Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to calculate this imaginary part of a complex square root? The calculation of the square root of a complex number $a + ib$ involves solving the equation
$$ (x + iy)^2 = a + ib$$
So far so good. One obtains the equations
$$ 4x^4 -4ax^2 - b^2 = 0, y = b/2x$$
and using the quadratic formula for $x^2$ one gets
$$ x = ... | First off, up to a sign (reflected by that $\mbox{sgn }(b)$ factor), the two expressions are equal. You can see this if you multiply the expression you got by
$$
\frac{\sqrt{\sqrt{a^2+b^2}-a}}{\sqrt{\sqrt{a^2+b^2}-a}}
$$
to get
$$
y = \pm \frac{b}{\sqrt{b^2}} \frac{\sqrt{\sqrt{a^2+b^2}-a}}{\sqrt{2}}
$$
And of course, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using c... | Generally:
$$\begin{aligned}
&\phantom{mm}(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\\
&=((a+b)+c)((a+b)-c) \times ( c+(a-b))(c-(a-b)) \\
&=((a+b)^2 -c^2)(c^2-(a-b)^2)\\
&=[2ab+(a^2+b^2 - c^2) ] [ 2ab - (a^2+b^2-c^2)]\\
&= 4a^2b^2 - (a^2+b^2-c^2)^2.
\end{aligned}$$
Here $a^2=5, b^2=6,$ and $c^2=7$, so the answer is $104$ as you sai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the roots of $z^4+z^3+z^2+z+1$ I need to find the roots of $z^4+z^3+z^2+z+1$. One approach I have is to guess the first linear solution, then use polynomial long division to find a 3rd degree polynomial, guess another from that, long division again, then use the quadratic formula to solve the quadratic.
I thin... | In addition to the multiply-by-$\left(z - 1\right)$ trick, there is also the "symmetric" method to solve for the roots directly:
$$z^4 + z^3 + z^2 + z + 1 = z^2 \left(z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2}\right) = 0 $$
Since $z^2 \ne 0$, we can cancel it out, and we recognize the almost-square $z^2 + \frac{1}{z^2}$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute this limit $\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}$ $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$
such that $2\alpha n\le n$
I tried to use Stirling formula and we get
$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\li... | The Stirling formula is the right way. But ignore the logarithm first. I get
$$\ln \frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
It comes from $\displaystyle \ln \left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e})^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}$:
$$\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Improper Integral $\int_0^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)dx = \frac{7\zeta(3)}{\pi^2} $ $\newcommand{\sech}{\operatorname{sech}}$
$\displaystyle \int_0^{\infty}{\left(\frac{\tanh(x)}{x^3} - \frac{\sech^2(x)}{x^2} \right)\ dx }= \frac{7\zeta(3)}{\pi^2} $
What I tried
I simplified it to -
$\... | Integrating by parts we get $$I=\int_{0}^{\infty}\left(\frac{\tanh\left(x\right)}{x^{3}}-\frac{1}{x^{2}\cosh^{2}\left(x\right)}\right)dx= $$ $$-\int_{0}^{\infty}\frac{1}{x}\left(-\frac{\tanh\left(x\right)}{x^{2}}+\frac{1}{x\cosh^{2}\left(x\right)}+2\frac{\tanh\left(x\right)}{\cosh^{2}\left(x\right)}\right)dx$$ so $$\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
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how to solve $x^2 = 3x + 4$ I am a programmer in the eighth grade taking algebra 1. I am only about a month into the school year, and I need to know how to solve something similar to this equation:
$x^2 = 3x + 4$
However, the problem is that whenever I try to get the square root of both sides of the equation to get rid... | The standard method is to multiply everything by $4$ and transfer the $x$ term on the left-hand side:
$$
4x^2-12x=16
$$
Now recall $(a+b)^2=a^2+2ab+b^2$ and observe that we can take $a=2x$, so from $2ab=-12x$ we obtain $4bx=-12x$, that's satisfied for $b=-3$. We need the $b^2$ term, so we add it on both sides:
$$
4x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| Note that (by dividing the equations by $16$ and $32$) this is equivalent to showing one of $a,b,c,d$ is $1$ and the others are $0$ if $a^4 + b^4 + c^4 + d^4 = 1$ and $a^5 + b^5 + c^5 + d^5 = 1$. If none of $a,b,c,d$ are $1$ then they must all have absolute value less than one (otherwise the sum of their fourth powers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find $x$ given remainder conditions The problem: Find the smallest positive integer $x$ such that
$x$ divided by $4$ has remainder $1$
$x$ divided by $5$ has reminder $2$
$x$ divided by $6$ has remainder $3$
Now, my first idea was to add to each divisor its the remainder and multiply the quantities obtained.
$x=(4+1)(5... | If $x \equiv 1\pmod{4}$, then $$x \equiv 1,5,9,13,17\pmod{20}$$ Of these, only $17$ is also $2 \pmod{5}$. Therefore
$$x \equiv 17 \pmod{20}$$
Next, the LCM of $20$ and $6$ is $60$.
If $x \equiv 17 \pmod{20}$, then $$x \equiv 17, 37, 57 \pmod{60}$$
Of these, only $57$ is $3\pmod{6}$. This means
$$x \equiv 57\pmod{60}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of Trigonometric Integral
Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$
$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\... | HINT:
If $f(x)=\sin^2x\cos^2x=\dfrac{\sin^22x}4,f\left(\dfrac\pi4+0-x\right)=\dfrac{\cos^22x}4=\dfrac{(1-2\sin^2x)^2}4$
If $g(x)=\sin^3x+\cos^3x,g\left(\dfrac\pi4+0-x\right)=\dfrac{2\cos x(\cos^2x+3\sin^2x)}{2\sqrt2}=\dfrac1{\sqrt2}\cdot\dfrac{(1-\sin^2x)(1+2
\sin^2x)}{\cos x}$
Now use $$\int_a^bf(x)\ dx=\int_a^bf(a+b-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number?
| You can prove by induction that the sum of the squares of the first $6n$ terms is divisible by $8$.
First, show that this is true for $n=1$:
$\sum\limits_{m=1}^{6}F_m^2=40$
Second, assume that this is true for $n$:
$\sum\limits_{m=1}^{6n}F_m^2=8k$
Third, prove that this is true for $n+1$:
$\sum\limits_{m=1}^{6(n+1)}F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$ How do I find this limit?
$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$
I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!
Is there... | Here we do not use Hopital or derivatives.
Note that
$$x^2+x+\sin({\pi \over 2}x)-3=(x+2)(x-1)+\sin({\pi \over 2}-{\pi \over 2}(1-x))-1\\=(x+2)(x-1)+\cos({\pi \over 2}(1-x))-1\\
=(x+2)(x-1)-2\sin^2(\frac{\pi}{4}(1-x))$$
because $1-\cos(t)=2\sin^2(t/2)$.
Hence, as $x\to 1$,
$$\frac{x^2+x+\sin({\pi \over 2}x)}{(x-1)}=x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{... | The second factor in $$\frac{\sin x-x\cos x}{x^3}\frac{\sin x+x\cos x}{x}$$ clearly tends to $2$.
Then by L'Hospital,
$$\frac{\cos x-\cos x+x\sin x}{3x^2}$$ tends to $1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
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Question on algebra used in induction proof Can someone take a look at this simplification... it is part of a proof by induction solution I'm examining... I'm interested in the result of this simplification as I'm not sure how the third to last step was achieved.
$$\begin{align*}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\fr... | That step should be
$$
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+(-1)\cdot\left(\frac{k+2}{(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-1}{(k+1)(k+2)}\\
=1-\frac{k+1}{(k+1)(k+2)} \\
=1-\frac{1}{k+2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the two planes which contains a given line and forms an angle of $60^\circ$ with a given plane I am given the following question:
Find the equations of the two planes that contain the line $r \{ x=z+1=y+2$ and form a $60^\circ$ angle with the plane $ \pi \{ x+2y-3z+2=0$
My solution:
Lets call the normal vecto... | Since as has been oft remarked the lenght of $(a,b,c)$ is variable let us chose the length to be $\sqrt{14}$ in order to cancel with the other $\sqrt{14}$
So we have the equations:
$$a+b+c=0$$
$$a+2b-3c=\frac{1}{2}\sqrt{14}\sqrt{a^2+b^2+c^2}=\frac{1}{2}14=7$$
$$a^2+b^2+c^2=14$$
If we take the first two
$$a+b+c=0$$
$$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that the harmonic function $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions? How can I prove that $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions?
| Hint. Prove by induction that
$$1+\frac{n}{2}\leq H_{2^n}\leq 1+n$$
by using the fact that
$$\frac{1}{2}=\frac{2^n}{2^n+2^n}\leq H_{2^{n+1}}-H_{2^n}=\frac{1}{2^n+1}+\frac{1}{2^n+2}+\cdots +\frac{1}{2^{n}+2^n}\leq \frac{2^n}{2^n+1}<1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - ... | $$(\csc B-\cot B)(\csc B+\cot B)=1\implies\dfrac{\csc B+\cot B}{\csc B-\cot B}=(\csc B+\cot B)^2$$
$$(\csc B+\cot B)^2+(\csc B-\cot B)^2=2(\csc^2B+\cot^2B)$$
Now $\cot^2B=\csc^2B-1$
Finally, $$\dfrac{1+\cos^2B}{1-\cos^2B}=\dfrac{1+\cos^2B}{\sin^2B}=\csc^2B+\cot^2B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Finding a sum as function of $n$ using summation formulas i need to find the sum as a function of $n$ using summation formulas of the following serie:
\begin{equation}
\sum_{k=1}^n\bigg(4+\cfrac{2k}{n}\bigg)^2\bigg(\cfrac{1}{n}\bigg).
\end{equation}
Any idea how to proceed? Any help will be appreciated!
| I would proceed as follow:
$$\frac{1}{n}\sum_{k = 1}^n\ \frac{(4n^2+2k)^2}{n^2} = \frac{4}{n^3}\sum_{k = 1}^n (2n^2+k)^2 = \frac{4}{n^3}\sum_{k = 1}^n\ 4n^4 + k^2 + 4k$$
Now split the sum into three pieces, knowing that those are well known series:
$$\sum_{k = 1}^n 4n^4 = 4n^4\sum_{k = 1}^n = 4\cdot n^5$$
$$\sum_{k = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence test for the following series The series is :-
$$ \frac{1}{1^2} + \frac{1+2}{1^2+2^2} + \frac{1+2+3}{1^2+2^2+3^2} + ... ∞ $$
I am unable to think for its general term.
Can anybody help me solve this.
Thanks in advance.
| hint
$$1+2+3+....n=\frac{n}{2}(n+1)$$
$$1+2^2+3^2+...n^2=\frac{n}{6}(n+1)(2n+1)$$
for the another series, use the ratio test
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n+1)!}\frac{n!}{n^2}=\frac{(n+1)^2}{(n+1)n^2}=\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}$$
the limit is zero when $n\rightarrow \infty$, so the series converg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of a symmetric $7\times 7$-matrix Is there any simple method to calculate the determinant of the following symmetric $7\times 7$-matrix
$$M:=\begin{pmatrix}
1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\
0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\
0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\
0 & 0 & 0 & 1 & -a_{3} & a_... | Here is a rather simple method.
Consider the block partition of matrix $M$:
$$M=\pmatrix{I_4&B^T\\B&a_0I_3} \ \ \text{of the form} \ \ \pmatrix{R&S\\T&U}.$$
Using a classical determinantal formula (https://en.wikipedia.org/wiki/Schur_complement) valid for any partition where the diagonal blocks are square and the upper... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Functional equation $f(r \cos \varphi)+f(r \sin \varphi)=f(r)$
Find all such monotonous function $f:[0,+\infty)\to \mathbb R$, such that for any real $r\ge0$ and $\varphi \in \left[\frac{\pi}6, \frac{\pi}4\right]$,
$$f(r \cos \varphi)+f(r \sin \varphi)=f(r)\text.$$
My work so far:
*
*If $\varphi \in \left[\frac{\pi... | Consider a monotonous function $ f : [ 0 , + \infty ) \to \mathbb R $ satisfying the functional equation
$$ f ( r \cos \phi ) + f ( r \sin \phi ) = f ( r ) \tag 0 \label 0 $$
for every $ r \in [ 0 , + \infty ) $ and every $ \phi \in \big[ \frac \pi 6 , \frac \pi 4 \big] $. You can show that there is a $ c \in \mathbb R... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square ... | The idea is to move the equation's terms around such that one and only one square root appears on one side. Squaring then removes that isolated square root; the other side may still have a square root, but there will be one less overall. Repeat until a polynomial is obtained.
Let $y=x^2$, then (using fractions instead ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding the set of strings over $\{a, b\}$ that do not contain the substring aaa
I need to identify the set of all possible strings over $\Sigma = \{a, b\}$ that do not contain the substring $aaa$.
I have becoming more familiar with regular-language and languages explored in Automata Theory, and have been defining ... | We can describe the set of valid strings as the strings built from the alphabet $V=\{a,b\}$ which contain at most one or two consecutive $a$'s.
These are strings
*
*starting with zero or more $b$'s:$$b^*$$
*followed by zero or more occurrences of $a$ or $aa$ each followed by one or more $b$'s: $$(ab^+|aab^... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\le... | The extreme values of $A\cos x +B\sin x$ are $\pm\sqrt {A^2+B^2}.$ This is obvious and trivial if $A=B=0.$ If $A,B$ are not both $0,$ there exists $y$ such that $\cos y=A/\sqrt {A^2+B^2}$ and $\sin y=B/\sqrt {A^2+B^2},$ so $$|A\cos x +B\sin x| =|(\sqrt {A^2+B^2}\;)(\cos x \cos y+\sin x \sin y)|$$ $$=(\sqrt {A^2+B^2}\;)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of rational numbers I am recently starting in the subject of pure mathematics and this problem has eluded me.
Any rational number $\frac{p}{q}$ can be expressed in a simple series form:
$$\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4},...$$... | First of all I think that the first number in the sequence should be $\frac 11$.
Now consider the sum of the numerator and the denominator and write it as a sequence and we have:
$$2,3,3,4,4,5,5,5,5,6,6,6,6,6...$$
Therefore if $p+q=r$ we have that the fraction $\frac{p}{q}$ might appear after $1+2+3+...+{r-2} = \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\c... | Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$(2\sin^2\theta)^2+(2\cos^2\theta)^2=(1-\cos2\theta)^2+(1+\cos2\theta)^2=2(1+\cos^22\theta)$$
Now $2\cos^22\theta=1+\cos2(2\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
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Finding $x$ such that $2x^2-2x+1=(2m+1)^2$ We need to find the smallest integral value of $x$ such that $2x^2-2x+1=(2m+1)^2$ and $x \ge 10^{15}$.
| HINT: $$2x^2 - 2x + 1 = (2m+1)^2 \iff x^2 + (x-1)^2 = (2m+1)^2$$
So the problem reduces to finding a specific Pythagorean triplet.
UPDATE:
We have that this is a primitive Pythagorean triplet, as $(x,x-1) = 1$. Therefore we either have $x = m^2 - n^2, x-1 = 2mn$ or $x=2mn, x-1 = m^2 - n^2$ for some positive integers $m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$? I need to find the limit of this sequence:
$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$
So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$
$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n... | $$\dfrac{n}{(\sqrt{n^2+9}+\sqrt{n^2-n+9})}=\dfrac{n/n}{\frac{1}{n}(\sqrt{n^2+9}+\sqrt{n^2-n+9})}$$
$$=\dfrac{1}{\sqrt{n^2/n^2+9/n^2}+\sqrt{n^2/n^2-n/n^2+9/n^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $ Given the followning limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $$
To find limit I have made following steps:
*
*Let $ x = y $ ,then limit equals $0$
*Let $ x... | For $x, y >1$, we have the fact that
\begin{align}
2\leq\frac{x}{y}+\frac{y}{x}
\end{align}
which means
\begin{align}
\left(\frac{xy}{x^2+y^2}\right)^{x^2}=\left(\frac{1}{\frac{x}{y}+\frac{y}{x}}\right)^{x^2} \leq \left(\frac{1}{2}\right)^{x^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifcation of a sum I’m unsure if I’m allowed to simplify like this:
$$\begin{align*}
\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} &= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{... | $$\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!}+\sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{8}{5^n(n-1)!}+\frac{1}{7}\sum_{n=2}^\infty \left(\frac{7}{10}\right)^n =$$
$$=\frac{8}{5}\sum_{n=2}^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative va... | Sorry for being late
By ‘Rationalisation’
$$
\begin{aligned}
\int_0^{2 \pi} \frac{d x}{5-4 \cos x}=&2 \int_0^\pi \frac{d x}{5-4 \cos x} \\
= & 2 \int_0^\pi \frac{5+4 \cos x}{25-16 \cos ^2 x} d x \\
= & 10 \int_0^\pi \frac{d x}{25-16 \cos ^2 x}+8 \int_0^\pi \frac{\cos x}{25-16 \cos ^2 x} d x \\
= & 20 \int_0^{\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the number of solutions of an expression Find the number of integers $1\leq x \leq 2010 $ such that the expression
$\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}}-\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}}$
is a rational number
NOTE -> I LITERALLY do not understand where to start !! HELP ME!!
| The number of integers $1≤x≤2010$, (such that the expression above is a rational number), is 26.
The value of $x$ is
$x=3n^{2}-6n+4$.
In this way,
$\sqrt{\frac{x-1}{27}}=\sqrt{\frac{3n^{2}-6n+3}{27}}=\sqrt{\frac{3(n-1)^{2}}{27}}=\frac{n-1}{3}$;
$(x+8)\sqrt{\frac{x-1}{27}}=(n^{2}-2n+4)(n-1)$;
$x+(x+8)\sqrt{\frac{x-1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
| Let $b$ be the base in question. Presumably $b>8$, since $8$ appears as a digit in $118$. We also know that $5x^2-50x+118$ is a multiple of $(x-4)(x-7)$, and since the $x^2$ term of $(x-4)(x-7)$ is simply $x^2$, we must have
$$5x^2-50x+118=5(x-4)(x-7)\;.$$
From this we deduce that $55$ in base ten is $50_b$. This clear... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all primes $p$ for which $\frac{2^{p-1}-1}p$ is a perfect square
Find all primes p for which the quotient $\frac{2^{p-1}-1}p$ is a perfect square.
It is a number theory problem. By guessing I found the value of $p$ is 3. But how do I prove it and how do I found the other primes?
| Put $p = 2q+1$, then the expression is $\dfrac{4^q-1}{2q+1} = \dfrac{(2^q-1)(2^q+1)}{2q+1}$.
Now, since $2^q-1$ and $2^q+1$ are co prime, one of them must be a perfect square. Let $(2^q-1)(2^q+1)=pm^2$. Then two cases arise, namely that :
1) $2^q-1 = px^2,2^q+1 = y^2$
2) $2^q-1 = x^2,2^q+1 = py^2$
For the first case, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$ $$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$$
I tried this
$\frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1} = 1-2\sqrt{\frac{\arctan(x)-\frac{\pi}{4}}{x-1}}.\frac{1}{\sqrt{x-1}}$
I am stuck here !
| Let's compute
$$
\lim_{x\to1^+}\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}
$$
One way could be to substitute
$$
\sqrt{\arctan x-\frac{\pi}{4}}=t
$$
so
$$
\arctan x=t^2+\frac{\pi}{4}
$$
and therefore
$$
x=\tan(t^2+\tfrac{\pi}{4})=\frac{\tan(t^2)+1}{1-\tan(t^2)},
\qquad
x-1=\frac{2\tan(t^2)}{1-\tan(t^2)}
$$
so we get
$$
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are re... | Based on hagnatural's comment: No, your proof is incorrect since one of your signs is wrong:
$$(x-y-z)^2 = x^2 + y^2 + z^2 - 2xy - 2xz \color{red}{+} 2yz.$$
And unfortunately there is no direct term $(\pm x \pm y \pm z)^2$ which gives you opposite signs on the cross terms, compared to the squares. (But as noted in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$
Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.
I tried to do it by comparing the two expressions but I can't proceed.
| Let $f(x)=(x+b)(x+c)$ with $b,c\in\Bbb Z$.
Wlog $b\le c$.
For $-c\le x\le b$, we have $f(x)\le 0$, hence $f(x)-1<0$.
If $x\ge -b+2$, we have $f(x)-1\ge 2\cdot 2-1>0$. By symmetry, $f(x)-1>0$ also for $x\le -c-2$.
Hence the only possibly integer roots of $f(x)-1$ are $-b+1$ and $-c-1$.
Moreover, $f(-b+1)-1=f(-c-1)=c-b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | Hint. Is it possible to find $A,B,C$ such that for all $x\in\mathbb{R}$,
$$(x+3)^2=A(x+2)^2+B(x+1)^2+Cx^2?$$
P.S. The answer is yes: $A = 3$, $B = -3$, $C = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 4
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Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
| Multiply the numerator and denominator by $x(1+x)(1-x)$. This clears away all the "inner" denominators, leaving $${xx(1+x)+(1+x)(1+x)(1-x)\over (1-x)(1-x)(1+x)+xx(1-x)}={(1+x)\over (1-x)}{x^2+1-x^2\over x^2+1-x^2}={1+x\over 1-x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Calculation of an infinite sum $\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26$ Could you give me some advice on how to calculate the following sum?
$$\begin{aligned}\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26\end{aligned}$$
Thank you!
| $$
\begin{aligned}
& S_{0} = \sum_{n=1}^{\infty} \frac{n^{0}}{2^{n}} = \sum_{n=1}^{\infty} \frac{1}{2^{n}} \Rightarrow \color{red}{S_{0} = 1} \\ \\
& S_{1} = \sum_{n=1}^{\infty} \frac{n^{1}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1 - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} - S_{0} \\
& \qquad \Rightarrow S_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$ Find the number of integer solution of $$\left\lfloor \dfrac{x}{100} \left\lfloor \dfrac{x}{100}\right\rfloor \right\rfloor=5$$
$\lfloor . \rfloor$ is Greatest Integer Function or floor function.
| As $y-1<\lfloor y\rfloor\le y $, we find that for $y\ge1$
$$\tag1y^2-y-1<\lfloor y\lfloor y\rfloor\rfloor\le y^2 $$
whereas for $y<0$, we have
$$\tag2y^2-y-1>\lfloor y\lfloor y\rfloor\rfloor\ge y^2 $$
Finally, for $0\le y<1$, we clearly have $\lfloor y\lfloor y\rfloor \rfloor=0$
Thus we cannot have $\lfloor y\lfloor y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 1... | **$$ \color{red} {2^{12}=4096=21*195+1}$$
$$2^{12n+4}+5\equiv\\ 2^4 .2^{12n}+5 \equiv\\16.4096^n+5 \equiv\\16 (4095+1)^n+5\equiv\\16.(21.195+1)^n+5 \equiv \\ 16(1)^n+5 \equiv \\21\equiv0$$ so $$21|2^{12n+4}+5\\n=0,1,2,3,4,5,...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simple way to calculate $n! \pmod p$ I have the exercise "Calculate $10! \pmod{13}$".
I have the following two approaches to solve the exercise:
*
*Brute force approach
$$
1! \equiv 1 \pmod{13} \\
2! = 2 \cdot 1! \equiv 2 \cdot 1 \equiv 2 \pmod{13} \\
3! = 3 \cdot 2! \equiv 3 \cdot 2 \equiv 6 \pmod{13} \\
\cdots \... | Start with your Wilson's Theorem approach but finish off differently. Note that $12\equiv -1$ and $11\equiv -2 \pmod{13}$ and that these two numbers have easy inverses $(-2)(-7) \equiv 1$ and $(-1)(-1) \equiv 1 \pmod{13}$, so
$$10! \equiv -(11)^{-1}(12)^{-1} \equiv -(-2)^{-1}(-1)^{-1} \equiv -(-7)(-1) \equiv 6 \pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | This solution may look different. :-)
Note that $f(x)=(x-\frac{a}{2})^2+1-\frac{a^2}{4}$,
$$2*\frac{1}{4} \ge |f(x) -f(x+2)| = |2*(2x+2-a)| \Rightarrow |(x+2-\frac{a}{2})+(x-\frac{a}{2})|\le \frac{1}{4},$$
so $x-\frac{a}{2}$ and $x+2-\frac{a}{2}$ must have different signs as their difference is 2. It must be $x-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 5
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Is this the shortest proof of Fermat's last theorem for exponent 3? To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be don... | $n^3 + (n+1)^2 = (n+2)^3 \implies $
$(m-1)^3 +m^3=(m+1)^3; m=n+1$
$2m^3-3m^2+3m-1=m^3+3m^2+3m+1$
$m^3=6m^2+2$
$m=6 + 2/m^2$ so $m^2|2$ and $m > 6$. So $m=1>6$. Clearly impossible.
Of course, the shortest proof would be "this is a single cubic equation with one variable and can easily be confirmed to have no integer s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Decimal representation of $1/n$ includes the string of digits $777$
Find an integer $n$ with the property that the decimal representation of $1/n$ includes the string of digits $777$.
For $\dfrac{m}{n}$ we can find a string of digits $777$ just by looking at $0.777$ and writing that as a fraction. But how do we do i... | $$\frac{7}{9}=0.777\ldots,\qquad \frac{7}{90}=0.0777\ldots $$
and if we consider
$$ \frac{7}{90000} = 0.0000777\ldots $$ we have
$$ \frac{7}{90000}+\frac{1}{10000} = 0.0001777\ldots $$
with
$$ \frac{7}{90000}+\frac{1}{10000}=\frac{7+9}{90000}=\frac{16}{90000}=\frac{1}{\color{red}{5625}} $$
since $7+9=16$ is a divisor o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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prove or disprove $\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}$ let $n\in N^{+}$ and such $n\ge 2$
prove or disprove
$$\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}\tag{1}$$
and I have use... | $$L=\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}$$
$$\log L=\sum_{k=1}^{\infty }\frac{\log(k+1)}{2^k}=\quad{\text{LerchPhi}^{0,1,0}}(\frac{1}{2},0,2)=1.0156678457368767......$$
so
$$L=e^{1.01566784573687........}=2.761206841957498033230454646......$$
hence
$$2.7612.....<3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $30|(a^3b-ab^3) $ Prove that if three distinct integers are chosen at random then there will exists two among them, say $a$ and $b$ such that $30|(a^3 b-ab^3)$
| 1) Divisibility by $5$.
Let $\{a,b,c\}$ the three numbers reduced modulo $5$. If one of them is $0$ or two of them are equal we clearly have
$$ab(a+b)(a-b)\equiv 0\pmod 5$$
So be all the three, $a,b,c$ distinct and non-zero modulo $5$ so we have
$$\{a,b,c\}=\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$$
In the four cases we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac... | $$\left( \frac { -1+i\sqrt { 3 } }{ 1+i } \right) ^{ 2016 }=\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2\pi }{ 3 } \right) +i\sin { \left( \frac { 2\pi }{ 3 } \right) } } } \right) }^{ 1013 } }{ { \left( { \left( 1+i \right) }^{ 2 } \right) ^{ 1013 } } } =\frac { { { 2 }^{ 2016 }\left( { \cos { \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Arithmetic Mean/Geometrix Mean Inequality of Degree 3 For $a,b$ and $c\geq0$;
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Is there a simple way to deduce this result from the degree 2 verson?
That is:
$$\frac{a+b}{2}\geq\sqrt{ab}$$
Some sort of substitution involving $a,b,c \ $ for one of the variables was what I had in min... | This is a nice method:
We know that $\frac{a+b}{2}\geq\sqrt{ab}$
and similarly $\frac{c+d}{2}\geq\sqrt{cd}$
so $\frac{a+b+c+d}{2}\geq\sqrt{ab}+\sqrt{cd}\geq2\sqrt[4]{abcd}$
$$\Rightarrow\frac{a+b+c+d}{4}\geq\sqrt[4]{abcd}$$
and by making the substitution $d=\frac{a+b+c}{3} \rightarrow$
$$\frac{a+b+c+\frac{a+b+c}{3}}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculate a given limit Calculate the following limit:
$$\lim_{n \rightarrow \infty} \frac{3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} - n}{\ln(n^2 + 1)}$$
I tried to calculate the sum $3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n}$ but got nothing. I currently have no idea what I should do with this limit.
Thank you!
| Observe
\begin{align} \frac{3+3^{1/2}+3^{1/3}+\ldots + 3^{1/n} -n}{\ln(n^2+1)}
= \frac{(e^{\ln 3}-1)+(e^{\frac{1}{2}\ln 3}-1)+\ldots + (e^{\frac{1}{n}\log 3}-1)}{\ln(n^2+1)}. \end{align}
Using the fact that
\begin{align} x\leq e^x-1 \leq x-2x^2 \end{align}
when $0 \leq x\leq 2$, you can show that
\begin{align}
\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve for $\alpha$ in $1+\sqrt3\tan\alpha-\sec\alpha=0$
Solve for $\alpha$, where $0^\circ\le\alpha\le360^\circ$, in
$$1+\sqrt3\tan\alpha-\sec\alpha=0$$
I solved this way. But the answer in my book is only $0^\circ,120^\circ,360^\circ$.
Please help me. Which answer is suitable?
| Squaring both sides is false way because in this case you can get false roots $$1+\sqrt { 3 } \tan { \alpha -\sec { \alpha } =0 } \\ 1+\sqrt { 3 } \frac { \sin { \alpha } }{ \cos { \alpha } } -\frac { 1 }{ \cos { \alpha } } =0\\ \cos { \alpha +\sqrt { 3 } \sin { \alpha } =1 } \\ \frac { 1 }{ 2 } \cos { \alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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When does $\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$ converge? I am required to find the values of $C$ for which the integral
$$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.
I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.
I get stuc... | You know an antiderivative, namely
$$
\frac{7}{6}\bigl(3\ln(x^2+1)-2C\ln(3x+1)\bigr)=
\frac{7}{6}\ln\frac{(x^2+1)^3}{(3x+1)^{2C}}=
\frac{7}{6}\ln
\frac{\bigl(1+\frac{1}{x^2\mathstrut}\bigr)^3x^6}
{(3+\frac{1}{x})^{2C}x^{2C}}
$$
It's quite easy now to see when the limit is finite.
On the other hand, doing an as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $ Let $x,y,z > 0$ and $xyz=8.$ Prove that
$$
\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12
$$
I have tried with AM–GM inequality but no result.
| Using the Cauchy-Scwarz inequality
$$ (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}) \geq (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2} $$ with $a_{1}=\frac{x+y-1}{\sqrt{z}} $, $ a_{2}=\frac{x+z-1}{\sqrt{y}} $ $ a_{3}=\frac{z+y-1}{\sqrt{x}} $ and $ b_{1}=\sqrt{z} $, $ b_{2}=\sqrt{y} $ and $ b_{3}=\sqrt{x} $, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ If $\frac{a}{b}$ < $\frac{c}{d}$ then
$$\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$$
I have been searching and trying and couldnt find a reliable proof.
One might do this like here which I think it is very wrong:
$\frac{a}{b}$ < $\frac{c}{d}$ the... | Here is a simple intuitive proof for the case when a, b, c and d are positive. Imagine a team that scores a points in its first b games and c points in the remaining d games. Its seasonal points per game is $\frac{a+c}{b+d}$, which can also be expressed as the weighted average of its performance in each part of the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Manipulating functions
Let $g(n)$ be a function, defined for all integers $n\ge 0$, as follows:
$$g(n) = \begin{cases} 0, & \text{if } n=0 \\ 1+g(n/2), & \text{if $n\gt 0$ and $n$ is even} \\ 1+g(n-1), & \text{if $n\gt 0$ and $n$ is odd}\end{cases}$$
What is $g(2^l+2^k)$ where $l\gt k\ge 0$ are integers?
I have have ... | Using each of the rules for $g$:
$$\begin{align}
\color{red}{g(2^l + 2^k)} &= g\big(2^k(2^{l-k} + 1)\big) \\ &= 1 + g\big(2^{k-1}(2^{l-k} + 1)\big) \\ & \qquad\vdots \\ &= \color{red}{k+g(2^{l-k}+1)} \\ &= k+1+g\big((2^{l-k}+1)-1\big) \\ &= \color{red}{k+1+g(2^{l-k})} \\ &= k+1 + \big(1+ g(2^{l-k-1})\big)\\ &\qquad \vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking ... | We want to calculate
$$
I=\int_0^1 dx\frac{\arcsin(x)\arcsin\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2-x^2}}
$$
Let us start by a substitution $x=\sqrt{2}\sin(y)$ and integrate by parts so we can write $I$ as
$$
\int_0^{\pi/4}dy y \arcsin(\sqrt{2}\sin(y))\underbrace{=
}_{i.b.p}\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| Hint: If $x_0=-1$ is a zero of $x^{2016}-1$, then $(x-(-1))$ is a factor of $(x^{2016}-1)$.
In order to find $G(x)=(x^{2016}-1)/(x+1)$ you need to apply polynomial long division (you should see the pattern).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Finding the average rate of change of $S(x) = -2x^2 + 14x - 12$ I have to show that the average rate of change of $S(x) = -2x^2 + 14x - 12$ in the interval $[x,x+h]$ is $-4x - 2h + 14$ and so far I did:
$$A(x) = \frac{S(x+h)-S(x)}{(x+h)-(x)} = \frac{(-2(x+h)^2+14(x+h)-12)-(-2x^2+14x-12)}{(x+h)-x} = \frac{-2(x+h)^2+14(... | $$
\frac{-2(x+h)^2+14(x+h)+2x^2-14x}{h} = \frac{-2h^2+14h-4xh}{h} = -2h-4x+14.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+... | $\sin(a) + \sin(b) > \sin(a+b)$ if $(a,b)\in (0, \pi)\implies $
$\sin(a + b +c ) <= \sin(a) + \sin(b + c) < \sin(a) + \sin(b) + \sin(c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots \cdots \cdots +\frac{1}{2n}>\frac{13}{24}$
Proving $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots \cdots \cdots +\frac{1}{2n}>\frac{13}{24}\;, n>1\;,n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using Limit as a sum $$\frac{1}{n}\left[\frac{1}{1+\frac{1}{n}}+\fra... | Let $$f(n)=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}=\sum_{r=1}^{2n}\frac1r-\sum_{s=1}^n\frac1s$$
So, $$f(n+1)-f(n)=\frac1{(2n+1)(2n+2)}>0$$ for integer $n\ge0$
$\displaystyle \implies f(n)$ is an increasing function.
Now, $f(2)=\frac13+\frac14=\frac7{12}>\frac{13}{24}$ as $7\cdot24>12\cdot13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2-x^2)}$ How can we solve the equation:
$$\frac{dy}{dx} = \frac{y^3}{2(xy^2-x^2)}$$
I get the idea of dividing by $y^2$, But it doesn't become any more solvable (not homogenous).
$$\frac{dy}{dx} = \frac{y}{2(x-\frac{x^2}{y^2})}$$
Substituting $\frac{x}{... | Nice question!
Given:
$$\frac{dx}{dy}=\frac{2x}{y}-\frac{2x^2}{y^3}$$
$$=>\frac{dx}{dy}-\frac{2x}{y}=-\frac{2x^2}{y^3}$$
On dividing by $-x^2$,
$$=>\frac{dx}{dy}\frac{(-1)}{x^2}+\frac{2}{xy}=\frac{2}{y^3}$$
Substituting $v=\frac{1}{x}$; $\frac{dv}{dy}=\frac{dx}{dy}\frac{(-1)}{x^2}$gives:
$$\frac{dv}{dy}+\frac{2v}{y}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$.
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root,
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$.
Since $w$ is ... |
since w is a root, I first substituted w and got w^4=1 Which is w * w^3 = 1. Since w is not equal to 1, w^3=1
This step is wrong. Compare to the obviously false: $(-1)(-1)=1$ and since $(-1) \ne 1$ then it must be that $(-1) =1$.
What you can derive from $w^4=1$ however is that $\frac{1}{w} = w^3$ and $\frac{1}{w^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \... | Hint:$$\left( \sin a+\cos a \right) \left( \sin ^{ 2 } a-\sin a\cos a+\cos ^{ 2 } a \right) =\\ =\left( \sin a+\cos a \right) \left( { \left( \sin a+\cos a \right) }^{ 2 }-\frac { 3 }{ 2 } \left( { \left( \sin { a } +\cos { a } \right) }^{ 2 }-1 \right) \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and ... | Let us make the problem more general, replacing $\frac 13$ by
$x$. So, we have $$S=1+2x+6x^2+10x^3+14x^4+\cdots=1+\sum_{n=1}^\infty(4n-2)x^n$$ that is to say $$S=1+4\sum_{n=1}^\infty nx^n-2\sum_{n=1}^\infty x^n=1+4\color{red}x\sum_{n=1}^\infty nx^{n-1}-2\sum_{n=1}^\infty x^n$$ $$S=1+4x\left(\sum_{n=1}^\infty x^n\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Deriving the equation of a circle I have two points, $A=(-1,3)$ and $B=(2,7)$. There is a third point $P=(x,y)$.
I have found $|AP|$ and $|BP|$ in terms of $x$ and $y$. These are:
$|AP| = \sqrt{(-1-x)^2 +(3-y)^2}$
$|BP| = \sqrt{(2-x)^2 +(7-y)^2}$
I have shown that the set $S = \{P:|AP| = |BP|\}$ is a straight line and ... | Change the coordinate system to map $A$ to the origin in order to facilitate algebraic manipulations.
$$\begin{align}
(\;\ x,\;y)&\rightarrow \;\;(x+1,y-3)\\
A(-1,3)&\rightarrow A'(\;\;\;\;\;0,\;\;0)\\
B(\;\;2,7)&\rightarrow B'(\;\;\;\;\;3,\;\;4)\\
P(\;\;x,y)&\rightarrow P'(x+1, y-3)=P(x', y')
\\\\
|A'P'|&=\alpha |B'P'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $x$ is a real and $0 < x < 4$, then $\frac{4}{x(4-x)} \geq 1$ I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:
Suppose x is a real and $0 < x < 4$, it follows that,
\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow 4 < (x - 2)^2 < 4\\... | We can do this proof by contradiction: We want to prove $P\rightarrow Q$ and we prove it by assuming $P$ and $\neg Q$, and getting a contradiction.
Suppose $0<x<4$ with $x\in\mathbb{R}$, and the inequality $\dfrac{4}{x(x-4)}\geq 1$ does not hold.
Then we have $0<x,0<4-x$ (both follow from $0<x<4$) and $\dfrac{4}{x(x-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Limit of product with cubes $\lim\limits_{n \to \infty}\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}\dots\frac{n^3-1}{n^3+1}$ I am trying to evaluate
$$\lim\limits_{n \to \infty}{2^3-1 \over 2^3+1}{3^3-1 \over 3^3+1}\dots{n^3-1 \over n^3+1}$$
It seems to be the perfect candidate for a factoring formula, however, I get stuck r... | One may use a telescoping product, as $n \to \infty$,
$$
\prod_{k=2}^n\frac{k^3-1}{k^3+1}=\prod_{k=2}^n\frac{k-1}{k+1} \cdot \prod_{k=2}^n\frac{k(k+1)+1}{k(k-1)+1}=\frac23 \cdot \frac{n(n+1)+1}{n(n+1)}\to \frac23.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \fra... | matrix(
[1,1,1],
[(3/4)*a+(1/4)*b,(1/2)*a+(1/2)*b,(1/4)*a+(3/4)*b],
[((3/4)*a+(1/4)*b)^2,((1/2)*a+(1/2)*b)^2,((1/4)*a+(3/4)*b)^2]
);
$$
\begin{pmatrix}1 & 1 & 1\\
\frac{b}{4}+\frac{3a}{4} & \frac{b}{2}+\frac{a}{2} & \frac{3b}{4}+\frac{a}{4}\\
{{\left( \frac{b}{4}+\frac{3a}{4}\right) }^{2}} & {{\left( \frac{b}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim \limits_{(x, y)\rightarrow (1, -1)}\frac{(x-y)^2-4}{x^2+y^2-2}$ I want to calculate $\lim\limits_{(x, y)\rightarrow (1, -1)}\dfrac{(x-y)^2-4}{x^2+y^2-2}$.
I already know that $\lim \limits_{x\rightarrow 1}\lim\limits_{y\rightarrow -1}\dfrac{(x-y)^2-4}{x^2+y^2-2} = \lim\limits_{y\rightarrow -1}\lim\limits_... | Note that for all $x,y$ such that $x^2+y^2\neq 2$ the following holds:$$\frac{(x-y)^2-4}{x^2+y^2-2}=\dfrac{x^2+y^2-2-2xy-2}{x^2+y^2-2}=1-\dfrac{2xy+2}{x^2+y^2-2}.$$
Conjecture: $\lim \limits_{(x,y)\to (1,-1)}\left(\dfrac{2xy+2}{x^2+y^2-2}\right)=-1$.
Note that $$\dfrac{2xy+2}{x^2+y^2-2}+1=\dfrac{2xy+2}{x^2+y^2-2}+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x}$ Here $f(x) = x + \cos x$ and $g(x)= x- \cos x$
We can't apply L'Hopital's Rule since $\forall c > 0\quad \exists x_c \in (c, + \infty): g'(x) = 1 + \sin x = 0$.
But I don't know how to proceed.
My attempt:
Because $\cos x$ is bounded I was taking an arbitrary number $A >... | For $x \gt 1$, you have $\dfrac{x+\cos x}{x - \cos x}=1 + \dfrac{2\cos x}{x - \cos x}$
Since $0 \le | \cos x | \le 1$ you have $-2 \le 2 \cos x \le 2$ and $x-\cos x \ge x-1$
meaning $1 - \dfrac{2}{x - 1} \le \dfrac{x+\cos x}{x - \cos x} \le 1 + \dfrac{2}{x - 1}$ and thus a limit of $1$ as $x$ increases without limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Summation of tan What is $$\sum _1 ^7 \tan^2 (\frac {n\pi}{16}) -1$$ so I used $\tan(x)=\cot(\frac {\pi}{2}-x)$ for last three angles ie $5\pi/16,6\pi/16,7\pi/16$ . Thus it gets converted to some symmetry of tan,cot but after that I am not sure how to proceed or which identity to use.Use of AM-GM seems to give a very... | $$S=\sum^{7}_{n=1}\tan^2 \left(\frac{n\pi}{16}\right) = \tan^2 \left(\frac{\pi}{16}\right)+\tan^2 \left(\frac{\pi}{8}\right)+\tan^2 \left(\frac{3\pi}{16}\right)+1+\tan^2 \left(\frac{5\pi}{16}\right)+\tan^2 \left(\frac{3\pi}{8}\right)+\tan^2 \left(\frac{7\pi}{16}\right)$$
And using $\displaystyle \tan^2 \frac{7\pi}{16} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$ Well, I was trying to find $\tan40$ in terms of $\tan25$=x.
So, I expanded 40 as 25+15 and got the value of $\tan40$ in terms of $x$ as $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$.
Now, I solved $\tan40$ alternatively as
$$\frac{\tan155-\tan115}{1+\tan15... | On rearrange we have
$$2+\sqrt3=\dfrac{3x-x^3}{1-3x^2}=\cdots=\tan(3\cdot25^\circ)$$
Now $$\tan(75^\circ)=\tan(45^\circ+30^\circ)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the reduction formula for the integral $I_n = \int e^{ax}\sin^nxdx$ My attempt:
Let $$u=\sin^nx \rightarrow \text{d}u=n\sin^{n-1}x\cos x\text{d}x $$ and
$$ v = \int{e^{ax}}\text{d}x=\frac{e^{ax}}{a}.$$
Now we have $u\cdot v- \int{v\cdot \text{d}u} $, so
$$I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n... | Hint. We assume $a \neq0, \pm i$. From the identity you have already obtained by parts,
$$
I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} \tag1
$$ you may rather perform another integration by parts as follows
$$
\begin{align}
&\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}
\\&= \frac{e^{ax}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
factrorize $-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3$ factrorize $-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3$
or
$$ -(a^3 + b^3 + c^3) + (a^2 b + a b^2 + b^2 c + b c^2 + c^2 a + c a^2) - 2(abc) $$
It is a so long polynomial I don't have any ideas for starting factoring what should I do?
| $$ ( a +b - c) ( b + c - a ) ( c + a - b) $$
This is one factor short of Heron's formula,
$$ (a + b + c) ( a +b - c) ( b + c - a ) ( c + a - b) = 2 (b^2 c^2 + c^2 a^2 + a^2 b^2) - (a^4 + b^4 + c^4) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$
If the Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$ in $(0,1)\;,$ Then $k\in $
$\bf{My\; Try::}$ Let $3^x=y>0\;,$ Then inequality is
$$y^2-ky-k+3\leq 0\Rightarrow y^2-ky-k+3\leq 0\forall y\in (1,3)$$
So $$y... | Set $t=\mathrm 3^x$. The given equation has at least one root in $(0,1)$ if and only if the equation $\;p(t)=t^2-kt+3-t=0\;$ has at least one root in $(1,3)$.
The first condition is this equation has real roots , i.e.
\begin{align}&\Delta=k^2-4(3-t)=k^2+4t-12=(k+2)^2-16\ge 0\\
&\iff k\ge 2\enspace\text{or}\enspace k\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to direc... | I put my solution as an alternative (a little bit simpler). Write the original determinant as
$\begin{vmatrix}
-bc & b(b+c) & c(b+c)\\
a(a+c) & -ac & c(a+c) \\
a(a+b) & b(a+b) & -ab
\end{vmatrix}$
Multiply the first column with $b+c$ and the second one with c and bring $\frac{1}{c(b+c)}$ in front of the determinant:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y ... | Here's a proof that avoids (or disguises) the AM-GM inequality.
Let's start by establishing a completely different inequality: If $r,s\ge1$, then
$${1\over s}+r^2s+s\ge1+r+rs$$
The proof comes from first rewriting it in equivalent form as
$$(r-1)rs^2+(s-1)s\ge rs-1$$
and then letting $r=1+a$ and $s=1+b$ with $a,b\ge0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\tag1$$
Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a produ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = ... | $$3\cdot 5^{ x^{ 2 } }+3^{ x^{ 2 }+1 }-8\cdot 3^{ x^{ 2 } }=0\\ 3\cdot { \left( \frac { 5 }{ 3 } \right) }^{ { x }^{ 2 } }=5\\ { \left( \frac { 5 }{ 3 } \right) }^{ { x }^{ 2 } }=\frac { 5 }{ 3 } \quad \Rightarrow { x }^{ 2 }=1\quad \Rightarrow \quad x=\pm 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
| Try to write $2(3+2\sqrt 2)=6+4\sqrt 2$ as a square of $a+b\sqrt 2$ for some rational numbers $a,b$:
$$(a+b\sqrt 2)^2=a^2+2b^2+2ab\sqrt2 \;=\; 6+4\sqrt 2$$
By $\Bbb Q$-linear independence of $1$ and $\sqrt 2$, this implies $$ab=2,a^2+2b^2=6$$
Can you take it from here?
Here are some hints, if needed:
If we try to fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equi... | I will prove $a^a\cdot b^b\cdot c^c \geq (a+b-c)^{c}\cdot (b+c-a)^{b}\cdot (c+a-b)^{b}$
Let $a+b-c=2z,\ b+c-a=2x,\ a+c-b=2y$. Without losing generality, assume $x \geq y \geq z$. Now, we need prove
$$\prod \left(x+y\right)^{(x+y)} \geq \prod \left(2z\right)^{(x+y)}$$
$$\prod \left(\frac{x+y}{2z}\right)^{(x+y)} \geq 1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Time complexity of variation on Coupon's collector problem I need to know the complexity of the following algorithm:
Draw a set of n numbers from a larger set of m numbers, one by one, randomly, with replacement. The result may be any set of numbers as long as the size is n and the elements are different.
This is a var... | By way of enrichment here is the complexity using Stirling numbers
of the second kind. Using the notation from this MSE
link we have $n$
coupons, and ask about the expected time until a multiset containing
instances of $j$ different coupons has been drawn.
First let us verify that we indeed h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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For 1/A = 1/B + 1/C, why is A $\frac{1}{a}$ =$\frac{1}{b}$ +$\frac{1}{c}$
I read that $a$ is always < the smaller of $b$ and $c$
In the case of $0$ < $B$ $<1$ and $0$ < $C$ $<1$,
I can understand the rule as:
Firstly,
$$\frac{1}{A}=\frac{1}{B} +\frac{1}{C}=\frac{C+B}{B*C}$$
Now looking at 3 cases:
Case 1
$$\fr... | As you said, $\frac{1}{A}=\frac{C+B}{C*B}$
which means $A=\frac{B*C}{B+C}$
Now clearly $B > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = B*\frac{C}{B+C}$ and $\frac{C}{B+C} < 1$.
Similarly, $C > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = C*\frac{B}{B+C}$ and $\frac{B}{B+C} < 1$.
Thus $B > \frac{B*C}{B+C} = A$ and $C > ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$. What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$? Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$ and $k\in \mathbb{R}$.
What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$?
| Hint:
$$
1-a+a^2=\frac{1+a^3}{1+a}
$$
Or use that with $a=x^5$
$$
a+\frac1a=1+\frac1k\implies a^3+3a+3\frac1a+\frac1{a^3}=\left(1+\frac1k\right)^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is ... | According to Wolfram Alpha the expression
$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(b-a)}+\frac{c^4}{(c-a)(c-b)} \tag{$\ast$}$$ is not always an integer (see https://www.wolframalpha.com/input/?i=%5Cfrac%7Ba%5E4%7D%7B(b-a)(c-a)%7D%2B%5Cfrac%7Bb%5E4%7D%7B(c-b)(b-a)%7D%2B%5Cfrac%7Bc%5E4%7D%7B(c-a)(c-b)%7D,+a%3D3,b%3D5,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Does anybody recognize this array of rational numbers related to (hyper)spherical harmonics? While writing a program to compute bases of spherical harmonics on $S^n$, I discovered that the following array of rational numbers naturally arises when one considers the inclusion $S^n \subset S^{n+1}$:
\begin{array}{ccccccc}... | By sheer luck, I managed to stumble upon the correct formula shortly after posting this question.
$$ a_{ij} = \frac{i!!}{(i-1)!!} \cdot \frac{(i+2j-1)!!}{(i+2j)!!} $$
Here, $n!!$ denotes the double factorial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $p_n$ denote the $n$th prime number. For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$ Let $p_n$ denote the $n$th prime number. For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$.
The book gives a hint as $p^2_{n+3} \lt 4p_{n+2}^2 \lt 8p_{n+1}p_{n+2}$, but I don't understand how to show that t... | For $n=3$, show it manually:
$(p_{n+3})^2=(p_{6})^2=13^2=169<385=5\cdot7\cdot11=(p_{3})(p_{4})(p_{5})=(p_{n})(p_{n+1})(p_{n+2})$
For $n=4$, show it manually:
$(p_{n+3})^2=(p_{7})^2=17^2=289<1001=7\cdot11\cdot13=(p_{4})(p_{5})(p_{6})=(p_{n})(p_{n+1})(p_{n+2})$
For $n\geq5$, use the fact that $p_{n+1}<2p_{n}$:
$(p_{n+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \fr... | The limit cannot be found in your way.
Multiplying $$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
by
$$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}\ (=1)$$
gives
$$\frac{(x^4+x^3\tanh(x)+x^2)-(x^2+x)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\frac{x^3(\tanh(x)-2)}{(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Inequality. Prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$$
I tried SOS, C-S, the $uvw$'s technique and more, but without success.
| Let $p = a+b+c, \ q = ab+bc+ca, \ r = abc$. We have $q^2 \ge 3pr$.
Using AM-GM, we have $\sqrt{3pr} \ge \frac{6pqr}{q^2 + 3pr}$. Thus, it suffices to prove that
$$\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} +
\frac{12(a+b+c)(ab+bc+ca)abc}{(ab+bc+ca)^2 + 3(a+b+c)abc} \ge 3(a^2+b^2+c^2).$$
I can solve it by the Buffal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$ I'm stuck on a trigonometry question. The formulae I have been given don't seem to help so I had to come here.
The question is:
Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$
If these formulae can be used I would ... | In the right side you can use:
\begin{equation}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\\
\end{equation}
In this case:
\begin{equation}
\cos(x-\frac{\pi}{6})=\cos x\cos\frac{\pi}{6}+\sin x\sin\frac{\pi}{6}\\
\cos(x-\frac{\pi}{6})=\frac{\sqrt 3}{2}\cos x+\frac{1}{2}\sin x
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| $$
\begin{cases}
\text{a}^2-\text{b}^2-\text{a}=0\\
\\
\text{b}-2\text{a}\text{b}=\frac{1}{2}
\end{cases}\space\space\space\Longleftrightarrow\space\space\space
\begin{cases}
\text{a}=\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\\
\\
\text{b}-2\times\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\times\text{b}=\frac{1}{2}
\end{cases}
$$
Now:... | {
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"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| Observe that $1$ and $-1$ are roots of your polynomial. You thus find that it is equal to
$$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
and is thus clearly non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding S,T in matrix equivalence problem So guys I could really need some help finding the two matrices S,T. I seem to have some problems getting all the elemantary row,colmun opperations done to get the result. Maybe there is also an easier way of finding S,T. Anything can help.
The rank of the matrix A over the fi... | This problem comes down to finding appropriate bases for the domain (input) and codomain (output). Recall that the columns of a transformation matrix are the images of the domain basis vectors. The last two columns are zero, so the last two basis vectors for the domain must be elements of the kernel. We can choose the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
determmine image of $f$ Let $f(x)=\dfrac{1}{x+1}\sqrt{x^2-1}$
$D_f={(-\infty,-1)}\cup {[1, +\infty)}$
How can i show algebraically that $$f\left( (1,+\infty) \right)=(0,1)$$
Indeed,
let $y\in \mathbb{R} $
$$
\begin{align}
y\in f\left( ]1,+\infty[ \right) &\iff f(x)=y\\
&\iff \dfrac{1}{x+1}\sqrt{x^2-1}=y\\
&\iff \sq... | for $x>1$
$$f(x)=\sqrt{1-\frac{2}{x+1}}$$
we see that $f$ is continuous and strictly increasing at $(1,+\infty)$.
but
$$\lim_{x\to 1^+}f(x)=0$$
and
$$\lim_{x\to+\infty}f(x)=1$$
thus
$$f((1,+\infty))=(0,1)$$
algebraically
let $x>1$
$$f(x)=y\implies 1-\frac{2}{1+x}=y^2$$
$$\implies x=\frac{2}{1-y^2}-1$$
*
*$$x>1\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate with Cauchy integral formula Can you please check if i did any mistakes.
a)
$$\int_{\gamma_{B_2(0)}} \frac{z^7+41}{1-z}$$
$\frac{z^7+41}{1-z}$ = $ \frac{-z^7-41}{z-1}$ = $\frac{f(z)}{z-1}$
with the Cauchy integral formula i get $2 \pi i \cdot (-42) = -84\pi i$
b)
$$\int_{\gamma_{B_r(\frac{i}{2})}} \frac{1}{z... | For $r=2$, both poles are enclosed. We can write
$$\frac{1}{z^2+1}=\frac{1/2i}{z-i}-\frac{1/2i}{z+i}$$
and apply Cauchy's Integral Formula as
$$\begin{align}
\oint_{B_{2}(i/2)}\frac{1}{z^2+1}\,dz&=\oint_{B_{2}(i/2)}\left(\frac{1/2i}{z-i}-\frac{1/2i}{z+i}\right)\,dz\\\\
&=\oint_{B_{2}(i/2)}\frac{1/2i}{z-i}\,dz-\oint_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Axes of Symmetry for a General Ellipse
Given a general ellipse
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?
It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellip... | The principal axes are:
$$\left( y-\frac{2AE-BD}{B^2-4AC} \right)=
\frac{C-A \color{red}{\pm} \sqrt{(A-C)^2+B^2}}{B}
\left( x-\frac{2CD-BE}{B^2-4AC} \right)$$
Also refer to another answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Finding the joint Probability distribution of $X$ and $Y$? If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to s... | Here is a completed table of the discrete probabilities:
$$ \begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & 1 & 1 & 2 & 2 & 3 & 3 \\ \hline
\end {arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Polar form is
$$
r^4\left(\cos^4(\theta)+\sin^4(\theta)\right)=r^2\tag{1}
$$
which gives
$$
r^2=\frac1{\cos^4(\theta)+\sin^4(\theta)}\tag{2}
$$
The area is
$$
\begin{align}
\frac12\int_0^{2\pi}r^2\,\mathrm{d}\theta
&=\frac12\int_0^{2\pi}\frac{\mathrm{d}\theta}{\cos^4(\theta)+\sin^4(\theta)}\tag{3}\\
&=\frac12\int_\gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Geometry problem about triangle orthocenter Let $ABC$ be a triangle with orthocenter $H$. Let $AB = 84$, $AC = 32\sqrt{3}$ and $\angle{BAC} = 60^\circ$. Let $D, E, F$ be points on $AH, BH, CH$, respectively, such that: $7AD = AH$, $7BE = BH$, $7CF = CH$. Find $AB\cdot CF + BC\cdot AD + AC\cdot BE$.
This problem is t... | On each point $A, B, C$ of the triangle, pass through it a line parallel to its opposite side. Then, these 3 new lines form a new triangle, similar to the original one. the orthocenter of $ABC$ coincide with the circumcenter of the new triangle. Notice that
$$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = AB \cdot CH + BC ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $\sum\limits_{i=1}^6a_i^2=6$, where $a_i>0$, $a_7=a_1$. Prove that $\sum\limits_{i=1}^6\frac{a_i^2}{a_{i+1}}\geq6$ Let $a_i$ be positive numbers such that $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2=6$. Prove that:
$$\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\frac{a_3^2}{a_4}+\frac{a_4^2}{a_5}+\frac{a_5^2}{a_6}+\frac{a_6^2}{a... | We can write:
$$ \frac{a_1^2}{a_2} + \frac{a_2^2}{a_2} + \ldots + \frac{a_6^2}{a_1} \geq
a_1^2 + \ldots + a_6^2$$
which is equivalent to this inequality.
after dividing both sides by $(a_1^2 + \ldots + a_6^2)$ we get:
$$ \frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} + \ldots + \frac{a_6^2}{a_1^2 + \ldots + a_6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"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.