x)-1}{x+1}=\frac{x^3}{1+x}>0$$ for $x>0$ thus our function is monotonously increasing and we get $$f(x)>0$$ for all $$x>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the limit: $\lim_{x\to\infty}x\sin(\tan\frac1x)$ How do I find the limit:
$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$
| Rewrite the limit as
\begin{align}
L:=\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right)=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}
\end{align}
Since both the numerator and the denominator tend to $0$ in the limit, this is an indeterminate form and we can apply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2156109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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integration by parts with $\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz$ Can someone please show how to calculate this with integration by parts $(\int udv = uv - \int vdu)$? I found an example in the book is not clear and confusing.
$$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz$$
They st... | With $\dfrac{z^2}{2}=u$ and
$$\frac{2}{\sqrt{2\pi}} \int^\infty_0 z^2e^{\frac{-z^2}{2}}dz=\frac{2}{\sqrt{\pi}} \int^\infty_0 u^\frac12e^{-u}du=\frac{2}{\sqrt{\pi}} \Gamma(\frac32)=\frac{2}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2156350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to integrate this $\int \frac{\sqrt{x}}{1+x^4}dx$? How to integrate this ? $$\int \frac{\sqrt{x}}{1+x^4}dx$$
Any hint or idea on how to proceed?
Edit: Here is the final answer using Wolfram Alpha
| You can use the substitution $x=t^2$, $dx=2t\ dt$, so you get
$$\int \frac{2t^2}{(1+t^8)} \ dt$$
although that one is a rational integral which involves many calculations.
To solve that rational integral, we first have to decompose $p(t)=1+t^8$ in irreducible polynomials. As it's clear $p(t)$ doesn't have real roots, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The following equation to solve :$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$ The following equation to solve :
$$ \tan x+\cot x=\sqrt{2}(\cos x+\sin x)$$
My try:
$$\frac{2}{\sin 2x}=\sqrt{2}(\cos x+\sin x)$$
$$\left(\frac{2}{\sin 2x}\right)^2=(\sqrt{2}(\cos x+\sin x))^2$$
$$\left(\frac{2}{\sin 2x}\right)^2=2(1+\sin 2x)$$
... | Excluding $\sin x\cos x=0$, you can rewrite
$$\sqrt2(\cos x+\sin x)\sin x\cos x=1$$
or
$$\sin\left(x+\frac\pi4\right)\sin(2x)=1.$$
For this product to be $1$, both factors must be $1$ or $-1$.
Then
$$x+\frac\pi4=k\pi+\frac\pi2,\\2x=l\pi+\frac\pi2,$$
where $k$ and $l$ have the same parity. Then as $l=2k$, $l$ and $k$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\time... | I give here an answer advocating the simplicity of using Newton (or Newton-Girard) formulas. Using the notations of the Wikipedia article in all generality:
Let $e_0=1, e_1=x+y+z, e_2=xy+yz+zx, e_3=xyz.$
Let $p_1=x+y+z$, $p_2=x^2+y^2+z^2$, $p_3=x^3+y^3+z^3$.
Then:
$$\begin{cases}
e_1&=&p_1\\
2e_2&=&p_1e_1-p_2\\
3e_3&=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Evaluate $1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$ Evaluate:
$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$
a_n are the individua... | Recall that the multiple zeta values are defined by the series
$$
\zeta(s_1,\ldots,s_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\ldots n_k^{s_k}}.
$$
The sum $S$ can be expressed as a linear combination of multiple zeta values. We have
$$
\begin{align*}
S&=\sum_{n=1}^\infty \sum_{k_1,k_2=1}^n\frac{1}{n^2k_1k_2}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Limit of a infinite series $$\frac{2}{2} + \frac{2\cdot 5}{2\cdot 9} + \frac{2\cdot 5\cdot 10}{2\cdot 9\cdot 28} + \cdots + \frac{2\cdot 5\cdot 10 \cdots (n^2+1)}{2\cdot 9\cdot 28\cdots (n^3+1)}\tag1$$
For this series $(1)$, how would one go about applying the comparison test to check for convergence or divergence?
| For $n\ge 3$ we get
$$\frac{n^2+1}{n^3+1}<\frac{n^2+1}{n^3}=\frac1n+\frac1{n^3}\le\frac13+\frac1{27}=\frac4{27}<\frac12$$
Then
$$\sum_{k=1}^n\prod_{j=1}^k\frac{j^2+1}{j^3+1}<1+\frac{5}{9}+\sum_{k=3}^n\frac59\left(\frac12\right)^{k-2}$$
So, by comparison test, we get that the given series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
equation in D'(\R) we consider in $\mathcal{D}'(\mathbb{R})$ the equation
$$2 x T'' - T' =\delta$$.
I begin to solve the homogeneous equation $2x T'' - T'=0$
i search an solution $T=x^r$ with $r \in \mathbb{R}$, then i found that:
$T_h(x)= C_1 x^{1/2} + C_2 x$
my question is: can you give me please an indication to ... | Let's pose $S = T'$. Then:
$$2 x T'' - T' =\delta \Rightarrow 2x S' - S = \delta \Rightarrow \\
\Rightarrow 2xS' = \delta + S \Rightarrow 2x \frac{dS}{dx} = \delta + S \Rightarrow \\
\frac{dS}{\delta + S} = \frac{dx}{2x}.$$
Integrating both side...
$$\int\frac{dS}{\delta + S} = \int \frac{dx}{2x} \Rightarrow \log(\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Number of ways in which 3 people can throw a normal die to have a total score of 11
Number of ways in which 3 people can throw a normal die to have a total score of 11
My approach:
The answer can be obtained by finding the coefficient of $x^{11}$ in the expansion of $(x+x^2+x^3+x^4+x^5+x^6)^3$.
General term $... | The approach using generating functions is also fine. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{11}](x^1+x^2+x^3+x^4+x^5+x^6)^3}&=[x^{11}]x^3(1+x+x^2+x^3+x^4+x^5)^3\\
&=[x^8]\left(\frac{1-x^6}{1-x}\right)^3\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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How to find an upper bound on the number of solutions of $y^3=x^2+4^k$ I have solved the first two parts of this question but I am struggling with the remaining section. I can't see any meaningful way to reuse what I did before and/or find a way forward.
Just to be clear it is part c) that I am stuck on.
Would anyone ... | The following is heavily based on Theorem 3.3 in Keith Conrad's excellent "Examples of Mordell’s Equation".
$$y^2+4^k=x^3$$
Observe that $4^k$ is always a square, so we factor in $\mathbb{Z}[i]$
$$(y-2^ki)(y+2^ki)=y^2+2^{2k}=x^3$$
If $y-2^ki$ and $y+2^ki$ are cubes then: $y-2^ki=(m+ni)^3$, so $2^k=n(3m^3-n^2)$ there ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why do I get this matrix the wrong way round? The question is: The set B = {${1+t^2, t+t^2, 1+2t+t^2}$} is a basis for P2. Find the coordinate vector of $p(t)=1+4t+7t^2$ relatvive to B.
I made a matrix:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 1 & 0 & 6
\end{bmatrix}
When I row reduce it I get:
\... | Here is one sequence of steps to row reduce the matrix.
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 1 & 0 & 6
\end{bmatrix}
Subtract the second row from the third:
\begin{bmatrix}
1 & 0 & 1 & 1\\ 0 & 1 & 2 & 4\\
0 & 0 & -2 & 2
\end{bmatrix}
Add the third to the second:
\begin{bmatrix}
1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\... | $I=\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$
Put $x=\frac{1}{t}\implies dx=-\frac{1}{t^2}dt \implies
I=-\int\frac{t}{(t^2+1)\sqrt{1-t^2}}dt
$ Put$ 1-t^2=y^2\implies -t dt=y dy. \implies
I=\int\frac{y}{(2-y^2)y}dy
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+y}{\sqrt2-y}+C
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+\sqrt{1-t^2}}{\sqrt2-\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum... | Your work is good. Here's another way to do it.
Let,
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$f'(x)=\sum_{n=1}^{\infty} x^{2n}=\frac{1}{1-x^2}-1$$
It follows,
$$f(x)=\tanh^{-1}(x)-x$$
Now let,
$$g(x)=\sum_{n=1}^{\infty} n\frac{x^{2n+1}}{2n+1}$$
Clearly we have,
$$2g(x)+f(x)=\sum_{n=1}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Maximum value of equation $ab+bc+ca$ if $a+2b+c=4$ If $a,b,c$ ${\in}$ $\mathbb{R} $ such that
$a+2b+c=4$, then find the $max$ value of
$ab+bc+ca$.
I always get stuck with max, min questions. We cannot apply AM:GM here, I have not studied much calculus yet.
Can you do this with graphs or by plane algebra?
I don't reall... | Substitution of $b=\frac{4-a-c}{2}$ into $ab+bc+ca$ gives:
$$\begin{align}-\frac{a^2}{2} + 2 a -\frac{c^2}{2} + 2 c
& = \tfrac{1}{2}\left( 4a-a^2+4c-c^2\right) \\[3pt]
& = \tfrac{1}{2}\left( 4-\left(2-a \right)^2+4-\left(2-c \right)^2 \right) \\[3pt]
& = 4-\tfrac{1}{2}\left(2-a \right)^2-\tfrac{1}{2}\left(2-c \right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Stewart's Theorem In triangle $ABC$, angle $\angle A = 90^\circ$. Let $M$ be a point on the hypotenuse $BC$. Prove that $MB^2\left(AC^2\right)+MC^2\left(AB^2\right)=MA^2\left(BC^2\right)$.
Can someone please provide guidance as to how to go about this proof?
|
Note that $AQ = AC\cdot \frac{BM}{BC}$ and $AP = AB\cdot \frac{CM}{BC}$
Then $AM^2 = AP^2 + AQ^2 = \left(AC\cdot \frac{BM}{BC}\right)^2 + \left(AB\cdot \frac{CM}{BC}\right)^2$
So $AM^2\cdot BC^2 = BM^2\cdot AC^2 + CM^2\cdot AB^2$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of transition probabilities of an infinite Markov chain I have a Markov chain with state space $S = \left\{ 1,2,\dots \right\}$ with transition matrix as follows
$\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 & \dots \\
0 & 0 & 1 & 0 & 0 & 0 & \dots \\
0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
... | Yes it is appropriate to consider the the state transition matrix
$$\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 & \dots \\
\frac{2}{3} & 0 & \frac{1}{3} & 0 & \dots \\
\frac{2}{3} & 0 & 0 & \frac{1}{3} & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
$$
since the original chain gets in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factorise the expression Factorise of the expression $$ a^4+b^4-c^4-2a^2b^2+4abc^2~ \\(S:(a + b - c) (a + b + c) (a^2+ b^2 + c^2 - 2 a b))$$
I tried to solve this problem:
$$ a^4+b^4-c^4-2a^2b^2+4abc^2\rightarrow\\\
(a^2-b^2)^2-c^4+4abc^2\rightarrow\\\
(a^2-b^2-c^2)(a^2-b^2+c^2)+4abc^2\\\
((a-b)(a+b)-c^2)((a-b)(a+b)+... | .You should actually do:
\begin{split}
a^4+b^4-c^4-2a^2b^2+4abc^2 & = (a^4+b^4 + 2a^2b^2) - (4a^2b^2 + c^4 - 4abc^2) \\
& = (a^2+b^2)^2 - (c^2-2ab)^2 \\
& = (a^2+b^2-c^2+2ab)(a^2+b^2+c^2-2ab) \\
& = ((a+b)^2 - c^2) (a^2+b^2+c^2-2ab) \\
& = (a+b+c)(a+b-c)(a^2+b^2+c^2-2ab)
\end{split}
In the first step, I added and subtr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $BAB^{-1}=J$ for matrix $B$ I have 3 Matrices : $A , B$ and $J.$ I know matrices $A$ and $J. J$ is the real jordan normal form of $A.$ The relationship between the three matrices is given by $BAB^{-1} =J.$
How can I find the matrix $B.$
For example,
$A= \begin{bmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
1 &-1 &... | $\det(A-\lambda I) = 0$
$-(\lambda-1)(\lambda+i)(\lambda-i)$
Complex eigenvalues
Here is a little trick
$(B^{-1} A B)^2 = J^2\\
B^{-1} A^2 B = J^2$
And $A^2$ has real eigenvalues and the same matrices $B, B^{-1}$
$A^2 = \begin{bmatrix} 0&0&1\\1&-1&1\\1&0&1\end{bmatrix}$
and $J^2 = \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\... | {
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"url": "https://math.stackexchange.com/questions/2181211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the the given equation: $\sqrt{3x^2+x+5} = x-3$ We have to find the number of solution for the given equation:
$$\sqrt{3x^2+x+5} = x-3.$$
There are two solution one is
By using graph we get one solution
By squaring both sides we get no solution
I want to know which solution is correct
| Hint:
The given equation is equivalent to the system:
$$
\begin{cases}
3x^2+x+5=(x-3)^2\\
x-3\ge 0
\end{cases}
$$
Note that the system has no solutions because the roots of the second degree equations :
$$
x=\frac{-7\pm\sqrt{79}}{4}
$$
are less than $3$.
And this is in accord with the fact that the graphs of the two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x... | We assume you want real roots (complex roots are 6 exactly).
$x=1$ is one of the root of the main equation so $t=2$ is the root of the second, write
$$\displaystyle t^3+at^2+(b-3)t+1-2a=0~~;~~~t\neq0~~(x\neq-1)$$
$$(t-2)(t^2+mt+n)=0~~;~~~t\neq0$$
this concludes that $n=a-\dfrac12$ and $m=a+2$.
$t\neq0$ says $n=a-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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geometric progression Help me to solve this question :
The sum of first two terms of a positive geometric progression is $9$ times the sum of the following two terms. Given that the sum of the first four terms is $\frac{320}{9}$.
Find
*
*the first term and the common ratio
*the sum of fifth term to infinit... | You're almost there. First, the common ratio $r$ must be positive (since all terms are positive), that is, $\color\red{r>0}$. Start with equation $(1)$ by dividing by $a$ to get $$1+r=9(r^2+r^3).$$ Next, apply factoring to get $$1+r=9r^2(1+r).$$ Then cancel $1+r$ (since $r>0$ and so $1+r\neq 0$) to get $$1=9r^2$$ so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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a surface integral problem $F=(y-z,z-x,x-y)$ $S$ be the portion of surface defined by $x^2+y^2+z^2=1$ and $x+y+z\ge 1$. We want to evaluate $\int_S curl(F)\cdot dS$.
I have found $curl(F)=(-2,-2,-2)$ and the normal vector of $S$ is $(x,y,z)$. Thus the integral becomes $\int_S -2(x+y+z) dS$. However, if I use the polar... | You can also use the divergence theorem here, it will save you the heavy computations you are not comfortable with:
If $S_2$ denotes the part of the plane $x+y+z=1$ inside the sphere $x^2+y^2+z^2=1$, and $E$ the 3D-region bounded by $S$ and $S_2$, then
$$
\iint_S \nabla \times F\cdot dS+ \iint_{S_2} \nabla \times F\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Explanation why $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ = $\frac{-\pi^2}{12}$ with Euler's $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ Can somebody explain why the sum is $\frac{-\pi^2}{12}$. Can you somehow use the zeta function? If so, how?
| Let's add them together:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n+1}{n^2}$$
and for odd $n$, $(-1)^n+1=0$ and for even $n$, $(-1)^n+1=2$, so this rewrites to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $1^2+3^2+...99^2$ given $1^1+2^2+...+100^2$ and $1^1+2^2+...+50^2$ We know that $1^1+2^2+...+100^2=338350$ and $1^1+2^2+...+50^2=42925$. Find $1^2+3^2+...99^2$.
I don't know really where to start. I tried to find a pattern in the sequences, but there was none. Can I substitute values for the equations?
| As mentioned in the comments the sum of the series can be derived in the following way
$$\sum_{i=1}^{2n}i^2=1^2+2^2+3^2+...+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}$$
Similarly,
$$\sum_{i=1}^{n}{(2i)}^2=2^2+4^2+...+(2n)^2=4\sum_{i=1}^{n}{i}^2=\frac{2n(n+1)(2n+1)}{3}$$
Subtacting gives required sum
$$\sum_{i=1}^{n}{(2i-1)}^2=1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Given 4 equations, is there a unique solution that can fit a model with 2, 3, or 5 variables? Given the a data set with 4 equations with x1, x2 as main drivers.
A) Can there be a unique multilinear model s(x1,x2) = B0+ B1x1+ B2x2 that perfectly fits the data.
Even before putting the equations into MAtLab, I assumed the... | A
Problem specification: Start with a sequence of $m=4$ measurements $\left\{ x_{k}, y_{k}, z_{k} \right\}_{k=1}^{m}$. Use the method of least squares to find the best trial function
$$
z(x,y) = b_{0} + b_{1}x + b_{2}y.
$$
That is, find the solution vector $b$ defined as
$$
b_{LS} = \left\{ b \in \mathbb{C}^{m} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why are vectors of a regular hexagon cos60 of each other? In a regular hexagon OABCDE $a$ , $b$ denote respectively the position vectors of $A$, $B$ with respect to $O$
I would have assumed that $\overrightarrow {OA} = \overrightarrow {AB} = \overrightarrow {BC} $ since it's a regular hexagon. And, therefore, $\overri... | The actual vectors are:
$$\overrightarrow{OA}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$
$$\overrightarrow{AB}=\begin{pmatrix}1\\0\end{pmatrix}$$
$$\overrightarrow{BC}=\begin{pmatrix}\frac12\\\frac{\sqrt{3}}2\end{pmatrix}$$
So we have $|\overrightarrow{OA}|=|\overrightarrow {AB}|=|\overrightarrow{BC}|$.
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Complex number equation to cartesian equation I need to find the cartesian equation of the loci of z given the below equation:
$\frac{(z-j)\cdot(z-j)^*}{|3-4j|}=5$
So if $z=x+jy$
Then $(z-j)^*=(x+jy-j)^*$
$=[x+j(y-1)]^*$
$=x-j(y-1)$
$=x-jy+y$
Substituting back into original equation
$\frac{(x+jy-j)\cdot(x-jy+j)}{|3-4j|... | For any complex number $w = a + bj$,
$$|w| = \sqrt{a^2 + b^2}$$
$$ww^* = |w|^2 = a^2 + b^2$$
hence if $z=x + yj$, then
\begin{align*}
z-j &= x +yj - j = x + (y-1)j\\[4pt]
\implies\; (z-j)(z-j)^*&=|z-j|^2\\[4pt]
&= |x + (y-1)j|^2\\[4pt]
&= x^2 + (y-1)^2\\[4pt]
\end{align*}
Also, $|3-4j| = \sqrt{3^2 + (-4)^2} = \sqrt{25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int{\sqrt{x^2+1}} \ \mathrm{d}x$ using integration by parts, given that $\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\ln(x+\sqrt{x^2+1})+k$. Here is the question once again:
Deduce the integral $\int{\sqrt{x^2+1}} \ \mathrm{d}x$ using integration by parts, given the following identity: $$\int{\frac{1}{\sqrt{x... | Notice that
$$\int \sqrt {x^2 + 1} \ \Bbb d x = \int (x')\sqrt {x^2 + 1} \ \Bbb d x = x \sqrt {x^2 + 1} - \int x \frac {x} {\sqrt {x^2 + 1}} \ \Bbb d x = \\
x \sqrt {x^2 + 1} - \int \frac {x^2 + 1 -1} {\sqrt {x^2 + 1}} \ \Bbb d x = x \sqrt {x^2 + 1} - \int \sqrt {x^2 + 1} - \frac 1 {\sqrt {x^2 + 1}} \ \Bbb d x = \\
- \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*... | $$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$
As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.
Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization ... | We have
$$
(x^2-12x+72)(x^2+12x-72)=x^4 - 144x^2 + 1728x - 5184=x^2(x^2-12^2)+1728(x-3),
$$
The factorization is of the form $(a^2+b)(a^2-b)=a^4-b^2$, with $a=x$ and $b=12x-72$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| Very short with Lil' Fermat:
for any $x\in\mathbf Z$, $\;x^3\equiv x\mod 3$. Hence $\;a^3+b^3+c^3\equiv a+b+c\mod 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \c... | $\sin^2(a+b)+\sin^2(a−b)
$
$=\sin^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b+\cos^2a\sin^2b
$
$=2(\sin^2a\cos^2b+\cos^2a\sin^2b)
$
$=2\left ({\left (\dfrac{(1-\cos(2a)}{2}\right )(\cos^2b)+\left (\dfrac{\cos(2a)+1}{2}\right )(\sin^2b)}\right )
$
$=1+\cos(2a)\sin^2b-\cos(2a)\cos^2b
$
$=1-\cos(2a)\cos(2b)
$
Hence proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
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Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$
Let $x,y$ be positive integers satisfying $2x^2-y^2 = 1$. Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$.
I wasn't sure how to use the fact that $x,y$ are positive integers satisfying $2x^2-y^2 = 1$. We could use the theory of Pell's e... | If $(x,y)$ is a solution of $2x^2-y^2=1$ such that $x>5$, $y>0$ and $x\equiv 5(10)$ then
$$(x',y')=(99 x-70 y,-140 x+99 y)$$
is also a solution with $0<x'<x$ , $x' \equiv 5(10)$, and $y' \equiv y(7)$
The process of repeatedly applying this transformation to a solution $(x,y)$ will stop at the solution $(5,7)$ thereby p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The minimum value of $x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$ is The minimum value of $$f(x)=x^8 – 8x^6 + 19x^4 – 12x^3 + 14x^2 – 8x + 9$$ is
(a)-1
(b)9
(c)6
(d)1
Apart from trying to obtain $1$, which in this case is simple and $f(2)=1$ is there a standard method to approach such problems.
Please keep in mind tha... | Let $a,b,c,d,e,k\in\Bbb R$,
$$(x^4-ax^2)^2+(bx^2-cx)^2+(dx-e)^2+k=\\ x^8- 2 a x^6+(a^2+ b^2) x^4 - 2 b c x^3 + (c^2 + d^2) x^2 - 2 d e x+e^2+k$$
By identification we easily obtain
$$a=4 \implies b = \sqrt{3} \implies c=\frac{6}{\sqrt{3}} \implies d= \sqrt{2} \implies e= \frac{4}{\sqrt{2}} \implies k=1$$
this implies th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ How can I prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ without induction? I think I have to use Taylor series at some point.
| We may notice that by setting $C_i=\frac{1}{i+1}\binom{2i}{i}$ and $A_i=\frac{C_i}{4^i}$ we have
$$ f(x) = \sum_{i\geq 0}A_i x^i = 2\frac{1-\sqrt{1-x}}{x}\tag{1} $$
hence
$$ \frac{f(x)}{1-x}=\sum_{i\geq 0}(A_0+\ldots+A_i)x^i = 2 \frac{1-\sqrt{1-x}}{x(1-x)} \tag{2} $$
and the wanted sum is the coefficient of $x^{n-1}$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\... | We put $t=\sqrt{x}$, so $dt=\frac{1}{2\sqrt{x}}dx$.
$$\int_0^12(1+3t^2+5t^6)$$
$$=2\left[t+t^3+\frac{5t^7}{7}\right]^1_0$$
$$=2\left[\sqrt{x}+\sqrt{x}^3+\frac{5\sqrt{x}^7}{7}\right]^1_0$$
$$=\frac{38}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Different method of proving $\int_{0}^{1}{ dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdots\right)=-\pi^2$ Given the integral $(1)$
$$\text{Prove that}\ \int_{0}^{1}{\mathrm dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdot{1-\sqrt[5... | Let us generalize to
$$\int^1_0 \frac{\mathrm d x}{x} \log \left(\frac{1-x^a}{1+x^a}\right)$$
Let $x^a = y $ which implies $\mathrm d x = \frac{1}{a}y^{\frac{1}{a}-1} \mathrm d y$
\begin{align}\int^1_0 \frac{\frac{1}{a}y^{\frac{1}{a}-1}\mathrm dy}{y^{\frac{1}{a}}} \log \left(\frac{1-y}{1+y}\right) &= \frac{1}{a}\int^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Prove: If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Proof by Contradiction:
If $a$ and $b$ are odd perfect squares then $a = (2k+1)^2$ and $b = (2r + 1)^2$.
Assume $a + b$ is a perfect square.
\begin{align... | You could stop at $4(k^2 + r^2) + 4(k + r) + 2$ and note that this is divisible by $2$, but not $4$. Therefore it cannot be a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The functional equation $\frac{f(x)}{f(1-x)} = \frac{1-x}{x}$ One set of solutions to this is $f(x) = \frac{c}{x}$ for constant $c$. Are these the only solutions?
| Let $f$ be a solution to the equation, and let $g(x):=(x+ \frac{1}{2})f(x + \frac{1}{2})$. Then $g$ is an even function (i.e.
$ g(x)=g(-x)$). Conversly, let $h$ be any even function. Then by setting $f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$, we get $xf(x)=(1-x)f(1-x)$. Thus the set of all solutions to this equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that if $gcd(n,42) = 1$ then $n^6 \equiv 1 \pmod {42} $ Show that if $gcd(n,42) = 1$ then $n^6 \equiv 1 \pmod {42} $
I notice that $a \equiv 1 \pmod{2}$, $a^2\equiv 1 \pmod{3}$, $a^6 \equiv 1 \pmod{7}$
In the first case $a=-1,1,-3,3$ In order for the second condition to hold $a=-1,1$ which would imply then that $a... | Fermat's little theorem gives immediately $n^6\equiv 1\mod 7$
Since $n$ is odd, we have $n^6\equiv 1\mod 2$
Since $n$ is not divisble by $3$, we have $n^6\equiv 1\mod 3$ because of $n^2\equiv 1\mod 3$
The chinese remainder theorem gives $n^6\equiv 1\mod 42$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4... |
By the above diagram, we can simplify the answer obtained by the author.
$$
\begin{array}{l}
\begin{aligned}
I &=\frac{1}{16}\left(\tan \theta+\frac{\tan ^{3} \theta}{3}\right)+C \\
&=\frac{1}{16}\left[\sqrt{4 x^{2}-1}+\frac{\left(4 x^{2}-1\right)^{\frac{3}{2}}}{3}\right]+C
\\&=\frac{\sqrt{4 x^{2}-1}}{48}\left[3+4 x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Winning strategy of Peter such that quadratic equations has Rational roots Peter picked three non zero real numbers $a,b,c$ and Alan arranges these numbers as coefficients of Quadratic equation $ax^2+bx+c=0$.
Peter wins if this Quadratic has distinct Rational roots, else Alan wins.
How can we prove that Peter always ... | It's enough to find one set of real numbers that Peter can choose to win. For example, $\{1, 2, -3\}$ will work. We can check that:
\begin{align}
x^2 + 2x - 3 &= (x+3)(x-1) \\
x^2 - 3x + 2 &= (x-2)(x-1) \\
2x^2 + x - 3 &= (2x+3)(x-1) \\
2x^2 - 3x + 1 &= (2x-1)(x-1) \\
-3x^2 + x + 2 &= -(3x+2)(x-1) \\
-3x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing a sequence in the Cantor set Let $\eta:2^\omega\to 2^\omega$ be such that:
If $x \neq$ all 1's and $n$ is least such that $x_n=0$, then $\eta(x)=y$ where $$y_i=\begin{cases}x_i &\text{for } i>n\\ 1-x_i&\text{for }i\leq n.\end{cases}$$
I am interested in starting with the particular $x$ defined below. I believ... | Applying $\eta$ to $y = 0000\dots$ instead would correspond to counting in binary, but backwards:
\begin{align}
y &= 0000\dots \\
\eta(y) &= 1000\dots \\
\eta^2(y) &= 0100\dots \\
\eta^3(y) &= 1100\dots \\
\eta^4(y) &= 0010\dots
\end{align}
This is closely related to sequence A030109 in the OEIS, and won't have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rationalise the Denominator Rationalise the denominator and simplify fully:
$$\dfrac{6}{\sqrt{7} + 2}$$
I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?
I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sq... | $$\frac{6}{\sqrt{7}+2}=\frac{6(2-\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})} = \frac{12-6\sqrt{7}}{4-7} = \frac{3(4-2\sqrt{7})}{-3} = 2\sqrt{7}-4$$
Note that in general $$\frac{a}{b+c\sqrt{d}} = \frac{a(b-c\sqrt{d})}{(b+c\sqrt{d})(b-c\sqrt{d})} = \frac{ab-ac\sqrt{d}}{b^2-c^2d}$$
Notice that the denominator is now rational, i.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| First, you can see that
$$f(x)=-f(x-2) + 4(x-1)^2-2(x-1)+10$$
$$f(x-2) = -f(x-4) + 4(x-3)^2-2(x-3)+10$$
$$f(x-4) = -f(x-6) + 4(x-5)^2 - 2(x-5)+10$$
$$\dots$$
You can see that you need a basis or some additional constraint to express $f(x)$, because now a solution is not unique. Once you find a solution $f_0(x)$, the g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
| $$x^2 + y^2 = kxy$$
Let $d =gcd(x,y)$. Then $x=da, y=db$ for some relatively prime $a,b$.
The equation becomes
$$a^2+b^2=kab $$
with gcd$(a,b)=1$. This implies $a=\pm 1$ and $b=\pm1$. Indeed, if $a$ or $b$ are not $\pm1$, then $a$ or $b$ is divisible by a prime $p$. Then the prime $p$ also divides $kab$ and hence di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find a,b and c. Any method? Given that $$\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}=\frac{16}{38}$$, find $a,b,c$.
I've been figuring for this quite some time? Is it possible to solve?
| Here is a solution ; sorry for hand writing
$$\begin{align*}
\frac{1}{a+\frac{1}{b+\frac{1}{c+1}}}&=\frac{16}{38}\\
a+\frac{1}{b+\frac{1}{c+1}}&=\frac{38}{16}= 2+\frac{3}{8}\\
\text{So}\quad a=2\\
\frac{1}{b+\frac{1}{c+1}}&=\frac{3}{8}\\
b+\frac{1}{c+1}&=\frac{8}{3}=2+\frac{2}{3}\\
\text{So}\quad b=2\\
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit of a sum 1: $\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$ $$\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k} \cdot \sin \frac{2k\pi }{n}$$
It looks like a Riemann sum, but I don't know how to approach it. Any hint would be appreciated. (Without Taylor expansion)
EDIT: The answer is $- \... | I don't have time to finish right now; I'll leave this as Community Answer, so that anyone can feel free to edit it; if no one does, I'll get back to it later.
The inhomogeneity between $n^4$ and $k$ will a priori prevent you from seeing this as a Riemann sum: since there is no way to massage $\sqrt{n^4+k}$ in order to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Rational points on a circle with a center having non rational coordinates Reflection of a rational point with respect to line having its equation with rational coefficient
gives a rational point
What is the number of rational points on a circle with centre $(\sqrt 2, \sqrt 3)$ and with any radius
In this the centre is ... | Let $(a,b)$ and $(c,d)$ be distinct rational points on the circle.
$$\begin{array}{rcl}
(a-\sqrt2)^2 + (b-\sqrt3)^2 &=& (c-\sqrt2)^2 + (d-\sqrt3)^2 \\
a^2 + b^2 - 2a\sqrt2 - 2b\sqrt3 &=& c^2 + d^2 - 2c\sqrt2 - 2d\sqrt3\\
a^2 + b^2 - c^2 - d^2 &=& 2(a-c)\sqrt2 + 2(b-d)\sqrt3\\
(a^2 + b^2 - c^2 - d^2)^2 &=& 8(a-c)^2 + 12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation calculus: $\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$ How can I solve this?
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$$
Actually I tried many direction, but failed.
Please give me some right direction.
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum_{k=1}^n \frac{2^{2^{k-1}}}{(1-2^{2^{k-1}})(1+2^{2^{... | Let $$S = \sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum^{n}_{k=1}\bigg[\frac{(1+2^{2^{k-1}})-1}{1-2^{2^k}}\bigg] = \sum^{n}_{k=1}\bigg[\frac{1}{1-2^{2^{k-1}}}-\frac{1}{1-2^{2^k}}\bigg]$$
which is nothing but Telescopic Sum
So $$S = -1-\frac{1}{1-2^{2^{n}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$ converges absolutely or conditionally?
Does the following series converges absolutely or conditionally? $$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$$
$$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}=\sum_{n=1}^{\infty} \frac{(-1)^n}{... | Hint:
$$(-1)^n-(-1)^{n+1}=(-1)^n+(-1)^{n+2}=(-1)^{n}\left[1+1^2\right]=2\cdot(-1)^n\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
ABC is Isosceles right triangle with $AB=BC$ $P$ , $Q$ are points on $AC$ such that $AP ^2 + CQ^2 = PQ^2$ What is the value of angle $PBQ (x) $?
ABC is Isosceles right triangle with $AB=BC$
$P$ , $Q$ are points on $AC$ such that
$$AP ^2 + CQ^2 = PQ^2$$
What is the value of angle $PBQ (x) $?
Than... | Let $AC=1, AP=x, CQ=y.$ Then, $x^2+y^2 = (1-x-y)^2\Rightarrow 2x+2y-1=2xy$ or $y = \dfrac{3}{2x+2}-1=\dfrac{1-2x}{2-2x}.$ By theorem of Cosine, $BP^2 = \dfrac{1}{2}+x^2 - 2x\dfrac{1}{\sqrt{2}}\cos\dfrac{\pi}{4} = x^2-x+\dfrac{1}{2}$ and similarly $BQ^2 = y^2-y+\dfrac{1}{2}.$ Therefore, $$\cos\angle PQB = \dfrac{BP^2+B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to solve the partial fraction decomposition $\frac{x^3+5x^2+3x+6}{2x^2+3x}$. I have the following integral:
$$\int\frac{x^3+5x^2+3x+6}{2x^2+3x}dx$$
I'm trying to use partial fraction decomposition but I'm getting stuck at the following formula:
$$\int\frac{(x+6)(1+5x+x^2)}{x(2x+3)}-\frac{x+27}{2x+3}dx$$
I can't n... | We first deal with the improper fraction by division and then resolve it into partial fractions, $$
\begin{array}{l} \displaystyle
\frac{x^{3}+5 x^{2}+3 x+6}{x(2 x+3)}=\frac{1}{2} x+\frac{7}{4}+\frac{-
\frac{9}{4} x+6}{2 x^{2}+3 x} \\
\displaystyle \frac{-\frac{9}{4} x+6}{x(2 x+3)} \equiv \frac{A}{x}+\frac{B}{2 x+3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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partial fractions for a function I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is:
$(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.
| The key to it is how to factor the denominator $2x^4+3x^2+1$. Denote $y=x^2$ and you're left with a quadratic $2y^2+3y+1$ whose roots are $-1$ and $-1/2$. So
$$2x^4+3x^2+1=(y+1)(2y+1)=(x^2+1)(2x^2+1)$$
Now we're looking for
$${x^2+x+1\over 2x^4+3x^2+1}={ax+b\over x^2+1}+{cx+d\over 2x^2+1}$$
Multiply both sides by $x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. B... | To make things more interesting, and slightly less susceptible to exhaustive search, we can show that $x^2\equiv 1\bmod 91$ has four incongruent solutions and find them.
The analogy is that both $15=3\cdot 5$ and $91=7\cdot 13$ are composite numbers, the product of two odd primes.
So in my example, $x^2\equiv 1\bmod ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know h... | Given the odd number $x$,
$$x^2=2y^2-1$$
which is just a Pell equation, then we get the Pythagorean triple,
$$\Big(\frac{x-1}2\Big)^2+\Big(\frac{x+1}2\Big)^2=y^2$$
The difference $d$ between the addends, of course, is $d=1$. This yields,
$$3^2+4^2=5^2\\20^2+21^2=29^2\\119^2+120^2=169^2$$
and so on.
P.S. If interested,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Factor $ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$
a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.
b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$
So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to... | The polynomial
$$f(a,b,c)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$$
is an alternating function of $a$, $b$, $c$. That is it changes sign
whenever you swap two variables:
$$f(a,b,c)=-f(b,a,c)=-f(c,b,a)=-f(a,c,b).$$
This means that $a-b$, $a-c$ and $b-c$ are all factors (if a polynomial
changes sign when you swap $a$ and $b$ then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$? What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ ?
I proved that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)\supset \mathbb Q(\sqrt 5)$, and thus that $\mathbb Q(\sqrt 3+\sqrt5 )/\mathbb Q(\sqrt 5)$ ha... | Its conjugate is $\sqrt5-\sqrt3$.
Therefore, the minimal polynomial is:
$$\begin{array}{rcl}
(x-(\sqrt3+\sqrt5))(x-(\sqrt5-\sqrt3))
&=& (x-\sqrt5)^2 - (\sqrt3)^2 \\
&=& x^2-2\sqrt5x+5-3 \\
&=& x^2-2\sqrt5x+2
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Intermediate Value theorem inequality problem Let $h:[0,1]\rightarrow \mathbb{R}$ be continuous. Prove that there exists $w\in[0,1]$ such that $$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$$
I tried several things with this problem. I first tried a new function
\begin{align*}
g(x)... | Let $g:[0,1]\rightarrow \mathbb{R}$ be defined by
$$
g(w)=h(w) - \left(\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left({\small{\frac{1}{2}}}\right)+\frac{w+1}{12}h(1)\right)
$$
It suffices to show that $g(0),g(1/2),g(1)$ cannot all be positive, and cannot all be negative.$\\[9pt]$
\begin{align*}
\text{Let}\;\, a &= h(0),\;b=h(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Hard limit to solve involving exponential and square root How to find this limit:
$$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}.$$
Where $a$ and $b$ are integer constants. I've tried substitution, L'Hôpital's rule, rationalizing the denominator and all the techniques I knew about limits, ... | (assuming limit from the right). Use Taylor Series: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\implies e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots$$
so that $$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}$$ $$=\lim_{x \to 0}\frac{a(x-\frac{x^2}{2}+\ldots)+b(x+\frac{x^2}{2}+\ldots)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with Induction to show an Equality of Linear Operators I am trying to solve the following:
Let $X$ be a finite dimensional complex inner product space. Let $A,B \in \mathcal{L}(X)$, the space of linear mappings from $X$ to $X$. Use induction to show for any $n\geq1$, $$A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$$
F... | We know that $A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$ holds for $n = 1$. Assume it holds for $n = k$, then for $n = k+1$, we have
\begin{align}
& A^{(k+1)}-B^{(k+1)} = A^{(k+1)}-B^{(k+1)}+A^kB-A^kB \\
=\ & A^kB-B^{(k+1)} + A^{(k+1)}-A^kB = (A^k-B^k)B + A^k(A-B) \\
=\ & \left(\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$? How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$?
I tried show that difference between... | Let $f(x)$
be a function such that,
for $x \ge 0$,
$f(x) \ge 0$
and
$f(x)$ is bounded for
$0 < x \le 1$.
Let
$M = \max_{0 < x \le 1} f(x)
$.
This problem is the case
$f(x) = 2^x$,
so $M = 2$.
Let
$g(n)
=\sum\limits_{k=1}^n \frac{f(k/n)}{n+\frac{1}{k}}
= \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{1 + \frac{1}{kn}}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cd... | Let $x_{k} = \cos^{2}(k\pi/n)$ where $k = 1, 2, \dots, n$. Then we can see that $z_{k} = 2x_{k} - 1 = \cos(2k\pi/n)$ and hence these are the roots of the equation $P_{n}(z) = 1$ where $P_{n}(z)$ is a polynomial such that $P_{n}(\cos x) = \cos nx$. These polynomials are famous by the name Chebyshev polynomials and satis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Have I correctly taken this function $f(x) = \frac{x^{2}}{1+2x}$ and $f ' (x) $and turned them into power series? $f(x) = \dfrac{x^{2}}{1+2x}$
To turn this into a power series I recall the similar looking geometric series,
$f(x) = \sum\limits_{n=1}^{\infty} x^{n} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^5 +\cdots$
And n... | Your approach based upon the geometric series expansion
\begin{align*}
\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots
\end{align*}
is fine and appropriate. Just the calculations are not correct in all aspects.
We obtain
\begin{align*}
f(x)&=\frac{x^2}{1+2x}\\
&=x^2\cdot\frac{1}{1-(-2x)}\\
&=x^2\left(1-2x+4x^2-8x^3+16x^4-\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given the square, calculate $\tan{\alpha}.$ Problem: Given the square $ABCD$, let $M$ be the midpoint on the side $|CD|$ and designate $\alpha=\angle AMB$. Calculate $\tan{\alpha}.$
Attempt: We can, without compromising generality, assume that the side of the square is equal to 1. So drawing a figure we get
I know th... | You have
$$
\tan\frac{\alpha}{2}=\frac{1/2}{1}=\frac{1}{2}
$$
Then use
$$
\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}
$$
With $\beta=\alpha/2$, we get
$$
\tan\alpha=\frac{1}{1-1/4}=\frac{4}{3}
$$
Using your tools it can be done as well; set $r=AM$, for simplicity. Then
$$
\frac{1}{2}=\frac{1}{2}r^2\sin\alpha
$$
from t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that fourier transformation is a circle I've calculated the fourier transformation of a function and got $\hat{f}(w) = \frac{1}{\rho+iw}$.
Now I must show, that this is equal to a circle with radius $r$ and middlepoint $m$, where $r = m = \frac{1}{2\rho}$
When I plot the function, I see the circles, but I'm not su... | Note that
$$
\hat{f}(\omega) = \frac{1}{\rho + i \omega}\frac{\rho - i \omega}{\rho - i \omega} = \frac{\rho - i\omega}{\rho^2 + \omega^2}
$$
Therefore
$$
x = \Re(\hat{f}(\omega)) = \frac{\rho}{\rho^2 + \omega^2} ~~~\mbox{and}~~~
y = \Im(\hat{f}(\omega)) = -\frac{\omega}{\rho^2 + \omega^2}
$$
We then have
\begin{eqnarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there three integers such the nontrivial sums are squares and three are squares of consecutives?
Do there exist such three different positive integers $a, b, c$ that the following numbers: $a+b, b+c, c+a, a+b+c$ are all squares of integers and three of them ($a+b, b+c, c+a, a+b+c$) are squares of consecutive integ... | In the previous post, i miss understood the question.
We need to find $a,b,c$ such that $a+b,a+c,b+c,a+b+c$ all are square number and three of them are consecutive square numbers.
Which means with $d,e$ integers $a+b,a+c,b+c,a+b+c$ are equal to $(d-1)^2,d^2,(d+1)^2 ,e^2$. since $a,b,c$ are positive so $a+b+c >a+b,a+c,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.