Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
| Note that $2^x+2^{x+1}=2^x+2\cdot 2^x=3\cdot 2^x$.
Similarly, $3^{x+2}+3^{x+3}=9\cdot 4\cdot 3^x$.
So our equation can be rewritten as $3\cdot 2^x=9\cdot 4\cdot 3^x$, or equivalently $(3/2)^x=1/12$, which can be solved using logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$
$(1)\;\;$ If $f(n+1)-f(n) = 3$ and $f(0) = 1\;,$ Then $f(n) = \;,$ Where $n\in \mathbb{W}$
$(2)\;\;$ If $f(n+2)-5f(n+1)+6f(n)=0$ and $f(0)=2$ and $f(1)=5\;,$ Then $f(n)=$
Where $n\in \mathbb{W}$
$\bf{My\; Try\;, (1)::}$ Let $f(n)=a_{n}\;,$ Then Our functional equation convert into $a_{n+1}-a_{n} = 3$ and $a_{0}=1$
So $$a_{n+1}=3+a_{n}\;,$$ Now Replcae $n\rightarrow (n-1)\;,$ We get
$$a_{n}=3+a_{n-1}.............(1)$$
Now in a Similar way, We get $$a_{n}=3+3+a_{n-2}$$
So in a Similar way, We get $$a_{n}=\underbrace{3+3+3+.........+3}_{\bf{n\; times}}+a_{n-n}=3n+a_{0}$$
So we get $$a_{n}=1+3n\;,$$ Where $n\in \mathbb{W}$
$\bf{(2)}$ Here our equation Convert into $a_{n+2}-5a_{n+1}+6a_{n}=0\;\;,$ and $a_{0}=2\;\;a_{1}=5$
Now I did not Understand How can I solve it, Plz explain me in detail,
Thanks
| Let the generating function of the sequence $a$ defined by
$$a_{n + 2} - 5a_{n + 1} + 6a_n= 0$$
be
$$g(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots$$
Then
$$-5xg(x) = -5a_0x -5a_1x^2 - 5a_2x^3 - 5a_3x^4 - \dots$$
and
$$6x^2g(x) = 6a_0x^2 + 6a_1x^3 + 6a_2x^4 + 6a_3x^5 + \dots$$
Adding all three up,
$$\begin{align}g(x) - 5xg(x) + 6x^2g(x) &= a_0 + (a_1 - 5a_0)x \\
&+(a_2 - 5a_1 + 6a_0)x^2\\
&+(a_3 - 5a_2 + 6a_1)x^3\\
&+\dots\\
&=2 -5x \end{align}$$
so that
$$(1 - 5x + 6x^2)g(x) = 2 - 5x$$
$$g(x) = \frac{2 - 5x}{(3x - 1)(2x - 1)}$$
It should be relatively more trivial to decompose this into partial fractions. You can then use geometric series and some algebra to determine the general term for the coefficient of $x^n$ in $g(x)$. It will correspond to $a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| Use the facts
$$(a^x)' = a^x \ln a, \quad (\log_a(x))' = \frac{1}{x\ln a}$$
for $a > 0$ and $a \neq 1$.
Then use the multiplication formula for differentiation to get
\begin{align}
& (10^x \log_{10}x)' \\
= & (10^x)'\log_{10}x + 10^x (\log_{10}x)' \\
= & 10^x \ln (10) \log_{10}x + 10^x \frac{1}{x\ln 10}.
\end{align}
If you want to express your final result in natural logs, then you can simplify by writing
$$\log_{10}x = \frac{\ln x}{\ln 10},$$
in this way the answer would agree with the wolfram output exactly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm trying to figure out the right test for this example. I haven't touched Calc II in a while, and I'm really not interested in getting a weird explanation from Wolfram. I was thinking that I might want to do some kind of ratio test between the $n$th and $n+1$th term.
| An elementary argument, though not the most simple or quick: let $a_n = (1+4n)^{-1}$. Then let $b_n = (4+4n)^{-1} = 4^{-1} (1+n)^{-1}$, so that we clearly have for all positive integer $n$, $$a_n > b_n.$$ Now note that $$S_n = \sum_{k=1}^{n-1} b_k = \frac{1}{4} \sum_{k=1}^{n-1} \frac{1}{k+1} = \frac{1}{4} \left( -1 + \sum_{k=1}^n \frac{1}{k} \right),$$ so it suffices to show that the sequence of partial sums $$H_n = \sum_{k=1}^n \frac{1}{k}$$ for $n = 1, 2, \ldots$ increases without bound. Indeed, the familiar proof of this fact is not hard to see: $$H_{2^{m+1}} = H_{2^m} + \sum_{k=2^m+1}^{2^{m+1}} \frac{1}{k} \ge H_{2^m} + \sum_{k=2^m+1}^{2^{m+1}} \frac{1}{2^{m+1}} = H_{2^m} + \frac{1}{2}. $$ Since $H_{2^0} = 1$, it follows that $$H_{2^m} > 1 + \frac{m}{2}$$ for every positive integer $m$, consequently $$S_{2^m} > \frac{m}{8};$$ thus $S_n$ has no upper bound, and $\sum_{k=1}^{n-1} a_k$, being larger than $S_n$, must also diverge as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$. Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x).$Find the value of $g(2).$
As $P(x)$ is a polynomial containing only terms of odd degree.Therefore it should pass through origin.
$P(0)=0$
When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.
$P(x)=Q_1(x)(x-3)+6....................(1)$
If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x)$.
$P(x)=Q_2(x)(x^2-9)+ax+b............(2)$,where $ax+b=g(x)$
From $(1),P(3)=6$
From $(2),P(3)=6=3a+b$
But we need to find $g(2)=2a+b$
I dont know how to solve further.Please help me.Thanks.
| Since it is an odd degree polynomial
P(–a)= –p(a)
Let P(x)=(x^–9)Q(x)+ax+b
(where Q is quotient and (ax+b) = g(x) = remainder)
Given P(3)=6
Therefore P(-3) = -6
Solve these two equations to get value of a and b
We get a=2 and b=0
G(x) =ax+b
G(2)= 2(2)+0 =4
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On proving $\lim \frac{n^2}{n^2+n+1} = 1$
Prove that $\lim \frac{n^2}{n^2+n+1} = 1$
Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$
But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n+1} = \bigg|\frac{n^2}{n^2+n+1}-1\bigg|.$
Therefore $\bigg|\frac{n^2}{n^2+n+1}-1\bigg| < \varepsilon$, hence $\displaystyle \lim \frac{n^2}{n^2+n+1} = 1.$
Could someone please verify whether the above is correct.
| Your stream of thought is right; just notice that $1/\varepsilon$ is not necessarily an integer, which matters if we are talking about a sequence, which is a map defined on a subset of the set of all integers. Instead, you may write $N := \lceil 1/\varepsilon \rceil + 1$, by which you can ensure the choice of $N$ to be an integer. However, this depends on how one defines convergence of sequence; and it is really not a big deal as long as you are consistent in what you are talking about.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
| Lagrange multipliers? So $f(x,y,z,\lambda) = x^2 + y^2 + z^2 - \lambda(x+y+z-1)$.
So critical points obey: $2x - \lambda = 0, 2y - \lambda = 0, 2z - \lambda = 0, x+y+z-1 = 0$. So $x = y = z = \frac{1}{3}$, which gives sum of squares $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Series expansion of integral Consider the function $I(y)=\int_0^\infty e^{-\sqrt{x^2+y^2}} \mathrm{d} x$. I'd like to see the leading order term of $I(y)$ about $y=0$, so I expand the integrand:
$$
e^{-\sqrt{x^2+y^2}}=e^{-x}-e^{-x}\frac{1}{2x}y^2+e^{-x}\frac{1+x}{8x^3}y^4+\dots
$$
However, only the integral of the $0^\mathrm{th}$ order term converges (to $1$ of course). How would one proceed here to find the second order term (or higher)?
| Let $I(y)$ be the integral of interest given by
$$\begin{align}
I(y)&=\int_0^\infty e^{-\sqrt{x^2+y^2}}\,dx\\\\
&=\int_0^{|y|} e^{-\sqrt{x^2+y^2}}\,dx+\int_{|y|}^\infty e^{-\sqrt{x^2+y^2}}\,dx \tag 1
\end{align}$$
For the first integral on the right-hand side of $(1)$, we note that $\sqrt{x^2+y^2}\le \sqrt{2}|y|$. Therefore, inasmuch as we are developing an asymptotic series for "small" $|y|$, we can write
$$\begin{align}
\int_0^{|y|} e^{-\sqrt{x^2+y^2}}\,dx&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^{|y|}\left(x^2+y^2\right)^{n/2}\,dx\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n|y|^{n+1}}{n!}\int_0^{\pi/4}\sec^{n+2}\theta \,d\theta\\\\
&=|y|-\left(\frac{\sqrt 2 +\log\left(1+\sqrt 2\right)}{2}\right)|y|^2+\frac23 |y|^3+O\left(|y|^4\right) \tag 2
\end{align}$$
To facilitate evaluation of the second integral on the right-hand side of $(1)$, we expand the exponential as
$$\begin{align}
e^{-\sqrt{x^2+y^2}}&=e^{-x}-\frac12 y^2\frac{e^{-x}}{x}+\frac18 y^4\frac{(x+1)e^{-x}}{x^3}+O\left(y^6\right)\\\\
&=e^{-x}-\frac12 y^2\frac{e^{-x}}{x}+\frac18 y^4\frac{e^{-x}}{x^2}+\frac18 y^4\frac{e^{-x}}{x^3}+O\left(y^6\right) \tag 3\\\\
\end{align}$$
Next, we write the integral of the first term on the right-hand side of $(3)$ as
$$\begin{align}
\int_{|y|}^\infty e^{-x}\,dx&=e^{-|y|}\\\\
&=\sum_{n=0}\frac{(-1)^n|y|^n}{n!}\\\\
&=1-|y|+\frac12 |y|^2-\frac16 |y|^3+O\left(|y|^4\right) \tag 4
\end{align}$$
We write the integral of the second term is
$$\begin{align}
-\frac12 |y|^2\,\int_{|y|}^\infty \frac{e^{-x}}x\,dx&=-\frac12 |y|^2\,\left(\int_{|y|}^1 \frac{e^{-x}}x\,dx+\int_1^\infty \frac{e^{-x}}x\,dx\right)\\\\
&=\frac12 \log (|y|)\,|y|^2-\frac12 \left(\int_1^\infty \frac{e^{-x}}x\,dx-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,n!}\right)\,|y|^2-\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+2}}{n\,n!}\\\\
&=\frac12 \log (|y|)\,|y|^2+\frac12 \gamma\,|y|^2-\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+2}}{n\,n!}\\\\
&=\frac12 \log (|y|)\,|y|^2+\frac12 \gamma\,|y|^2-\frac12 |y|^3+O\left(|y|^4\right) \tag 5
\end{align}$$
Using integration by parts, we find the third integral on the right-hand side of $(3)$ as
$$\begin{align}
\frac18 |y|^4\,\int_{|y|}^\infty \frac{e^{-x}}{x^2}\,dx&=\frac18 |y|^4\left(\frac{e^{-|y|}}{|y|}\right)-\frac18 |y|^4\,\int_{|y|}^\infty \frac{e^{-x}}{x}\,dx\\\\
&=\frac18 e^{-|y|}|y|^3+\log (|y|)|y|^4+\frac18 \gamma |y|^4-\frac18 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+4}}{n\,n!}\\\\
&=\frac18 |y|^3+O\left(\log(|y|)\,|y|^4\right) \tag 6
\end{align}$$
Using integration by parts twice, we find the fourth integral on the right-hand side of $(3)$ as
$$\frac18\,|y|^4\int_{|y|}^\infty \frac{e^{-x}}{x^3}\,dx=-\frac1{16}|y|^2-\frac18|y|^3+O\left(\log(|y|)\,|y|^4\right)\tag 7$$
Putting together results from $(2)$ and $(4)-(7)$, we obtain so far
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-(x^2+y^2)}\,dx=1+\left(\frac{\log(|y|)+\gamma -\log(1+\sqrt 2)-(1-\sqrt 2)}{2}\right)\,y^2+O(y^4)}\tag 8$$
IMPORTANT NOTE:
In $(3)$, we have omitted terms beyond $\frac18 y^4\frac{e^{-x}}{x^3}$. These terms will have a component that is of order $|y|^2$ and as such, the expansion in $(8)$ as written, is not correct.
However, the expansion is correct up to order $\log(|y|)\,|y|^2$. On way to complete the expansion up to order $|y|^2$ is to use the Faa di Bruno Formula to write the full expansion of $(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility of $2^n-n^2$ by 7 How many positive integers $n<10^4$ are there such that $2^n - n^2$ is divisible by 7?
| $10^4$ is small enough that it is possible to write a computer program to find the answer, but it is also possible to solve explicitly as follows:
We note that the possible remainders of $n^2$ modulo $7$ are $0, 1, 2$ and $4$. Since $2^n$ is never divisible by $7$, the possibilities that we have for which $2^n-n^2$ is divisible by $7$ are
$$ 2^n \equiv 1 \pmod 7 \quad\text{ and }\quad n \equiv \pm 1 \pmod 7$$
or
$$ 2^n \equiv 2 \pmod 7 \quad\text{ and }\quad n \equiv \pm 3 \pmod 7$$
or
$$ 2^n \equiv 4 \pmod 7 \quad\text{ and }\quad n \equiv \pm 2 \pmod 7$$
Now $2^n \equiv 1 \pmod 7$ if and only if $n \equiv 0 \pmod 3$, $2^n \equiv 2 \pmod 7$ if and only if $n \equiv 1 \pmod 3$, and $2^n \equiv 4 \pmod 7$ if and only if $n \equiv 2 \pmod 3$.
So our possibilities are
$$
\begin{align*}
n & \equiv 0 \pmod 3 && \text{and} & n & \equiv 1 \pmod 7 \\
n & \equiv 0 \pmod 3 && \text{and} & n & \equiv -1 \pmod 7 \\
n & \equiv 1 \pmod 3 && \text{and} & n & \equiv 3 \pmod 7 \\
n & \equiv 1 \pmod 3 && \text{and} & n & \equiv -3 \pmod 7 \\
n & \equiv 2 \pmod 3 && \text{and} & n & \equiv 2 \pmod 7 \\
n & \equiv 2 \pmod 3 && \text{and} & n & \equiv -2 \pmod 7
\end{align*}
$$
We can use the Chinese Remainder Theorem to find the solution to each pair of equations. We find that the possibilities for $n$ modulo $21$ are $15, 6, 10, 4, 2$ and $5$. Or, in order: $2, 4, 5, 6, 10, 15$ and $21$.
We now just need to count how many numbers there are with each remainder up to $10^4$. We have that $10^4 = 476 \times 21 + 4$, and so there are $476$ complete sets of $21$ numbers up to $10^4$, and $4$ extra. We find that the answer is $476 \times 6 + 1$ (Not $+2$ since we exclude $10^4$ since $n$ is strictly less than $10^4$) This is equal to $2857$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$
a) Solve $x^2\equiv 2y^2\pmod{3}$
b) Use part a to prove that $x$ and $y$ are divisible by $3$ when $x^2+6u^2=2y^2+3v^2$
My attempt:
$x^2\equiv 2y^2\pmod{3}\iff $$(y^{-1}x)^2 \equiv 2\pmod{3}$ but $\left(\frac{2}{3}\right)=-1$ since $3 \equiv 3\pmod{8}$. This means that 2 can't be Q.R. but at the same time I showed that $(xy^{-1})^2$ is clearly a square. Does it mean that the inverse doesn't exist?
| Once you have $x \equiv y \equiv 0 \pmod 3, $ you get a little extra, since $x^2 \equiv y^2 \equiv 0 \pmod 9. $ This says something about the other two letters, $ 6 u^2 \equiv 3 v^2 \pmod 9. $ Then $ 2 u^2 \equiv v^2 \pmod 3. $ Finally, this part of the conclusion is $u \equiv v \equiv 0 \pmod 3. $
Stated carefully, the result is simply that if
$$ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2 \pmod 9, $$
then
$$ x \equiv u \equiv y \equiv v \equiv 0 \pmod 3. $$ Which means that if
$ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2 \pmod 9, $ then $$ \gcd(x,u,y,v) \neq 1. $$
Now assume we have any integer solution to $$ X^2 + 6 U^2 = 2 Y^2 + 3 V^2 $$
where the only requirement is that at least one of $X,U,Y,V$ is not equal to $0.$ We may then find
$$ T = \gcd(X,U,Y,V); $$
in turn, we may define
$$ x = X/T, \; u = U/T, \; y = Y/T, \; v = V/T. $$
It follows that $ \gcd(x,u,y,v) = 1. $
It is still true that t least one of $x,u,y,v$ is not equal to $0.$ It is also still true that $x^2 + 6 u^2 = 2 y^2 + 3 v^2.$ However, this means $ x^2 + 6 u^2 \equiv 2 y^2 + 3 v^2 \pmod 9, $ from which we proved that $ \gcd(x,u,y,v) \neq 1. $ This contradicts the assumption that there is any solution in integers $X,U,Y,V$ with at least one of them nonzero.
The final result is that $x^2 + 6 u^2 = 2 y^2 + 3 v^2$ in integers implies
$x=u=y=v=0.$ Only trivial integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to prove that a solution exists for a fourth order ode Does there exists a solution of the initial value problem
$$(x^2 − 4)\frac {d^4y}{dx^4} + 2x \frac {d^2y}{dx^2} + (\sin x)y(x) = 0$$
where $y(0) = 0, y'(0) = 1, y''(0) = 1, y'''(0) = −1$
So far I have $y_1=y$, $y_2= \frac {dy}{dx}$ and so on til $y_n = \frac {d^{n-1}y}{dx^{n-1}}$ from the Existence and Uniqueness Theory. i dont know how to proceed further. i tried to solve it through the existence and uniqueness theorem but this is all i have right now
| I do not guarantee that no mistakes have been made.
Let
\begin{equation*}
y_{n}=\partial _{x}^{n-1}y
\end{equation*}
Then
\begin{eqnarray*}
\partial _{x}y_{1} &=&y_{2} \\
\partial _{x}y_{2} &=&y_{3} \\
\partial _{x}y_{3} &=&y_{4} \\
\partial _{x}y_{4} &=&y_{5}
\end{eqnarray*}
and
\begin{eqnarray*}
(x^{2}-4)\partial _{x}^{4}y+2x\partial _{x}^{2}y+\sin xy &=&0 \\
(x^{2}-4)y_{5}+2xy_{3}+\sin xy_{1} &=&0 \\
y_{5}+\frac{2x}{x^{2}-4}y_{3}+\frac{\sin x}{x^{2}-4}y_{1} &=&0 \\
\partial _{x}y_{4} &=&-\frac{2x}{x^{2}-4}y_{3}-\frac{\sin x}{x^{2}-4}y_{1}
\end{eqnarray*}
Now let
\begin{equation*}
\mathbf{y}(x)=\left(
\begin{array}{c}
y_{1} \\
y_{2} \\
y_{3} \\
y_{4}%
\end{array}
\right) ,\;\mathbf{y}(0)=\left(
\begin{array}{c}
0 \\
1 \\
1 \\
-1
\end{array}
\right)
\end{equation*}
Then
\begin{equation*}
\partial _{x}\mathbf{y}=\left(
\begin{array}{c}
y_{2} \\
y_{3} \\
y_{4} \\
-\frac{2x}{x^{2}-4}y_{3}-\frac{\sin x}{x^{2}-4}y_{1}%
\end{array}
\right) =\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-\frac{\sin x}{x^{2}-4} & 0 & -\frac{2x}{x^{2}-4} & 0
\end{array}
\right) \mathbf{y}=\mathsf{M}\cdot \mathbf{y}
\end{equation*}
or
\begin{equation*}
\mathbf{y}(x)=\exp [x\mathsf{M}]\cdot \mathbf{y}(0)
\end{equation*}
Suppose that $\mathsf{M}\cdot \mathbf{u}=0$. Then
\begin{equation*}
\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-\frac{\sin x}{x^{2}-4} & 0 & -\frac{2x}{x^{2}-4} & 0
\end{array}
\right) \left(
\begin{array}{c}
u_{1} \\
u_{2} \\
u_{3} \\
u_{4}
\end{array}
\right) =\left(
\begin{array}{c}
u_{2} \\
u_{3} \\
u_{4} \\
-\frac{\sin x}{x^{2}-4}u_{1}-\frac{2x}{x^{2}-4}u_{3}
\end{array}
\right) =0
\end{equation*}
and it follows that $\mathbf{u}$ must vanish. Thus $\mathsf{M}$ has an empty
null-space and is well-behaved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$ $\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx$
My Work:
By the comparison principle: $$\lim_{n\to\infty}\int_0^1\frac{x\sin{nx}}{1+x^2n^6}dx\le\lim_{n\to\infty}\int_0^1\frac x{1+x^2n^6}dx$$ I think that the way to compute this integral has to do with uniform convergence, but I cannot figure how to determine it.
| Hint: For $n\ge1$, $$\left|\frac{x\sin(nx)}{1+x^2n^6}\right|\le x$$ so Dominated Convergence should apply if the domain of integration is $[0,1]$.
If the Domain of Integration is $\boldsymbol{[0,\infty)}$
Integration by Parts gives
$$
\begin{align}
\int_0^\infty\frac{x\sin(nx)}{1+x^2n^6}\,\mathrm{d}x
&=-\frac1n\int_0^\infty\frac{x}{1+x^2n^6}\,\mathrm{d}\cos(nx)\\
&=\frac1n\int_0^\infty\frac{1-x^2n^6}{\left(1+x^2n^6\right)^2}\cos(nx)\,\mathrm{d}x\\
&=\frac1{n^4}\int_0^\infty\frac{1-x^2}{\left(1+x^2\right)^2}\cos\left(\frac{x}{n^2}\right)\,\mathrm{d}x
\end{align}
$$
Now Dominated Convergence applies since
$$
\begin{align}
\int_0^\infty\left|\frac{1-x^2}{\left(1+x^2\right)^2}\right|\,\mathrm{d}x
&\le\int_0^\infty\frac{1+x^2}{\left(1+x^2\right)^2}\,\mathrm{d}x\\
&=\int_0^\infty\frac1{1+x^2}\,\mathrm{d}x\\
&=\frac\pi2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$
$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$\bf{My\; Try::}$ Using $$\sin 3x =3\sin x -4\sin^3 x\Rightarrow 4\sin^3 x=3\sin x-\sin 3x$$ and
$$\cos 3x = 4\cos^3 x-3\cos x\Rightarrow 4\cos^3 x=\cos 3x+3\cos x$$
So we get $$A(4\sin^3 x) +B(4\cos^3 x)+4C=0$$
So $$A(3\sin x-\sin 3x)+B(\cos 3x+3\cos x)+4C=0$$
I did not understand How can I solve that question, Help me
Thanks
| Here, I found a restriction for the constants $A$, $B$, $C$ that wanted to share. We know by the well-known inequalities $|\sin x| \le 1$ and $|\cos x| \le 1$ that
$$-(|A|+|B|) \le A\sin^3 x+B\cos^3 x \le (|A|+|B|)$$
So, if we wish the equation to have a root we must have
$$-(|A|+|B|) \le -C \le (|A|+|B|)$$
Or equivalently
$$|C| \le |A|+|B|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
The chart-problem; problem solving In how many ways can we construct a $6\times 6$ chart with only $1$ and $-1$ such that in every row and column, the product is always positive?
| There are $\color{red}{2^{25}}$ ways of filling a $6\times 6$ grid with $0$ or $1$ in such a way that along every row or column there is an even number of $1$s. We may first fill a $5\times 5$ subgrid in any way we like, then complete those rows and columns in the only way they fit our constraints. The only square we still have to fill has to be filled with the parity of the number of $1s$ in the chosen $5\times 5$ subgrid:
$$\begin{pmatrix} 0 & 0 & 1 & 0 & 0 & ? \\
0 & 1 & 1 & 1 & 1 & ? \\
1 & 1 & 0 & 0 & 1 & ? \\
1 & 0 & 1 & 1 & 0 & ? \\
0 & 1 & 1 & 0 & 1 & ? \\
? & ? & ? & ? & ? & ? \end{pmatrix}\mapsto
\begin{pmatrix} 0 & 0 & 1 & 0 & 0 & \color{red}{1} \\
0 & 1 & 1 & 1 & 1 & \color{red}{0} \\
1 & 1 & 0 & 0 & 1 & \color{red}{1} \\
1 & 0 & 1 & 1 & 0 & \color{red}{1} \\
0 & 1 & 1 & 0 & 1 & \color{red}{1} \\
\color{red}{0} & \color{red}{1} & \color{red}{0} & \color{red}{0} & \color{red}{1} & \color{blue}{0} \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
I would like to calculate $\lim_ {n \to \infty} {\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}$ I would like to calculate the following limit: $$\lim_ {n \to \infty} {\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}$$
where $\lfloor x \rfloor$ is floor of $x$ and $x ∈ R$.
Now I know the result is $2$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
| You may observe that, as $n \to \infty$,
$$
\begin{align}
{\frac{n+\lfloor \sqrt{n} \rfloor^2}{n-\lfloor \sqrt{n} \rfloor}}&={\frac{2n+(\lfloor \sqrt{n} \rfloor-\sqrt{n})(\lfloor \sqrt{n} \rfloor+\sqrt{n})}{n-\lfloor \sqrt{n} \rfloor}}\\\\
&={\frac{2+(\lfloor \sqrt{n} \rfloor-\sqrt{n})(\lfloor \sqrt{n} \rfloor+\sqrt{n})/n}{1-\lfloor \sqrt{n} \rfloor/n}}
\\\\& \to 2
\end{align}
$$ since, as $n \to \infty$,
$$
\left|\frac{\lfloor \sqrt{n} \rfloor}{n}\right|\leq\frac{\sqrt{n}}{n} \to 0
$$ and
$$
\left|\frac{(\lfloor \sqrt{n} \rfloor-\sqrt{n})(\lfloor \sqrt{n} \rfloor+\sqrt{n})}{n}\right|\leq\frac{2\sqrt{n}}{n} \to 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n})$ I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$
I know it is $1$ but I don't understand why this is wrong :
$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.
So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) = \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.
Thanks.
| First of all, notice that the function $f (\theta) = \sin^2 \theta$ is periodic with a period $\pi$. Then, for any integer $n$ we should have $\sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \left (\pi \left(\sqrt {n^2 + n} - n\right)\right)$. Since $$\lim_{n \to \infty} \left(\sqrt {n^2 + n} - n\right) = \frac {1} {2},$$ it becomes obvious that $$\lim_{n \to \infty} \sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \frac {\pi} {2} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$. Who can help me to prove this equation?
$$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$
| Notice, here is the correct proof $$LHS=\sqrt{2a+2\sqrt{a^2-b}}$$
$$=\sqrt{(a-\sqrt{b})+(a+\sqrt b)+2\sqrt{a^2-(\sqrt b)^2}}$$
$$=\sqrt{\underbrace{\left(\sqrt{a-\sqrt{b}}\right)^2}_{A^2}+\underbrace{\left(\sqrt{a+\sqrt b}\right)^2}_{B^2}+2\underbrace{\sqrt{(a-\sqrt b)}}_{A}\underbrace{\sqrt{(a+\sqrt b)}}_{B}}$$
using identity, $A^2+B^2+2AB=(A+B)^2$,
$$=\sqrt{\left(\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right)^2}$$
$$=\left|\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right|$$
since, $a\ge 0,\ b\ge 0 $
$$=\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}$$
$$=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inverse Laplace problem using partition fraction Hello I am solving inverse Laplace transform using partial fraction. The question is:
$$ X(s) = \frac{10(s+1)}{s(s^2+4s+8)} => \frac{10(s+1)}{s((s+2)^2+4)} $$
$$ \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$
for finding C1
$$ \frac{10(s+1)}{s((s+2)^2+4)} = \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$
Multiplying both side by $s$ and set $s=0$, we get
$C1=5/4$
for finding C3
$$ \frac{10(s+1)}{s((s+2)^2+4)} = \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$
Multiplying both side by $((s+2)^2+4)$ and set $s=-2$, we get
$C3=5$
For C2 I will take first derivative $X(s)$and Multiplying both side by $((s+2)^2+4)$
$$ \frac{d}{ds}[((s+2)^2+4) X(s)] = \frac{d}{ds} \frac{10s}{s}$$
$$ \frac {s(10) - (10s + 10)*1} {s^2} $$
We set $s=-2$
I got $C2=-5/2$...
but the solution I have they said $C2 = -5/4$
Please can you tell where I did mistake?
| Notice, for factorization of denominator you should find the roots of quadratic equation: $s^2+4s+8=0$ as $$s=\frac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}=-2\pm 2i$$
$$\implies s^2+4s+8=(s+2+2i)(s+2-2i)$$
Now, one should have the following partial fractions $$X(s)=\frac{10(s+1)}{s(s^2+4s+8)}=\frac{10(s+1)}{s(s+2+2i)(s+2-2i)}$$
$$\frac{10(s+1)}{s(s+2+2i)(s+2-2i)}=\frac{A}{s}+\frac{B}{s+2+2i}+\frac{C}{s+2-2i}$$
Now, you can solve for $A, B$ & $C$ & then apply formula for inverse Laplace for each partial fraction $$L^{-1}\left[\frac{1}{s+a}\right]=e^{-at}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to integrate $\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$? As my title says, I need help integrating with floor functions,
$$\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$$
What does this even mean exactly? How would approach this?
| Essentially, you have this:
$$\int_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx=\int_\frac{1}{2}^1\left(-1\right)^1dx+\int_\frac{1}{3}^\frac{1}{2}\left(-1\right)^2dx+\int_\frac{1}{4}^\frac{1}{3}\left(-1\right)^3dx+\int_\frac{1}{5}^\frac{1}{4}\left(-1\right)^4dx+\ldots$$
$$=\left(-1+\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+ \left(-\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots$$
$$=-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots\right)$$
$$=-1+2(1-\log2)=1-\log 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Find one basis of kernel, one basis of image of the linear transformation and it's defect Find one basis of kernel, one basis of image of the linear transformation
$T:\mathcal{M}_{2\times 2}(\mathbb{R})\rightarrow \mathcal{M}_{2\times 3}(\mathbb{R})$ and defect of $T$ where $T(X)=X \begin{bmatrix}
0 & 0 & 0 \\
2 & -2 & 4 \\
\end{bmatrix}+3\cdot trace(X) \begin{bmatrix}
3 & -3 & 6 \\
-1 & 1 & -2 \\
\end{bmatrix}$
I don't know how to find the matrix of $T$. Is it $6\times 4$ matrix?
Could someone show how to find matrix of $T$?
Also, what is the defect of a linear transformation?
I am familiar with the idea of defective matrix. Defective matrix is a square matrix that is not diagonalizable. What would be the defect of a matrix? What if a matrix is not squared (like in this case)?
| The space of $2\times 2$ matrices has as basis the elementary matrices $e_{11},e_{12},e_{21},e_{22}$ with a $1$ in the $ij$ position and $0$ elsewhere, so you can calculate the matrix of $T$ relative to this basis and with the corresponding basis of the space of $2\times 3$ matrices. Note that $e_{ij}$ has trace $0$ if $i\neq j$ and trace $1$ if $i=j$, so that
\begin{align}
T(e_{12})&=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}
0 & 0 & 0 \\
2 & -2 & 4 \\
\end{pmatrix}+0 \begin{pmatrix}
3 & -3 & 6 \\
-1 & 1 & -2 \\
\end{pmatrix}=\begin{pmatrix}2&-2&4\\0&0&0\end{pmatrix}\\
T(e_{11})&=\begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix}
0 & 0 & 0 \\
2 & -2 & 4 \\
\end{pmatrix}+3 \begin{pmatrix}
3 & -3 & 6 \\
-1 & 1 & -2 \\
\end{pmatrix}=\begin{pmatrix}9&-9&18\\-3&3&-6\end{pmatrix}\\
T(e_{21})&=\begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix}
0 & 0 & 0 \\
2 & -2 & 4 \\
\end{pmatrix}+0 \begin{pmatrix}
3 & -3 & 6 \\
-1 & 1 & -2 \\
\end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&0\end{pmatrix}\\
T(e_{22})&=\begin{pmatrix}0&0\\0&1\end{pmatrix} \begin{pmatrix}
0 & 0 & 0 \\
2 & -2 & 4 \\
\end{pmatrix}+3 \begin{pmatrix}
3 & -3 & 6 \\
-1 & 1 & -2 \\
\end{pmatrix}=\begin{pmatrix}9&-9&18\\-1&1&-2\end{pmatrix}\end{align}
By looking at matrices as row vectors, say, you can consider $T$ as a transformation $T:\Bbb R^4\to\Bbb R^6$ using the isomorphisms that send canonical elementary matrices to canonical row vectors. If we use the order $e_{11},e_{12},e_{21},e_{22}$ and send it to $e_1,e_2,e_3,e_4$ and use the order $e_{11},e_{12},e_{13},e_{21},e_{22},e_{23}$ and send it to $e_1,\cdots,e_6$, we get the matrix $$\begin{pmatrix} 2&9&0&9\\-2 &-9&0&-9 \\4&18&0&18\\0&-3&0&-1\\0&3&0&1\\0&-6&0&-2\end{pmatrix}$$
Suppose that we have linear transformations $f:V\to W$ and $g:W\to Z$ for which $gf=0$, over finite dimensional vector spaces, and that $gf=0$. One then calls $$d=\dim W-\operatorname{rank}{f}-\operatorname{rank}{g}$$
the "defect". If one knows some homological algebra, $d$ is the dimension of the quotient space $\ker g/{\rm im}\, f$ which is the homology of $$ V\longrightarrow W\longrightarrow Z$$
at $W$, since $\dim\ker g=\dim W-\operatorname{rank}g$ by the rank-nullity theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality exercise (olympiad) For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$
From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But clearly, 15 can be less than $5(a+b+c)$. So this technique won't work. I fail to see some other inequalities to prove this. Any other ideas/hints? Thanks
| Lemma: for any postive $x,y,z>0$,and $xyz=1$ we have
$$(xy+yz+xz)^2+3\ge 4xyz(x+y+z)$$
proof:since Use Schur inequality we have $$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)\ge 2((ab)^{3/2}+(bc)^{3/2}+(ac)^{3/2})$$
take $a'^2=a^3,b'^2=b^3,c'^2=c^3$ and Assmue that $a'b'c'=1$so we have
$$a'^2+b'^2+c'^2+3\ge 2(a'b'+b'c'+c'a'),$$
so
$$(a'+b'+c')^2+3\ge 4(a'b'+b'c'+c'a')$$
Let $a'=xy,b'=yz,c'=zx$ by done.
let $a+b+c=p,ab+bc+ac=q,abc=r=1$,your inequality equal to
$$pq+6\ge 5p$$
Use Lemma we have
$$q^2+3\ge 4p$$
$$\Longleftrightarrow p\sqrt{4p-3}+6\ge 5p,p\ge 3$$
it is clear .Because
$$p^2(4p-3)-(5p-6)^2=4(p-3)^2(p-1)\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}$ by trigonometric function ?
The signs : + - - + - - + - - ... .
Thanks for helping.
| I. Solution
You wish to find the exact value of $x,y$ for,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-...}}}}}}\tag1$$
$$y=\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}\tag2$$
and specify that both have period $3$. So all you do is substitute at the correct point,
$$x=\sqrt{7+2\sqrt{7-2\sqrt{7-2x}}}\tag3$$
$$y=\sqrt{4+\sqrt{4+\sqrt{4-y}}}\tag4$$
One can easily form a non-radical equation from $(3)$ and $(4)$ by repeated squaring as H.R. has done in his answer. Since $2^3 = 8$, then $x,y$ are at most roots of octics. Fortunately, both octics factor, and the correct $x,y$ are roots of just cubics, given by,
$$x^3 - x^2 - 9x + 1 = 0\tag5$$
$$y^3 + y^2 - 6y - 7 = 0\tag6$$
Using the positive case of the square roots of $(1),(2)$, then the correct values are,
$$x = 4\cos \tfrac{2\pi}{7} +1 =3.4939\dots$$
$$y = 2\left(\cos{\tfrac{4\pi}{19}}+\cos{\tfrac{6\pi}{19}}+\cos{\tfrac{10\pi}{19}}\right) = 2.50701\dots$$
II. Some Nice Things
The three roots of $(5)$ are,
$$ 4\cos \tfrac{2\pi}{7} +1,\quad 4\cos \tfrac{4\pi}{7} +1,\quad 4\cos \tfrac{8\pi}{7} +1$$
Ramanujan found the beautiful relation,
$$\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+
\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+
\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}=
\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$$
However, we can generalize that. Using the three roots of $(6)$, then,
$$\sqrt[3]{1+\cos\bigl(\tfrac{4\pi}{19}\bigr)+\cos\bigl(\tfrac{6\pi}{19}\bigr)+\cos\bigl(\tfrac{10\pi}{19}\bigr)}+\\
\sqrt[3]{1+\cos\bigl(\tfrac{2\pi}{19}\bigr)+\cos\bigl(\tfrac{14\pi}{19}\bigr)+\cos\bigl(\tfrac{16\pi}{19}\bigr)}+\\
\sqrt[3]{1+\cos\bigl(\tfrac{8\pi}{19}\bigr)+\cos\bigl(\tfrac{12\pi}{19}\bigr)+\cos\bigl(\tfrac{20\pi}{19}\bigr)}=\\ \sqrt[3]{\frac{-1+3\sqrt[3]{19}}{2}}=1.51867\dots$$
Note the third multiplier is the sum of the first two. One can find relations like these for all cubics. If interested, see this post.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$
This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?
| Since $1=abc$, we have
\begin{align}
\dfrac{1+ab}{1+a} + \dfrac{1+bc}{1+b} + \dfrac{1+ca}{1+c} & = \dfrac{abc+ab}{1+a} + \dfrac{abc+bc}{1+b} + \dfrac{abc+ca}{1+c}\\
& = ab \dfrac{1+c}{1+a} + bc \dfrac{1+a}{1+b} + ca \dfrac{1+b}{1+c}\\
& \geq 3\sqrt[3]{ ab \dfrac{\color{red}{1+c}}{\color{blue}{1+a}} \cdot bc \dfrac{\color{blue}{1+a}}{\color{orange}{1+b}} \cdot ca \dfrac{\color{orange}{1+b}}{\color{red}{1+c}}}\,\,\,\, \left(\because\text{AM-GM}\right)\\
& = 3\sqrt[3]{(abc)^2}\\
& = 3
\end{align}
| {
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Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me?
| $$2008^2=2000^2+2*2000*8+8^2=>4 (mod 10)$$
$2^m=6 (mod 10) $ (m = $2008$)
$$4+6=0 (mod 10)$$
$$ k=0 (mod 10)$$
$$ k^2=0 (mod 10)$$
So, it is obvious that last digit of $2^k$ is $2^4=6 (mod 10)$ and last digit of $k^2$ is $0$.
So the answer is $6+0=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality olympiad For all positive numbers $a,b,c$, prove that
$$\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq 3 \frac{(ab+bc+ac)}{a+b+c}$$
Note that both side are homogeneous of degree 1, so I think it is safe to assume $a+b+c=1$ but this does not go very far.
Any ideas/hint?
Thanks
| By Cauchy-Schwarz inequality:
$$\sum_{\text{cyc}}\dfrac{a^3}{b^2-bc+c^2}\left(\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)\right)\ge \left(a^2+b^2+c^2\right)^2$$
In fact, you can prove the following stronger inequality:
$$\frac{\left(a^2+b^2+c^2\right)^2}{\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)}\ge a+b+c\ge3\dfrac{ab+bc+ac}{a+b+c}$$
This holds:
$$\iff \left(a^2+b^2+c^2\right)^2\ge (a+b+c)\sum_{\text{cyc}}a\left(b^2-bc+c^2\right)$$
$$\iff a^4+b^4+c^4+abc(a+b+c)\ge ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ac\left(a^2+c^2\right)$$
The last step is true by Schur's inequality, where $t=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ is The number of real roots of the equation $1+a_1x+a_2x^2+.....+a_nx^n=0$ where $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ is
$(a)n$ if $n$ is even
$(b)1$ if $n$ is odd.
$(c)0$ for any natural number $n$
$(d)$none of these
As the degree of the polynomial is $n$,so it should have $n$ real roots.
But the answer in the book says number of real roots are $0$ for any natural number $n$.I dont understand why?May be $|x|<\frac{1}{3}$ and $|a_i|<2 \forall i=1,2,3....,n$ affects the number of real roots.But i dont know how?May be some Cauchy's theorem is applicable here.But i am not sure which theorem.Please help me.Thanks.
| Clarifying something the original poster said: When the degree of the polynomial is $n$, there are $n$ complex roots (counting multiplicity). Hence there are at most $n$ real roots.
Motivation: Noting how the range of $x$ and $a_i$ are restricted, we want to show that all the terms of $a_ix^i$ combined cannot "overpower" the $1$ in order to sum to $0$. We want a bound to prove that $1+a_1x+a_2x^2+.....+a_nx^n>0$, or equivalently, $a_1x+a_2x^2+.....+a_nx^n>-1$.
By triangle inequality,
$$|1+a_1x+a_2x^2+\dots+a_nx^n|+|a_1x|+|a_2x^2|+\dots+|a_nx^n|\\
\geq1+a_1x+a_2x^2+\dots+a_nx^n+(-a_1x)+(-a_2x^2)+(-a_3x^3)+\dots+(-a_nx^n)\\
=1$$
Let's try to substitute $x$ into the equation.
$$\begin{align*}
|1+a_1x+a_2x^2+\dots+a_nx^n|&\geq1-|a_1x|-|a_2x^2|-\dots-|a_nx^n|\\
&=1-|a_1||x|-|a_2||x|^2-\dots-|a_n||x|^n\\
&>1-2\times\frac{1}{3}-2\times\frac{1}{3^2}-2\times\frac{1}{3^3}-\dots-2\times\frac{1}{3^n}\\
&=\frac{1}{3^n}>0
\end{align*}$$
Hence, there are no real roots with magnitude $<\frac{1}{3}$.
The last equality is true as:
$$\begin{align*}
1&=3\times\frac{1}{3}\\
&=2\times\frac{1}{3}+3\times\frac{1}{3^2}\\
&=2\times\frac{1}{3}+2\times\frac{1}{3^2}+3\times\frac{1}{3^3}\\
&=\dots\\
&=2\times\frac{1}{3}+2\times\frac{1}{3^2}+2\times\frac{1}{3^3}+\dots+3\times\frac{1}{3^n}
\end{align*}$$
Rearranging gives the required expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$ Let $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$ where $\alpha\in(0,90^\circ)$,then find $\alpha$(in degree.)
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
$\frac{2\cos8^\circ}{\sin 16^\circ}+\frac{2\cos16^\circ}{\sin 32^\circ}+\frac{2\cos32^\circ}{\sin 64^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
$\frac{2^2\cos8^\circ\cos16^\circ}{\sin 32^\circ}+\frac{2^2\cos16^\circ\cos32^\circ}{\sin 64^\circ}+\frac{2^2\cos32^\circ\cos64^\circ}{\sin 128^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
In this way this series is getting complicated at each stage,is there any way to simplify it?Please help me.Thanks.
| HINT:
For $\sin A\ne0\iff A\ne m\pi$ where $m$ is any integer,
$$\cot A-\cot2A=\dfrac{\sin(2A-A)}{\sin2A\sin A}=\csc2A$$
Do you recognize the Telescoping Series?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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} |
Solve the functional equation $f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$ If $f: \Bbb R ^+ \to \Bbb R$ satisfies $$f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$$ for all $x, y \ge 0$ and if $f'(1)=e$, determine $f(x)$.
I'm a beginner. Can someone give me some hints for this sum?
I'm getting $f(1)=0$. Is that right?
| \begin{align*}
f(xy)&=e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right) \\
f(1) &= e^0 f(1) + e^0 f(1) \\
f(1) &= 0
\end{align*}
If $x>0$, set $y = \frac{x+h}{x}$, then $xy=x+h$.
\begin{align*}
f(xy)&= f(x+h) \\
&= e^{x\frac{x+h}{x}-x-\frac{x+h}{x}} \left( e^\frac{x+h}{x} f(x)+e^xf(\frac{x+h}{x}) \right) \\
&= e^{h-\frac{x+h}{x}} \left( e^\frac{x+h}{x} f(x)+e^xf(\frac{x+h}{x}) \right) \\
&=e^h f(x) + e^{x+h-\frac{x+h}{x}} f(1+\frac{h}{x})
\end{align*}
\begin{align*}
\frac{f(x+h)-f(x)}{h} &= \frac{(e^h - 1) f(x) + e^{x+h-\frac{x+h}{x}} f(1+\frac{h}{x})}{h} \\
&= \frac{e^h - 1}{h} f(x) + \frac{e^{x+h-\frac{x+h}{x}}}{x} \frac{f(1+\frac{h}{x}) - f(1)}{\frac{h}{x}}
\end{align*}
Since $f'(1) = e$, take limit on both sides as $h \to 0$.
\begin{align*}
f'(x) &= f(x) + \frac{e^{x-1}}{x} f'(1) \\
&= f(x) + \frac{e^x}{x} \\
&= f(x) + \frac{e^{x}}{x}
\end{align*}
Let $y = f(x)$.
\begin{align*}
y'-y &= \frac{e^{x}}{x} \\
(y e^{-x})' &= \frac{1}{x} \\
y e^{-x} &= \ln x + C \\
y &= e^{x} \ln x + C e^x
\end{align*}
Since $y = f(1) = 0, C = 0$. Hence $y = e^{x} \ln x \forall x > 0$.
Alternative method
I followed Sanchayan Dutta's method.
Why can we do a partial differentiation w.r.t. $x$?
We keep $y$ constant, and treat $x$ as the only independent variable.
Looking at the calucations Sanchayan Dutta's answer, I was puzzled and I wondered where the term $e^{xy}$ was. Why are our calculations so different? Why can we get the right answer despite different calculations? As a result, I decided to do it myself and to add this section.
\begin{equation*}
f(xy)=e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right)
\end{equation*}
Partial differentiation with respect to $x$ gives
\begin{align*}
y f'(xy) &= (y-1) e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right) + e^{xy-x-y} \left( e^yf'(x)+e^xf(y) \right) \\
&= e^{xy-x-y} \left( (y-1) e^yf(x)+(y-1) e^xf(y) + e^yf'(x)+e^xf(y) \right) \\
&= e^{\color{red}{x}(y-1)-y} \left( e^yf'(x) + (y-1) e^yf(x)+y e^xf(y) \right)
\end{align*}
Now, I know why we got different calculations but the same result: replace $x$ and $y$ by $1$ and $x$ respectively, so the missing $\color{red}{x}$ doesn't affect the result, and we still have
\begin{equation*}
f'(x)-f(x) = \frac{e^{x}}{x}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Which quantity is greater, $\frac{x^2}{y+\frac1y}$ or $\frac{y^2}{x+\frac1x}$? $x \gt y$, $ xy \neq 0$
A= $ x^2\over {y+{1\over y}}$
B= $ y^2\over {x+{1\over x}}$
Options:
1) Quantity A is greater.
2) Quantity B is greater.
3) The two quantities are equal.
4) The relationship cannot be determined from the given information.
By taking $x=2, y=1 $, I get $A\gt B$ , Thus options 2 and 3 are eliminated.
By taking$ x=2, y=-1$ , I get $B\gt A$ , Thus option 1 is eliminated.
So answer is option 4.
But I am not satisfied with this solution by taking particular values of $x$ and $y$.
Is there any other method to deal with this question?
What should be proper tag for this?
I think it should be comparision but that is not available in tag list.So please edit it.
| Suppose I compute the difference $A-B$ and see if it is always of one sign.
$$\begin{align}A - B &=\frac{x^2}{y+1/y} - \frac{y^2}{x+1/x}\\
&=\frac{yx^2}{y^2+1} - \frac{xy^2}{x^2+1}\\
&=\frac{xy\left(x(x^2+1)-y(y^2+1)\right)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)(x^2+xy+y^2+1)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)\left((x+y/2)^2+3y^2/4+1\right)}{(x^2+1)(y^2+1)}\end{align}$$
Every term in the last expression is positive but the term $xy.$ Therefore $$A > B \text{ if and only if } xy > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| Induction on k.
Base Case: 1=1
Induction Hypothesis: $(1+2+{...}+k)^2$=$1^3 + {...} + k^3$
Based on the hypothesis:
$(1+2+{...}+k+(k+1))^2$
=$(1+2+{...}+k)^2$+2$(1+2+{...}+k)(k+1)$+$(k+1)^2$
=$1^3 + {...} + k^3$+$k(k+1)^2$+$(k+1)^2$
=$1^3 + {...} + k^3 + (k+1)^3$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Using the sum and product formulae we have,
$q=tanAtanB, $ $-p=tanA+tanB$
And,
$tan(A+B)=\frac{tanA+tanB}{1-tanAtanB} \Rightarrow tan(A+B)=\frac{-p}{1-q}$
Now,
$\frac{p^{2}-p^{2}}{1-q}=0 \Rightarrow (\frac{p}{1-q})^{2}(1-q)+p(\frac{-p}{1-q})=0 \Rightarrow [tan(A+B)]^{2}(1-q)+p[tan(A+B)]=0 \Rightarrow [\frac{sin(A+B)}{cos(A+B)}]^{2}(1-q)+p[\frac{sin(A+B)}{cos(A+B)}]=0 \Rightarrow sin^{2}(A+B)+p(sin(A+B)cos(A+B))-q(sin^{2}(A+B))=0 \Rightarrow sin^{2}(A+B)+p[sin(A+B)cos(A+B)]+qcos^{2}(A+B))=q$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How do I find the roots of this polynomial of degree $4$? I am studying for finals and in the review packet is shown this problem:
$$P(x)=2x^4 + 5x^3 + 5x^2 + 20x - 12$$
I don't know what to do, I have already tried looking in the textbook and Khan academy.
| Simple factorisation solves your problem.
$$P(x)=2x^4+5x^3+5x^2+20x-12$$ $$=2x^4+5x^3-3x^2+8x^2+20x-12$$
$$=x^2(2x^2+5x-3)+4(2x^2+5x-3)$$ $$=(x^2+4)(2x^2+5x-3)$$ $$=(x^2+4)(2x-1)(x+3)$$
Hence the $4$ roots are $x=\frac{1}{2}$,$x=-3$,$x=2i$ and $x=-2i$.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Is there a formula for computing the number of arrangements from a set of N elements groupped in K groups? Eg: N=3, K=2
There will be two groups in each solution. We need to calculate the number of such possible solutions. Consider the set S={1,2,3}. The possible solutions are:
{1} {2,3}
{1} {3,2}
{2} {1,3}
{2} {3,1}
{3} {1,2}
{3} {2,1}
{1,2} {3}
{2,1} {3}
{1,3} {2}
{3,1} {2}
{2,3} {1}
{3,2} {1}
The result for this example is: 12.
Eg: N=4, K=3
{1} {2} {3,4}
{1} {2} {4,3}
{1} {2,3} {4}
{1} {3,2} {4}
....
Can we generalize this formula?
| We assume that the example for $N=3$, $K=2$ is correct, meaning that the internal order of the elements in individual groups matters.
There are $N!$ ways of lining up our $N$ objects. That produces $N-1$ interobject "gaps." We choose $K-1$ of them to put a separator into.
That can be done in $\binom{N-1}{K-1}$ ways, for a total of $N!\binom{N-1}{K-1}$.
Remark: If internal order within groups does not matter, the solution is quite different, and uses Stirling numbers of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Simultaneous Equations (Stuck on the algebra)
Question: Solve the following simultaneous equations for real values of x and y
$$
\left\{
\begin{array}{l}
9^{2x+y} - 9^x \times 3^y = 6 \\
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
\end{array}
\right.
$$
What I have attempted; for the first equation
$$
9^{2x+y} - 9^x \times 3^y = 6 \\
9^{2x+y} - 3^{2x} \times 3^y = 6 \\
9^{2x+y} - 3^{2x+y} - 6 = 0
$$
Let $z = 3^{2x+y}$. Then
$$
z^2 - z - 6 = 0
\iff (z-3)(z+2) = 0
\iff z = 3, z \ne -2
$$
where
$$
3^{2x+y} = 3 \iff y = 1-2x.
$$
Now using the second equation I get
$$
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
$$
Substituting $y = 1-2x$
$$
\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3.
$$
Now this is the part I am stuck on , how do I solve for $x$ algebraically?
| Writing the last equation
$$ \log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3 $$
as
$$ \log_{x+1}[(4-2x)(5-x)] = 3 $$
you can get both term as exponents of $x+1$, so that
$$ (4-2x)(5-x) = (x+1)^3 $$
This equation can be easily solved, leading to $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The value of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$ The value of series
$\sum_{n=1}^{\infty}\frac{n}{2^n}$
I try to write some terms,but of no use. Is there any general method to approach such questions.
Thanks
| Notice that
\begin{align*}
\sum_{n=1}^\infty \frac{n}{2^n}
&= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^{n-1}}
= \frac{1}{2} \sum_{n=0}^\infty \frac{n+1}{2^n}
= \frac{1}{2} \left( \sum_{n=0}^\infty \frac{n}{2^n}
+ \sum_{n=0}^\infty \frac{1}{2^n} \right) \\
&= \frac{1}{2} \sum_{n=1}^\infty \frac{n}{2^n} + 1,
\end{align*}
thus $\left( \sum_{n=1}^\infty \frac{n}{2^n} \right) = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How do I show $f(x+2)-f(x)>2 \forall x$? For the function
$f(x)=x\cos{\frac{1}{x}}$, $x\geq1$,
How do I show that $f(x+2)-f(x)>2 \forall x$?
| $\cos{\frac{1}{x}}=1-\frac{1}{x^22!}+\frac{1}{x^44!}-....$
$x\cos{\frac{1}{x}}=x-\frac{1}{x2!}+\frac{1}{x^34!}-....$
So $(x+2)\cos{\frac{1}{x+2}}=(x+2)-\frac{1}{(x+2)2!}+\frac{1}{(x+2)^34!}-....$
So $f(x+2)-f(x)=2+\{\frac{1}{2x}-\frac{1}{2(x+2)}\}+...$
$=2+\frac{1}{x(x+2)}+...>2$
NOTE: As the number of terms in the summation increases the magnitude of the terms decreases in particular less than the term before and since $x>0$ the terms are always positive .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate the integral $\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx,\alpha\in(0,\pi)$ For the equation $x^2-2x\cos\alpha+1=0$ solutions are
$$x_1=\cos\alpha-\sqrt{\cos^2\alpha-1},x_2=\cos\alpha+\sqrt{\cos^2\alpha-1}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\int_{-1}^1 \frac{1}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}\mathrm dx$$
$\frac{1}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}=\frac{A}{(x-\cos\alpha+\sqrt{\cos^2\alpha-1})}+\frac{B}{(x-\cos\alpha-\sqrt{\cos^2\alpha-1})}\Rightarrow$
$$A=\frac{-1}{2\sqrt{\cos^2\alpha-1}},B=\frac{1}{2\sqrt{\cos^2\alpha-1}}\Rightarrow$$
$$\int_{-1}^1 \frac{1}{x^2-2x\cos\alpha+1}\mathrm dx=\frac{-1}{2\sqrt{\cos^2\alpha-1}}\left(\int_{-1}^1 \frac{1}{x-\cos\alpha+\sqrt{\cos^2\alpha-1}}\mathrm dx - \int_{-1}^1 \frac{1}{x-\cos\alpha-\sqrt{\cos^2\alpha-1}}\mathrm dx\right)$$
How to evaluate these partial integrals?
| If we let $t = \cos \alpha$, we have $$\frac{1}{x^2 - 2t x + 1} = \frac{1}{(x-t)^2 + (1-t^2)} = \frac{1}{1-t^2} \cdot \frac{1}{\frac{(x-t)^2}{1-t^2} + 1}.$$ This suggests using the substitution $$u = \frac{x-t}{\sqrt{1-t^2}}, \quad du = \frac{1}{\sqrt{1-t^2}} \, dx.$$ This results in the indefinite integral $$\frac{1}{\sqrt{1-t^2}} \int \frac{1}{u^2 + 1} \, du = \csc \alpha \tan^{-1} u + C = \csc \alpha \tan^{-1} \frac{x-\cos \alpha}{\sin \alpha} + C. $$ Now evaluating this integral at the endpoints gives $$\int_{x=-1}^1 \frac{dx}{x^2 - 2x\cos \alpha + 1} = \csc \alpha \left(\tan^{-1} \frac{1 - \cos \alpha}{\sin \alpha} + \tan^{-1} \frac{1+\cos \alpha}{\sin \alpha} \right),$$ and because $$\frac{1 - \cos \alpha}{\sin \alpha} \cdot \frac{1+\cos \alpha}{\sin \alpha} = 1,$$ it follows that the term in parentheses is $\pi/2$ for $\alpha \in (0,\pi)$; hence the answer is $$\frac{\pi}{2 \sin \alpha}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
| Steve explained where you made a mistake in your calculations. And Patrick explained how you can save computations by judiciously choosing the rows/ columns you expand along. Just for fun, I'll explain a different way of evaluating the determinant. I'm just going to use the relationship between the elementary row/ column operations and the determinant.
Here are those relationships:
*
*Swapping two rows/ columns of a matrix will give a factor of $-1$ to the determinant. Let $a_k$ be the $k$th row (or column) of the matrix $A$. Then $$\det(A) = \det(a_1,\dots, a_i, \dots, a_j, \dots, a_n) = -\det(a_1,\dots, a_j, \dots, a_i, \dots, a_n)$$
*A common factor can be "pulled out" of a row/ column. $$\det(a_1,\dots,ka_i,\dots,a_n) = k\det(a_1,\dots,a_i,\dots,a_n)$$
*Adding a scalar muliple of one row/ column to another will not change the determinant at all. $$\det(a_1,\dots, a_i, \dots, a_j, \dots, a_n) = \det(a_1,\dots, a_i, \dots, a_j+ka_i, \dots, a_n)$$
Let's use these properties of the determinant to calculate the determinant of your matrix:
$$\begin{align}\begin{vmatrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{vmatrix} &= \enspace\ \frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & 6 & 0 & -8 \\ 0 & -15 & 2 & 5 \\ 0 & -1 & 0 & 2\end{vmatrix} &{\begin{pmatrix}R_2 \to 2R_2 \\ R_3 \to R_3+R_1 \\ R_4 \to R_4-R_2\end{pmatrix}} \\ &= -\frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & -1 & 0 & 2 \\ 0 & -15 & 2 & 5 \\ 0 & 6 & 0 & -8\end{vmatrix} & \begin{pmatrix}R_2 \leftrightarrow R_4\end{pmatrix} \\ &= -\frac 12\begin{vmatrix} 5 & -7 & 2 & 2 \\ 0 & -1 & 0 & 2 \\ 0 & 0 & 2 & -25 \\ 0 & 0 & 0 & 4\end{vmatrix} & \begin{pmatrix}R_3\to R_3-15R_2 \\ R_4\to R_4+6R_2\end{pmatrix} \\ &\stackrel{(*)}= -\frac 12(5)(-1)(2)(4) \\ &= \enspace\ 20\end{align}$$
where $(*)$ is due to the fact that the determinant of a triangular matrix is the product of the diagonal elements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\lt 0$ and substituing $t=x^2,$ $2x$ appears.That's a problem.
| Hint :
$$x^4+x^2+1=(x^2+1)^2−x^2=(x^2−x+1)(x^2+x+1)$$
So $$∫ \frac {1}{x^4+x^2+1}dx=\int\frac {1}{(x^2−x+1)(x^2+x+1)}dx$$
$$=\int\frac {1-x}{2(x^2−x+1)}+\frac {1+x}{2(x^2+x+1)}dx$$
$$=-\int\frac {x-1}{2(x^2−x+1)}dx+\int\frac {x+1}{2(x^2+x+1)}dx$$
$$=-\int\frac {2x-2}{4(x^2−x+1)}dx+\int\frac {2x+2}{4(x^2+x+1)}dx$$
$$=-\int\frac {2x-1}{4(x^2−x+1)}dx+\int\frac {1}{4(x^2−x+1)}dx+\int\frac {2x+1}{4(x^2+x+1)}dx+\int\frac {1}{4(x^2+x+1)}dx$$
$$=\int\frac {2x+1}{4(x^2+x+1)}dx-\int\frac {2x-1}{4(x^2−x+1)}dx+\int\frac {1}{4(x^2−x+1)}dx+\int\frac {1}{4(x^2+x+1)}dx$$
$$=\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\int\frac {1}{4(x^2−x+1)}dx+\int\frac {2x+1}{4(x^2+x+1)}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Which of the following matrices has the same determinant as matrix $G$ Which of the following matrices has the same determinant as matrix:
$G$ = $ \begin{bmatrix}
1 & 2 & 3 \\
-2 & -3 & 2 \\
1 & 2 & 1
\end{bmatrix}$
I could just calculate each individual determinant; however I am preparing for the CSET exam with no calculators and need to be efficient with my time.
Is there another way to identify the solution?
A. $ \begin{bmatrix}
-2 & -3 & 2 \\
1 & 2 & 3\\
1 & 2 & 1
\end{bmatrix}$
B. $ \begin{bmatrix}
1 & 2 & 3 \\
-2 & -3 & 2 \\
2 & 4 & 4
\end{bmatrix}$
C. $ \begin{bmatrix}
2 & 4 & 6 \\
-2 & -3 & 2 \\
1 & 2 & 1
\end{bmatrix}$
D. $ \begin{bmatrix}
2 & 1 & 3 \\
-3 & -2 & 2 \\
2 & 1 & 1
\end{bmatrix}$
| Note that the matrix $B$ is obtained from $G$ by adding $\DeclareMathOperator{Row}{Row}\Row_1(G)$ to $\Row_3(G)$. Elementary row operations of this form leave the determinant unchanged. Thus $\det B=\det G$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding product of odd/even integers This is a question that came to mind and I was trying multiple ways from using combinatorial methods, graphical approaches, etc. The question I have is, how can I find the product of all the odd or even numbers less than some positive integer $n$.
For example:
If $n=6$, then the product of all odd numbers less than 6 would be $15$ and all even numbers would be $8$. But this gets impossible after you reach larger integers.
| Case for Odd Product:
Since you want the product less than $n$, you are looking for the product of,
$$ 1\cdot 3\cdot 5\cdot 7\cdot 9\cdot \ldots \cdot (n-1) $$
Now, notice that you don't want the even numbers, so just divide by them,
$$\frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {2\cdot 4\cdot 6\cdot 8\cdot 10\cdot \ldots \cdot n} = \frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {(2\cdot 1) \cdot (2\cdot 2) \cdot (2\cdot3) \cdot (2\cdot 4) \cdot (2\cdot 5)\cdot \ldots \cdot \left(2\cdot \frac n2\right)} = \displaystyle\frac{n!}{2^{\frac n2}\left(\frac n2\right)!} $$
Now take a similar approach to find even products.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$
Given $$R=(8+3\sqrt{7})^{20}, $$ if $\lfloor R \rfloor$ is Greatest integer less than or equal to $R$, then which of the following option(s) is/are true?
*
*$\lfloor R \rfloor$ is an even number
*$\lfloor R \rfloor$ is an odd number
*$R-\lfloor R \rfloor=1-\frac{1}{R}$
*$R(R-\lfloor R \rfloor-1)=-1$
My try: I wrote $R$ as $$R=8^{20}\left(1+\sqrt{\frac{63}{64}}\right)^{20} \approx8^{20}\left(1+\sqrt{0.98}\right)^{20} \approx8^{20}\left(1.989\right)^{20} .$$
Now, $8^{20}\left(1.989\right)^{20}$ is slightly less than $8^{20} \times 2^{20}=2^{80}$,
$$\lfloor 2^{80}\rfloor=2^{80}$$
hence
$$\lfloor R \rfloor=2^{80}-1,$$
so option $2$ is correct.
How does one figure out whether options $3$ and $4$ are correct or wrong?
| Building on Anomaly's answer, let $a_n = (8 + 3\sqrt{7})^n + (8 - 3 \sqrt{7})^n$. Prove that $a_n$ can be defined by induction by $a_0 = 2$, $a_1 = 16$, $a_n = 16a_{n-1} - a_{n-2}$. This will help in obtaining information about $a_{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving $x = y$ or $x = -y$ when $x^n = y^n$ and $n$ is even I'm currently going through Spivak's calculus, and after a lot of effort, i still can't seem to be able to figure this one out.
The problem states that you need to prove that $x = y$ or $x = -y$ if $x^n = y^n$
I tried to use the formula derived earlier for $x^n - y^n$ but that leaves either $(x-y) = 0$ or $(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$ and i'm not sure how to proceed from there.
| Let $n=2p$. For convenience let us denote $y=a$. From the algebraic identities
\begin{eqnarray}
x^{2p}-a^{2p} &=&(x-a)\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k}, \tag{1} \\
\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k} &=&(x+a)\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k},\tag{2}
\end{eqnarray}
we conclude that
\begin{equation}
x^{2p}-a^{2p}=(x-a)(x+a)\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k}. \tag{3}
\end{equation}
Since for $a\neq 0$ the polynomial $\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k}$ on the right-hand side of (3) has no real
roots, it follows that the equation $x^{2p}-y^{2p}=0$ is equivalent to $(x-y)(x+y)=0$, thus proving that if $x^{n}=y^{n}$ and $ n $ is even, then $x=y$ or
$x=-y$.
The identities $(1)$ and $(2)$ can be justified by applying Ruffini's Rule twice: for identity $(1)$
$$
\begin{array}{c|cccccccc}
& 1 & 0 & 0 & \ldots & 0 & 0 & & -a^{2p} \\
a & & a & a^2 & \ldots & a^{2p-2} & a^{2p-1} & & a^{2p} \\
\hline
& 1 & a & a^{2} & \ldots & a^{2p-2} & a^{2p-1} & | & 0
\end{array}
$$
\begin{equation*}
x^{2p}-a^{2p}=(x-a)(x^{2p-1}+ax^{2p-2}+a^{2}x^{2p-3}+\cdots
+a^{2p-2}x+a^{2p-1}),
\end{equation*}
and for identity $(2)$
$$
\begin{array}{c|cccccccc}
& 1 & a & a^{2} & a^{3} & \ldots & a^{2p-2} & & a^{2p-1} \\
-a & & -a & 0 & -a^{3} & \ldots & 0 & & -a^{2p-1} \\
\hline
& 1 & 0 & a^{2} & 0 & \ldots & a^{2p-2} & | & 0
\end{array}
$$
$x^{2p-1}+ax^{2p-2}+\cdots +a^{2p-2}x+a^{2p-1}$
$$=(x+a)(x^{2p-2}+a^{2}x^{2p-4}+a^{4}x^{2p-6}+\cdots +a^{2p-4}x^{2}+a^{2p-2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$ Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$
This exercise is from the chapter of my book regarding the Cauchy-Schwarz inequality but I don't know if I've applied it correctly .
The problem asks for the maximum value of the expression so it's clear that it must be on the lesser side of the Cauchy Inequality ,i.e.
\begin{array}
RR&\ge \cfrac{(3x+4y)^2}{x^2+y^2} \\
R(x^2+y^2) &\ge (3x+4y)^2 \\
\left(\left(\sqrt{R}\right)^2 \right) (x^2+y^2) &\ge (3x+4y)^2 \\
\end{array}
Now this would be true iff
$$ \left(\left(\sqrt{R}\right)^2 +(4)^2\right) (x^2+y^2) \ge (3x+4y)^2 $$ where $R=9$ ,which would imply that this is the maximum.
However I am not really sure if I am correct.
| I think ,you can consider as taking $x=\frac{4y}{3}$;
$\left(\left(\sqrt{R}\right)^2 +(\frac{4}{3})^2\right) (x^2+y^2) \ge (3x+4y)^2$
which R=25 for maximum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$ To solve the equation
$$x^2-5=0$$
There exitsts a iteration method
$$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$
I know that it is cubic convergence but I don't know how to prove it.
I have tried the following
$$
\begin{align}
e_{k+1}&=x_{n+1}-\sqrt{5} \\ &=\frac{x_n(x_n^2+15)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{\frac{1}{3}x_n(3x_n^2+5+40)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{1}{3}x_n(1+\frac{40}{3x_n^2+5})-\sqrt{5} \\ &=\frac{\frac{40}{3}x_n}{3x_n^2+5}+\frac{1}{3}x_n-\sqrt{5}
\end{align}
$$
I'm trying to extract $e_n=x_n-\sqrt{5}$ from $e_{n+1}$ in an effort of proving $lim \frac{e_{n+1}}{e_n^3}\rightarrow C$ . But it seems that I'm not on the right way.
Do you have any idea? Any discussion is appreciated. Thanks in advance.
| In general, Halley's method for the square root reads as
$$
x_{n+1}=x_n·\frac{x_n^2+3a}{3x_n^2+a}
$$
The difference to the square root develops then as
$$
x_{n+1}-\sqrt{a}=\frac{x_n^3+3x_n\sqrt a^2 - 3x_n^2\sqrt a-\sqrt a^3}{3x_n^2+a}
=\frac{(x_n-\sqrt a)^3}{3x_n^2+a}
$$
which gives you the third order convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } = \frac{65}{8}; a \neq 0
$$
| First question:
Let $y=(x-2)(x-3)$. So it's equivalent to
$$\frac{y(y-2)}{(y+10x)(y+10x-2)}=1$$
$$y^2-2y=y^2+20xy+100x^2-2y-20x=0$$
$$20x=100x^2+20xy$$
For $x\neq 0$,
$$1=5x+(x-2)(x-3)$$
$$x^2=-5$$
$$x=\pm\sqrt{5} i$$
Otherwise $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself? If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself?
For example the square of $2$ is $2^2=2 \cdot 2=4 $ .
But square root of $2$ is not $\frac{2}{2}=1$ .
| Assuming $x > 0$
Algebraically:
$$
\begin{matrix}
x \cdot x = x^2 & \rightarrow & \sqrt{x \cdot x} = \sqrt{x^2} & \rightarrow & \sqrt{x} \cdot \sqrt{x} = x \\
\downarrow & & & & \downarrow \\
x^2 \div x = x & \rightarrow & \sqrt{x^2 \div x} = \sqrt{x} & \rightarrow & x \div \sqrt{x} = \sqrt{x}
\end{matrix}
$$
Visually (expanding on dkeck's answer):
Graphically:
Another way of wording the question is:
"If the curve for $y=x^2$ intersects with the line for $y = x \cdot z$ at $x=z$, then shouldn't the curve for $y=\sqrt{x}$ intersect with the line for $y=x \div z$ at $x=z$?"
See for yourself, where $z=2$:
$y = \sqrt{x}$ intersects with $y = x \div z$ at $x = z^2$, not at $x = z$. $y = \sqrt{x}$ intersects with $y = x \div \sqrt{z}$ at $x = z$.
The question comes from the very subtle logical error of confusing the function $\lambda x \space \space x \cdot x$ with the function $\lambda a \space \space a \cdot x$. The above graph calls attention to the difference between these functions.
If squaring a number meant multiplying that number by $z$, and square root were defined as the inverse of square, then yes, taking the square root of a number would mean dividing it by $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 25,
"answer_id": 16
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
| $$\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
$$=e^2\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{x^2-3\sqrt2 x+4}\cdot\dfrac1{\left(\lim _{x\to \sqrt{2}}\dfrac{\ln(1+x^2-3\sqrt2 x+4)}{x^2-3\sqrt2 x+4}\right)^2}$$
Now,
$$\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{(x^2-3\sqrt2 x+4)^2}=\lim _{x\to \sqrt{2}}\dfrac{e^{x^2-2}-1-(x^2-2)}{(x^2-2)^2}\cdot\lim _{x\to \sqrt2}\dfrac{(x+\sqrt2)^2}{(x-2\sqrt2)^2}$$
$$=(-2)^2\lim_{y\to0}\dfrac{e^y-1-y}{y^2}$$
Finally use How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?
to get the limit to be $$e^2\cdot4\cdot\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express a general vector $(x, y, z)$ in $R^3$ as a linear combination of $(1, 2, 1),(1, 0, -1),(1, -2, 1)$ $a + b + c = x$
$2a + 0 - 2c = y$
$a – b + c = z$
Then, $(a + b + c, 2a - 2c, a - b + c) = (x, y, z)$, but this looks horrendous unless they mean I should simplify this somehow or something entirely different. Am I not understanding the question correctly?
| hint: $a+c=\dfrac{x+z}{2}, a-c = \dfrac{y}{2}\Rightarrow 2a = \dfrac{x+y+z}{2}\Rightarrow a = \dfrac{x+y+z}{4}$, and similarly $b = \dfrac{x-z}{2}$. Can you solve for $c$ ?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
| It's easiest to use Taylor expansions about $x = 0$. First the denominator to leading order, so we can tell how many terms are needed:
$$
\sin\left(\frac{x}{2}\right)^4 = \frac{x^4}{16} + O(x^6)
$$
So we need expansions to this order also for the numerator:
$$
\sin(\ln(x+1)) = x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{12}+O\left(x^6\right)
$$
and
$$
\ln(sin(x) + 1) = x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{12}+\frac{x^5}{24}+O\left(x^6\right)
$$
You see that the two terms in the numerator agree through order $x^3$. Hence these terms all cancel. Then the leading term in the numerator is $\frac{x^4}{12}$. Therefore, the limit equals $\frac{1/12}{1/16} = \frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$ Let $x$ and $y$ be two positive numbers:
Prove that $$ \left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right).$$
I answered this one by squaring the two expressions. And therefore finding the difference after squaring the formulas. I don't even know if it's right but I wanted to find another way to answer this question?
| An alternative approach:
$\left( \frac {x+y}{x^2+y^2}\right) \leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right) \Longleftrightarrow 4x^2y + 4xy^2 \leq (x^2 + y^2)(2y + 2x) \Longleftrightarrow x^2y + y^2x \leq x^3 + y^3$
We assume WLOG $x \leq y \Longrightarrow \exists c\in \mathbb{R}_{≥0}$ such that $x+c = y$
Hence the rightmost inequality can be simplified to $0 \leq (x^3 + (x+c)^3) -(x^2(x+c) + x(x+c)^2) = c^2(c+2x)$ which is clear.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The sequence $\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$ is bounded above . A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$
Show that the sequence is convergent.
Because this sequence is monotonic, proving it is bounded above will be sufficient to prove that it is convergent.
So how to show that this sequence is bounded above?
| The following approach will also give a reasonable bound:
$$1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+\frac{1}{8^4}+\frac{1}{9^4}+\ldots+\frac{1}{n^4} \\ < 1+\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{8^4}+\frac{1}{8^4}+\ldots+\frac{1}{n^4} \\ < 1+\frac{2}{2^4}+\frac{4}{4^4}+\frac{8}{8^4}+\frac{16}{16^4}+\ldots = 1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}+\frac{1}{16^3}+\ldots \\ = 1+\frac{1}{8}+\frac{1}{8^2}+\frac{1}{8^3}+\frac{1}{8^4}+\ldots < \sum_{k=1}^{\infty} \frac{1}{8^k} = \frac87$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Positive correlation with the sequence $\sqrt{ij}/2-\min(i,j)$ Is there a sequence of positive real numbers $x_1,\ldots,x_n$ for which
$$
\sum_{1\leq i,j\leq n}\left[\frac{\sqrt{ij}}{2}-\min(i,j)\right]x_ix_j> 0?
$$
| Alternative solution:
There exists such $n$ and a sequence of positive real numbers $x_1, x_2, \cdots, x_n$.
Let $A$ denote the matrix whose $(i,j)$-th entry is $\frac{\sqrt{ij}}{2}$.
Then, $A = uu^\mathsf{T}$ where
$u = \frac{1}{\sqrt{2}}[\sqrt{1}, \sqrt{2}, \cdots, \sqrt{n}]^\mathsf{T}$.
Let $B$ denote the matrix whose $(i,j)$-th entry is $\min(i, j)$.
Note that $B^{-1}$ is a symmetric tridiagonal matrix of the form
\begin{align}
B^{-1} = \begin{pmatrix} 2 & -1 & 0 & \ldots & 0 & 0 \\ -1 & 2 & -1 & \ldots & 0 & 0 \\ 0 & -1 & 2 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 2 & -1 \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{pmatrix}.
\end{align}
(Note: The diagonal entries are $2, 2, \cdots, 2, 1$. The subdiagonal and superdiagonal entries are all $-1$. )
Let $v = B^{-1}u$.
Fact 1: $v_i > 0$ for $i=1, 2, \cdots, n$.
Fact 2: $u^\mathsf{T}B^{-1}u = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$.
Fact 3: $v^\mathsf{T}(A - B)v = (u^\mathsf{T}B^{-1}u)^2 - u^\mathsf{T}B^{-1}u$.
The proofs are easy and thus omitted.
Now, from Facts 1, 2, 3, if $u^\mathsf{T}B^{-1}u > 1$ or equivalently $\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$, then $v^\mathsf{T}(A - B)v > 0$.
It is easy to prove that there exists $n$ such that $\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$.
Indeed, by using
$\sqrt{k(k+1)} \le k + \frac{1}{2} - \frac{1}{16k}$ for $k \ge 1$, we have
$$\sum_{k=1}^{n-1} \sqrt{k(k+1)} \le \frac{n^2}{2} - \frac{1}{2} - \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}$$
which results in
$$\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} \ge \frac{1}{2} + \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}.$$
The desired result follows.
For such $n$ and $v$, we have
$$\sum_{1\le i, j\le n} \left[\frac{\sqrt{ij}}{2} - \min(i, j)\right] v_iv_j = v^\mathsf{T}(A - B)v > 0.$$
Remark: Denote $F(n) = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$. By Maple, $F(55) = 0.99974443 < 1$ and $F(56) = 1.00199623$.
| {
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Find the matrix of linear transformation with respect to different basis Let $f:P_2[x]\rightarrow M_{2\times 2}(\mathbb{R})$ is a linear transformation defined as $f(a+bx+cx^2)= \begin{bmatrix}
b+c & a \\
b & c \\
\end{bmatrix}$
Find the matrix of $f$ with respect to basis $ ({\begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}})$ and $(1,x,x^2)$. Find a basis and dimension of $Im (f)$ and $Ker(f)$.
In a standard basis for $P_2[x]$, the mapping is:
$$f \begin{bmatrix}
a \\
b \\
c \\
\end{bmatrix}= \begin{bmatrix}
0 + b + c \\
a + 0 + 0 \\
0 + b + 0 \\
0 + 0 + c
\end{bmatrix}\Rightarrow f=\begin{bmatrix}
0 & b & c \\
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{bmatrix}$$
Question: Is the matrix of $f$ the same for both basis $ ({\begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}})$ and $(1,x,x^2)$? If not, what is the method for finding the matrix of $f$ in both cases?
| Your question is fuzzy: $f$ has one matrix with respect to the two given bases. So there is only one case.
Let's call the matrix we want to find $M_f$ and write the coefficients of the two given vectors down:
The vector $a + bx + cx^2$ with respect to the basis $(1,x,x^2)$ is $(a,b,c)$.
The vector $\begin{bmatrix}
b+c & a \\
b & c \\
\end{bmatrix}$ with respect to the basis $({\begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
0 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix},\begin{bmatrix}
1 & 1 \\
1 & 1 \\
\end{bmatrix}})$ is $(b+c-a,a-b,b-c,c)$.
So, $M_f (a,b,c)^T = (b+c-a,a-b,b-c,c)^T$.
If you denote the $3$ columns of $M_f$ by $c_1, c_2, c_3$ then you want $c_i$ to be such that
$$ a c_1 + b c_2 + c c_3 = (b+c-a,a-b,b-c,c)$$
How to find the $c_i$?
Note that a linear map is uniquely determined by what it maps basis vectors to. In this case, we have the basis vectors
$$ 1 = (1,0,0)^T = b_1, x = (0,1,0)^T=b_2, x^2 = (0,0,1)^T = b_3$$
and
$$ M_f(b_1) = (-1,1,0,0), M_f(b_2) = (1,-1,1,0), M_f(b_3)=(1,0,-1,1)$$
Can you read the columns of $M_f$ off of this?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is giving the $\frac{0}{0}$ form after plugging in $3$.
| Set $\sqrt{x+6}-3=y\implies x=(y+3)^2-6=y^2+6y+3$
$$\implies\lim_{x\to3}\dfrac{\sqrt{x+6}-3}{x^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y+3)^2-9}$$
$$=\lim_{y\to0}\dfrac y{(y^2+6y)^2+6(y^2+6y)}$$
As $y\to0,y\ne0$
So, cancel out $y$ from N & D to get
$$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solving integral using trig substitution $\tan(x/2)=t$ I have problems with solving the following integral:
$$ \int{{\sin x - \cos x}\over {\sin x + \cos x}} \, dx$$
Could anybody please help me to find the solution and show me the method how it can be solved? I already tried to solve similar ones but I get always stuck when trying the technique with partial fraction decomposition.
Thanks very much in advance!
| \begin{align}
\tan \frac x 2 & = t \\[8pt]
\frac x 2 & = \arctan t \\[8pt]
x & = 2\arctan t \\[8pt]
\sin x & = \sin(2\arctan t) = \sin(2 \, \bullet) = 2 \sin(\bullet)\cos(\bullet) \\
& = 2\sin(\arctan t)\cos(\arctan t), \\[8pt]
\cos x & = \cos(2\arctan t) = \cos(2\,\bullet) = \cos^2(\bullet) - \sin^2(\bullet) \\
& = \cos^2(\arctan t) - \sin^2(\arctan t).
\end{align}
Now let us consider what $\sin(\arctan t)$ and $\cos(\arctan t)$ are.
$\tan = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac t 1$ so $\text{hypotenuse} = \sqrt{t^2+1^2}$, and so we have
$$
\sin(\arctan t) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac t {\sqrt{t^2+1}}.
$$
Similarly
$$
\cos(\arctan t) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac 1 {\sqrt{t^2+1}}.
$$
Consequently
$$
2\sin(\arctan t)\cos(\arctan t) = 2\cdot \frac 1 {\sqrt{t^2+1}} \cdot \frac t {\sqrt{t^2+1}} = \frac{2t}{1+t^2}.
$$
Similarly
$$
\cos^2(\arctan t) - \sin^2(\arctan t) = \frac 1 {1+t^2} - \frac t {1+t^2} = \frac{1-t^2}{1+t^2}.
$$
So now we have
$$
\left.
\begin{align}
\sin x & = \frac{2t}{1+t^2}, \\[10pt]
\cos x & = \frac{1-t^2}{1+t^2}.
\end{align}
\right\} \tag 1
$$
Finally, since $x = 2\arctan t$, we have
$$
dx = \frac{2\,dx}{1+t^2}. \tag 2
$$
Your ultimate answer comes from $(1)$ and $(2)$, followed by actually evaluating the resulting integral of a rational function.
| {
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Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$. Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$
$\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$.Its graph look like the graph of $\frac{1}{x^2+1}$ but peak at $(1,1)$ buti do not know how does the graph of $f(f(x))$ behave and hence i am not able to find the number of points of intersection of $f(f(x))$ and $\frac{1}{x^2-2x+2}$.
| Hint: If $f(x)$ is monotonic, then $f(f(x))$ is (possibly non-strictly) increasing, so…
Full solution: Clearly if $f$ is increasing, so is $f\circ f$. Furthermore, if $f$ is decreasing, then $x \leq y$ implies $f(x) \geq f(y)$, which in turn implies $f(f(x)) \leq f(f(y))$. Either way, $f \circ f$ is increasing.
Since $f$ is $1$ at $x=1$, it follows that $f \circ f$ must have minimum $1$ at $x=1$. But it is easy to see that $(x^2-2x+2)^{-1}$ has maximum $1$ at $x=1$, thus $N=1$.
In particular, note that you don't need the hypothesis that $f$ is differentiable.
| {
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How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is $$\frac{\sqrt{2}\sqrt{3}}{9}$$ which isn't equal to $$\frac{2\sqrt{2}-\sqrt{3}}{5}$$
What did I do wrong?
| $$(2\sqrt2-\sqrt3)(2\sqrt2+\sqrt3)=(2\sqrt2)^2-(\sqrt3)^2=?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$i^4 = (i^2)^2 = 1$$
$$ \frac{(1+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-i+1+1+i)}{(-i+1+1+i)}$$
$$ \frac{-4-4i + 2}{2}$$
$$ \frac{-2-4i}{2} = -1-2i$$
Where is the error?
| Following your line of algebraic manipulation,
\begin{align}
\frac{i^4+3}{i-1} &= \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i} \\
&= \frac{(i^4+3)(-1-i)}{(i-1)(-1-i)} \\
&= \frac{(1+3)(-1-i)}{-i-i^{2}+1+i} \\
\end{align}
I'll leave to you to finish the details. I think the mistake was in your first line of algebra in the numerator.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $p$'s such that $p^4 + p^3 + p^2 + p +1$ is a perfect square.
Let $p$ be a prime number. Find all $p$'s such that $p^4 + p^3 + p^2 + p +1$ is a perfect square.
I tried rewriting the expression as $p^4 + p^3 + p^2 + p +1 = x^2 \iff (p^2 + p)(p^2 + 1)=(x - 1)(x + 1)$. Then I think I need to bound this using $x$ but am not sure how to.
| Note that $p^2=q^4+q^3+q^2+q+1$ has solutions only for $p=11$ and $q=3$.
Indeed we can write $$\left(q^2+\frac{q}{2}\right)^2={q^4+q^3}+\frac{q^2}{4}<{q^4+q^3}+q^2+q+1 \\ \frac{q^2}{4}<q^2+q+1 $$ and on the other hand $$ \left(q^2+\frac{q+2}{2}\right)^2=q^4+q^3+2q^2+\frac{q^2+4q+4}{4}>q^4+q^3+q^2+q+1 \\ {q^4+q^3}+\frac{9}{4}q^2{+q+1}>{q^4+q^3}+q^2{+q+1} \\ \frac{9}{4}q^2>q^2.$$ From here, $q$ cannot be even, and for some odd $q$ we must have $$\left(q^2+\frac{q+1}{2}\right)^2={q^4+q^3+q^2}+\frac{q^2+2q+1}{4}={q^4+q^3+q^2}+q+1 \\ q^2+2q+1=4q+4 \\ q^2-2q-3=(q-3)(q+1)=0,$$ from here $q=3$. In particular, $$3^4+3^3+3^2+3+1=11^2$$
therefore the only solutions are $p=11$, $q=3$
| {
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What is the probability that two numbers between 1 and 10 picked at random sum to a number greater than 5? We have the numbers $1$ through $10$ in a box, we pick one at random, write it down and put it back in the box. We pick another of those numbers at random and write it down again. If we add the two numbers, what is the probability that it will be greater than $5$?
At first I though that I could count the number of ways we could add two numbers to get six, i.e. $2+4$ and see what are the chances to get numbers bigger than those choices. Then adding all the probabilities that relate to each way. However, I get numbers greater than $1$ which is impossible. I also thought about the chance of getting a $1$ and then a number equal to or bigger to $5$, $P(x \ge 5) = \frac 12$ multiplying them together and repeating until all numbers run out. Again, wrong answer.
My question is: how do we get to the correct answer? Is it possible to generalize? Say that the probability of $n$ numbers picked at random from $N$ choices add to something greater than $k$.
| You could solve this with generating functions. The generating function for this situation, equivalent to rolling a fair 10-sided die twice, is:
$$(x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^2$$
which expands to
$$x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 5 x^6 + 6 x^7 + 7 x^8 + 8 x^9 +
9 x^{10} + 10 x^{11} + 9 x^{12} + 8 x^{13} + 7 x^{14} + 6 x^{15} + 5 x^{16} +
4 x^{17} + 3 x^{18} + 2 x^{19} + x^{20}$$
The coefficient of each $x^n$ is the number of ways of getting a sum of $n$ from the two random draws. There are 100 total possibilities ($10 \cdot 10$) with two draws from 1...10 with replacement. Looking at the polynomial above, the coefficients of the monomials $x^2$ through $x^5$ show that there are a total of 10 ways to get a sum of 5 or less. Thus, the probability of a sum greater than 5 is 90/100 or 0.9.
Perhaps a simpler way is to first eliminate the cases where at least one draw is 5 or greater, since these guarantee a sum greater than five. There are 84 of these, leaving only the 16 cases in which both draws are 4 or less. This can also be done with a generating function (though counting manually is easy too):
$$(x + x^2 + x^3 + x^4)^2 = x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 3 x^6 + 2 x^7 + x^8$$
once again giving 10 out of 100 cases where the sum is 5 or less.
| {
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Solutions to $x+y+z-2 = (x-y)(y-z)(z-x)$ Show that the equation $$x+y+z-2 = (x-y)(y-z)(z-x)$$ has infinite solutions $(x,y,z)$ with $x, y,z$ distinct integers.
In my attempt to solve the problem only found solutions form $x=y, z=2-2x$.
There are solutions as required by the statement?
| Set $x=y+a$ and $z=y-b$. The equation then becomes
$$ 3y+a-b-2 = -ab(a+b) $$
which simplifies to
$$ 3y-2 = -a^2b - ab^2 - a + b $$
If we set $a=2$, then every $b$ that is a multiple of $3$ will lead to a solution for $y$.
In general an integral $y$ is possible exactly when $a+1\equiv b\pmod 3$ or $a\equiv b\equiv 1\pmod 3$, as can be seen by solving
$$ a-b-2 \equiv -ab(a+b) \pmod 3 $$
by brute force, trying each of the 9 possibilities.
This produces every integral soltution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this
$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$
Someone can help me ?
Thanks.
| For $x\neq \frac{\pi}{4}+k\pi/2$ and $x\neq \pi/2+k\pi$ ($k\in\mathbb{Z}$), we have
$$\tan(2x)=\frac{2\tan(x)}{1-\tan^{2}(x)}$$
and
$$\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\left(\frac{\pi}{4}\right)-\tan(x)}{1+\tan\left(\frac{\pi}{4}\right)\tan(x)}$$
Combining the two expressions and using the fact that $\tan(\pi/4)=1$, we get
\begin{align*}
\tan(2x)\tan\left(\frac{\pi}{4}-x\right) &=\frac{2\tan(x)}{1-\tan^{2}(x)}\cdot\frac{1-\tan(x)}{1+\tan(x)}\\
&=\frac{2\tan(x)}{(1+\tan(x))^{2}}\\
\end{align*}
Hence,
$$\lim_{x\to\frac{\pi}{4}}\tan(2x)\tan\left(\frac{\pi}{4}-x\right)=\lim_{x\to\frac{\pi}{4}}\frac{2\tan(x)}{(1+\tan(x))^{2}}=\frac{2}{(1+1)^{2}}=\frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.
For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$, and I supposed that $(x,y,z)$ is a primitive solution for this equation. Thus we have $(x^2)^2 +z^2 = (2y^2)^2$.
Now $(2y^2)^2$ is even, it tells us that $x$ and $z$ must have the same parity. But we know by Diophante's equation that there exist $a$ and $b$ with $a > b$, $(a, b) = 1$. If $a \not \equiv b \pmod 2$, we have $x^2 = a^2 - b^2, z = 2ab$ and $2y^2 = a^b + b^2$. But this is impossible, because of $2y^2 = a^2 + b^2$, $a$ and $b$ should have the same parity, which is not our hypothesis.
If $a$ and $b$ have the same parity, they are both odd (otherwise they are not coprime) and we have $x^2 = \frac{a^2 - b^2}{2}$, $z = ab$ and $2y^2 = \frac{a^2 + b^2}{2}$. But once again, as $(a, b) = 1$ and are both odd, they have no factor "2" in common thus $2y^2 = \frac{a^2 + b^2}{2}$ is impossible.
Therefore the equation $X^4 - 4Y^4 = -Z^2$ has no solution in $\mathbb{N} \setminus \{0\}$.
What do you think of this argument? My first idea was to use the infinite descent, but in this case I did not manage to do it.
My second idea was to prove that it was equivalent to another equation such as $X^4 + Y^4 = Z^2$ which we know has no solutions such that $XYZ \neq 0$ but I did not have more success.
| For another way of showing your required conclusion, I give you the following:
Hint:
$$Z^2 = 4Y^4 - X^4 = (2Y^2 + X^2)(2Y^2 - X^2)$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with restrictions on $x_1,x_2,x_3,x_4$? How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$? How many solutions with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
So for the first part I did this:
We set $x_i = y_i + 1$ for $i = 1, 2, 3$ and obtain $y_1 + y_2 + y_3 = 12 − 3 = 9$.
So, there is one-to-one correspondence between the positive integer solutions of $x_1 + x_2 + x_3 = 12$ and the non-negative integer solutions of $y_1 + y_2 + y_3 = 9$.
Hence, both equation have the same number of solutions. Since $y_1 + y_2 + y_3 = 9$ has ${9+3-1 \choose 3-1} = {11 \choose 2} = 55$ integer solutions.
Is this correct?
For the second part I do not understand how to figure out how many solutions with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
| I am not sure why you have four variables in the statement of your question and three variables in your answer. I will assume you meant to work with four variables.
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers.
Method 1: We reduce the problem to one in the non-negative integers. Let $y_k = x_k - 1$ for $1 \leq k \leq 4$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 1$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to placing three addition signs in a row of eight ones. For instance,
$$1 1 1 1 1 + 1 + 1 1 1$$
corresponds to the solution $y_1 = 5$, $y_2 = 1$, and $y_3 = 3$, while
$$1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 6$. Thus, the number of solutions of equation 2 is the number of ways three addition signs can be inserted into a row of eight ones, which is
$$\binom{8 + 3}{3} = \binom{11}{3}$$
since we must choose which three of the eleven symbols (eight ones and three addition signs) will be addition signs.
Method 2: A particular solution of equation 1 in the positive integers corresponds to inserting three addition signs in the eleven spaces between successive ones in a row of $12$ ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance,
$$1 1 1 + 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 5$. Thus, the number of solutions of equation 1 in the positive integers is the number of ways three addition signs can be inserted into the eleven gaps between successive ones in a row of $12$ ones, which is
$$\binom{11}{3}$$
How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
We reduce the problem to one in the non-negative integers. Since $x_1$ is an integer, $x_1 > 1 \implies x_1 \geq 2$. Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \implies x_2 \geq 2$ and $x_3 > 3 \implies x_3 \geq 4$. Let
\begin{align*}
y_1 & = x_1 - 2\\
y_2 & = x_2 - 2\\
y_3 & = x_3 - 4\\
y_4 & = x_4
\end{align*}
Then each $y_k$, $1 \leq k \leq 4$, is a non-negative integer. Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields
\begin{align*}
y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with
$$\binom{4 + 3}{3} = \binom{7}{3}$$
solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try:}$ For lower bound $$\sum^{99}_{k=1}\frac{1}{(k+1)\sqrt{k}}>\sum^{99}_{k=1}\frac{1}{(k+1)k}=\sum^{99}_{k=1}\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{99}$$
Now I didn't understand how can I solve it, any help?
Thanks
| Hint. Observe that, for $x\in [k,k+1],\, k=1,2,3,\cdots$,
$$
\frac1{(k+2)\sqrt{k+1}} \leq\frac1{(x+1)\sqrt{x}}\leq \frac1{(k+1)\sqrt{k}}.
$$
Integrating both sides from $x=k$ to $x=k+1$ gives
$$
\frac1{(k+2)\sqrt{k+1}} \leq\int_k^{k+1}\frac1{(x+1)\sqrt{x}}dx\leq \frac1{(k+1)\sqrt{k}}
$$ then, summing from $k=1$ to $k=99$,
$$
\sum_{k=1}^{100}\frac1{(k+1)\sqrt{k}}-\frac12\leq\int_1^{100}\frac1{(x+1)\sqrt{x}}dx\leq \sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}.
$$ We have
$$
\int_1^{100}\frac1{(x+1)\sqrt{x}}dx=2\arctan (10) -\frac{\pi}2.
$$
It follows that
$$
2\arctan (10) -\frac{\pi}2\leq\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}\leq2\arctan (10) -\frac{\pi}2+\frac12-\frac1{1010}.
$$ or
$$
\color{red}{1.371\cdots}<\sum_{k=1}^{99}\frac1{(k+1)\sqrt{k}}<\color{blue}{1.870\cdots}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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the minimum value of $a^2+b^2-a-\frac{2b}{3}$ using known standard inequalities If $a,b$ are real numbers satisfying $a+2b=3,$ then the minimum value of $a^2+b^2-a-\frac{2b}{3}$
Here $a+2b=3\implies a=3-2b$
$a^2+b^2-a-\frac{2b}{3}=(3-2b)^2+b^2-(3-2b)-\frac{2b}{3}$
$=9+4b^2-12b+b^2-3+2b-\frac{2b}{3}$
I diiferentiated it with respect to $b$ and put it equal to $0$ and then found $a$ by using the relation $a=3-2b$ to get the minimum value.
But this is a long method.Can we find its minimum value by using known inequalities like AM-GM etc.
| Hint:
$$ a^2+b^2-a-\frac{2b}{3} = (a - \frac12)^2 + (b - \frac13)^2 - \frac{13}{36}$$
and the first $2$ terms at the RHS can be seen as the square of the distance from a point to the point $(\frac12, \frac13)$. So the problem becomes finding the distance from a certain point to a line.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find numbers whose sum of digits equals a value How do I find all of the numbers in a given range whose sum of digits equal to a given value?
For example:
Range : 100 - 9000
Value : 4
Result : 103, 112, 121, 130, 202, 211, 220, 301, 310, ..., 4000, ...
| Observe that any four digit number larger than $4000$ will have a digit sum greater than $4$. Thus, the problem reduces to finding the number of three or four digit numbers with digit sum $4$.
For three digit numbers, let $x_2$ denote the hundreds digit, $x_1$ denote the tens digit, and $x_0$ denote the units digit. Then the number of three digit numbers whose digit sum is $4$ is the number of solutions of the equation
$$x_2 + x_1 + x_0 = 4 \tag{1}$$
in the non-negative integers subject to the constraint that $x_2 \geq 1$. Let $x_2' = x_2 - 1$. Then $x_2'$ is a non-negative integer. Substituting $x_2' + 1$ for $x_2$ in equation 1 yields
\begin{align*}
x_2' + 1 + x_1 + x_0 & = 4\\
x_2' + x_1 + x_0 & = 3 \tag{2}
\end{align*}
which is an equation in the non-negative integers. A particular solution of equation 2 corresponds to placing two addition signs in a row of three ones. For instance,
$$1 + 1 1 +$$
corresponds to the solution $x_2' = 1$, $x_1 = 2$, $x_3 = 0$, while
$$1 + 1 + 1$$
corresponds to the solution $x_2' = x_1 = x_0 = 1$. Thus, the number of solutions of equation 2 is the number of ways two addition signs can be inserted into a row of three ones, which is
$$\binom{3 + 2}{2} = \binom{5}{2}$$
since we must choose which two of the five symbols (three ones and two addition signs) will be addition signs.
For four digit numbers, let $x_2$, $x_1$, and $x_0$ be defined as above. Let $x_3$ denote the thousands digit. Then the number of four digit numbers whose digit sum is $4$ is the number of solutions of the equation
$$x_3 + x_2 + x_1 + x_0 = 4 \tag{3}$$
in the non-negative integers subject to the constraint that $x_3 \geq 1$. Let $x_3' = x_3 - 1$. Then $x_3$ is a non-negative integer. Substituting $x_3' + 1$ for $x_3$ in equation 3, then simplifying yields
$$x_3' + x_2 + x_1 + x_0 = 3 \tag{4}$$
Equation 4 is an equation in the non-negative integers. The number of solutions of equation 4 is the number of ways of inserting three addition signs in a row of three ones, which is
$$\binom{3 + 3}{3} = \binom{6}{3}$$
since we must choose which three of the six symbols (three ones and three additions signs) will be addition signs.
In total, there are
$$\binom{5}{2} + \binom{6}{3}$$
numbers between $100$ and $9000$ with digit sum $4$.
| {
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Simplifying the arc length integral for a segment of a circle (Calc 2) So I know the length L of the curve $y=\sqrt{R^{2}-x^{2}}$ from $x=0$ to $x=a$ where $|a| < R$ is given by:
$$L= \int_0^a \frac{R}{\sqrt{R^{2}-x^{2}}}dx $$
Now I must set up the arc length integral and simplify it so that it is in the form listed above.
$$L= \int_0^a \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx$$
and
$$\frac{dy}{dx}=-\frac{x}{\sqrt{R^{2}-x^{2}}}$$
$$\left(\frac{dy}{dx}\right)^{2}=\frac{x^{2}}{R^{2}-x^{2}}$$
so
$$L= \int_0^a \sqrt{1+\frac{x^{2}}{R^{2}-x^{2}}}dx$$
I am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks
| Assume $R>a$ and $a,R\in\mathbb{R^+}$:
$$\text{L}=\int_{0}^{a}\sqrt{1+\left(\frac{\partial}{\partial x}\left[\sqrt{R^2-x^2}\right]\right)^2}\space\text{d}x=\int_{0}^{a}\sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}\space\text{d}x=$$
$$\int_{0}^{a}\sqrt{1+\frac{x^2}{R^2-x^2}}\space\text{d}x=\int_{0}^{a}\sqrt{\frac{R^2-x^2}{R^2-x^2}+\frac{x^2}{R^2-x^2}}\space\text{d}x=$$
$$\int_{0}^{a}\sqrt{\frac{R^2-x^2+x^2}{R^2-x^2}}\space\text{d}x=\int_{0}^{a}\sqrt{\frac{R^2}{R^2-x^2}}\space\text{d}x=$$
$$\int_{0}^{a}\sqrt{\frac{\frac{R^2}{R^2}}{\frac{R^2}{R^2}-\frac{x^2}{R^2}}}\space\text{d}x=\int_{0}^{a}\frac{1}{\sqrt{1-\frac{x^2}{R^2}}}\space\text{d}x=$$
Substitute $u=\frac{x}{R}$ and $\text{d}u=\frac{1}{R}\space\text{d}x$.
This gives a new lower bound $u=\frac{0}{R}=0$ and upper bound $u=\frac{a}{R}$:
$$R\int_{0}^{\frac{a}{R}}\frac{1}{\sqrt{1-u^2}}\space\text{d}u=R\left[\arcsin(u)\right]_{0}^{\frac{a}{R}}=R\left(\arcsin\left(\frac{a}{R}\right)-\arcsin(0)\right)=R\arcsin\left(\frac{a}{R}\right)$$
| {
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Is this relationship already known? I like math because it's a puzzle to me, but am really not very good at it. But I figured out the relationship below myself. Just curious, is this already pretty common knowledge? Kind of proud of myself for figuring it out, but my son who's getting math minor had never heard of it. Apologies in advance for any poor explanation.
$b^2 = a^2 + a + b$ for positive integers where $b - a = 1$
e.g.
$2^2 = 4$
$3^2 = 9$
$4 + (2 + 3) = 9$
$4^2 = 16$
$9 + (3 + 4) = 16$
and so on.
Edit in response to @fleablood comment:
Where b = a + 2
Even numbers...
2^2 = 4 -> a^2 = a2
4^2 = 16 -> b^2 = b2
16 - 4 = 12 -> b2 - a2 = c
(a + 1) * b = 3 * 4 = c
Odd Numbers...
3^2 = 9 -> a^2 = a2
5^2 = 25 -> b^2 = b2
25 - 9 = 16 -> b2 - a2 = c
(a + 1) * b = 4 * 5 = 20
c != 20
| This is another way of stating that $n^2 = 1 + 3 + 5 + ...$; the sum of the first n-1 odd numbers.
$a - b = 1$ means $a = b + 1$ means $a + b = 2b + 1$ means $a + b$ is the (b-1)-th odd number.
So $a^2 = b^2 + a + b$ is the same as saying $(b + 1)^2 = b^2 + (b + 1) + b = b^2 + 2b + 1$; which can be shown by expansion: $(b + 1)(b+1) = b(b+1) + (b + 1) = b^2 + b +b + 1 = b^2 + 2b + 1$.
Expanding it is $a^2 = a + b + b + (b-1) + (b-1) + .... + 1 + 1$ = $a + 2(a-1) + 2(a - 2) + .... + 4 + 2$.
... which is another way of saying $a^2 = 1 + (1+2) + (2+3) + .... + ((a - 2) + (a-1)) + ((a-1) + a)$.
... which is another way of saying $a^2 = 1 + 3 + 5 + ... [2(a-1) + 1]$
... which is another way of saying $a^2 = (a -1)^2 + 2(a-1) + 1$
I remember being surprised when I first discovered it, too.
| {
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Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$ is a root of the cubic.
Let $a$ be a root of the cubic $x^3-21x+35=0$. Prove that $a^2+2a-14$
is a root of the cubic.
My effort
Working backwards I let $P(x)$ be a polynomial with roots $a,a^2+2a-14$ and $r$.
Thus, $$P(x)=(x-a)(x-r)(x-(a^2+2a-14))$$
Expanding, I get
$$P(x) =(x^2-(a+r)x+ar)(x-(a^2+2a-14)) $$
$$P(x) =x^3+x^2[-(a^2+2a-14)-(a+r)]+x[(a+r)(a^2+2a-14)+ar]-ar(a^2+2a-14)$$
Equating coefficients of $P(x)$ with the given cubic $x^3-21x+35=0$ I have the following system of equations :
\begin{array}
\space (a^2+2a-14)+(a+r)&=0 \\
(a+r)(a^2+2a-14)+ar&=-21 \\
-ar(a^2+2a-14)&=35 \\
\end{array}
From the first equation I have $(a^2+2a-14) =-(a+r) $ which, substituted in the other two equations ,it yields
\begin{array}
\space -(a+r)^2+ar &=-21 \\
ar(a+r) &=35 \\
\end{array}
Rearranging the second equation for $ar$ I have $ar=\cfrac{35}{(a+r)}$ which I now substitute into the first eq. to get:
\begin{array}
\space -(a+r)^2+\cfrac{35}{(a+r)}&=-21 \\
-(a+r)^3+35 +21(a+r) &=0 \\
\end{array}
My problem now is that the last equation looks pretty darn close to
$x^3-21x+35=0$ but some signs are not in the right place,which makes
me wonder if I have made some careless mistake(I have already checked
but I don't see it) or if I have left some algebraic manipulations to do.
| It's quite a lot easier to simply plug in $x=a^2+2a-14$ into $P(x)$ and see \begin{align}
x^3-21x+35&=(a^2+2a-14)^3-21(a^2+2a-14)+35\\
&=(35-21a + a^3) (-69 - 9 a + 6 a^2 + a^3)\\
&=0\cdot (-69 - 9 a + 6 a^2 + a^3)\\
&=0
\end{align}
Note that factorizing isn't too hard since you already "know" that one factor will be $35-21a + a^3$.
Now where did you go wrong? You had $(a+r)^3-21(a+r)-35=0$. This indeed looks very much alike $P(x)$, but not quite. You didn't make any mistake in your algebra, you assumed $a+r$ would be a root of $P$, which it is not (necessarily). But, when writing \begin{align}
Q(x)&=x^3+px^2+qx+r\\
&=(x-\alpha)(x-\beta)(x-\gamma)\\
&=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma
\end{align} then you'll see, when $p$, the coefficient for $x^2$, is $0$, (which it is, in your case of $P(x)$), then the roots add up to $\alpha+\beta+\gamma=-p=0$, so $a+r$ is the negative of a root! Now it becomes clear that $(a+r)^3-21(a+r)-35=0$, since we know $(-(a+r))^3-21(-(a+r))+35=0$.
| {
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Convergence of $a_{n+1}=\sqrt{2-a_n}$ I'm attempting to prove the convergence (or divergence, though I strongly suspect it converges) of a sequence defined as $a_{n+1}=\sqrt{2-a_n}$ with $a_1=\sqrt{2}$.
I cannot use the monotonic sequence theorem as the sequence is not monotonically increasing. In fact, the first few values of the sequence are:
$a_1 =\sqrt{2}\approx 1.4142$
$a_2 =\sqrt{2-\sqrt{2}}\approx .7653$
$a_3 =\sqrt{2-\sqrt{2-\sqrt{2}}}\approx 1.1111$
Thus, it seems that $a_{n \to \infty} \to 1$
It seems that the sequence is behaving similarly to $\frac{\sin x}{x}$, leading me to think that the squeeze theorem may be useful. Still, I cannot seem to make any progress besides numerical computation of successive terms.
| You have surely proved that $a_n\le 2$ for all $n$.
Consider the sequences $b_n=a_{2n-1}$ and $c_n=a_{2n}$. The recursions are
$$
b_{n+1}=a_{2n+1}=\sqrt{2-a_{2n}}=\sqrt{2-\sqrt{2-a_{2n-1}}}=
\sqrt{2-\sqrt{2-b_n}}
$$
Let's show $(b_n)$ is decreasing:
\begin{gather}
b_{n+1}\le b_n\\
\sqrt{2-\sqrt{2-b_n}}\le b_n\\
2-\sqrt{2-b_n}\le b_n^2\\
2-b_n^2\le \sqrt{2-b_n}\\
4-4b_n^2+b_n^4\le 2-b_n\\
(b_n+2)(b_n-1)\Bigl(b_n-\frac{1+\sqrt{5}}{2}\Bigr)\Bigl(b_n-\frac{1-\sqrt{5}}{2}\Bigr)\le0
\end{gather}
and we just need to show $1\le b_n\le\sqrt{2}$ (work out why).
The basis of the induction is obvious, as $b_1=\sqrt{2}$. Suppose $1\le b_n\le \sqrt{2}$; then
\begin{gather}
1\le b_{n+1}\le\sqrt{2}\\
1\le 2-\sqrt{2-b_n}\le 2\\
0\le\sqrt{2-b_n}\le 1\\
0\le 2-b_n\le 1\\
1\le b_n\le 2
\end{gather}
Since this is true, we are done.
Therefore $(b_n)$ is a decreasing and bounded sequence, so it converges to $L$ such that
$$
L=\sqrt{L-\sqrt{2-L}}
$$
The only nonnegative values for $L$ are $1$ and $(1+\sqrt{5})/2$, which however is greater than $b_1$, so we see $L=1$.
Now do the same in order to prove $(c_n)$ is increasing; actually you have to see that $0<c_n\le 1$, so
$$
(c_n+2)(c_n-1)\Bigl(c_n-\frac{1+\sqrt{5}}{2}\Bigr)\Bigl(c_n-\frac{1-\sqrt{5}}{2}\Bigr)\ge0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Maximum value of $\lambda$ It is given that a,b,c are be of same sign and a,b,c are in Harmonic progression i.e. $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ and also
$\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\geq \sqrt{\lambda \sqrt{\lambda \sqrt{\lambda.............\infty}}}$.
Find the greatest possible value of $\lambda$.
How should I begin? I think we somehow need to find value of R.H.S. when equality holds but how to achieve that? Also using the result $b=\frac{2ac}{a+c}$ doesn't give any fruitful result.
| Let $a=1/x,\; b=1/y, \;c=1/z.$ We have $y=(x+z)/2$. The LHS of the inequality is $$\frac {y+x}{2 y-x}+\frac {y+z}{2 y-z}=\frac { z/2+ 3 x/2}{z}+\frac { x/2+ 3 z/2}{x}=1+\frac {3}{2} (x/z+z/x).$$ Since $x$ and $z$ (which are $1/a$ and $1/c$) have the same sign, $x/z$ and $z/x$ are positive, so $$x/z+z/x=(\;(x/z)^{1/2}- (z/x)^{1/2}\;)^2+2\geq 2.$$ Therefore $1+\frac {3}{2}(x/z+z/x)\geq 1+\frac{3}{2}(2)=4.$ Equality occurs here iff $a=b=c.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$ = $\begin{bmatrix}
a^2 + c^2 & ab + cd \\
ab + cd & b^2 + d^2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle .
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
| If you had $A =\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ or $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$ instead of $A =\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, would it then be true that $A^T A = I \vphantom{\frac{\int^\int}\int} $?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Divisors of $75600$ of the type of $4n+2$
Find the total no. of divisors of $75600$ which of the type of $4n+2$
where $n\in \mathbb{N}$ and
$75600=2^4 \cdot 3^3 \cdot 5^2 \cdot 7^1$
Now I think divisors of type $(4n+2)$ should be of type
$2^1\cdot 3^a\cdot 5^b\cdot 7^c$. Am I correct?
| First of all, you are right.
Therefore, you need to count the number of divisors of $3^3\cdot5^2\cdot7^1$:
*
*In each divisor, the factor $3$ can appear between $0$ and $3$ times, i.e., $4$ different combinations
*In each divisor, the factor $5$ can appear between $0$ and $2$ times, i.e., $3$ different combinations
*In each divisor, the factor $7$ can appear between $0$ and $1$ times, i.e., $2$ different combinations
Hence there are $4\cdot3\cdot2=24$ divisors:
*
*$3^0\cdot5^0\cdot7^0$
*$3^1\cdot5^1\cdot7^1$
*$3^2\cdot5^2\cdot7^0$
*$3^3\cdot5^0\cdot7^1$
*$3^0\cdot5^1\cdot7^0$
*$3^1\cdot5^2\cdot7^1$
*$3^2\cdot5^0\cdot7^0$
*$3^3\cdot5^1\cdot7^1$
*$3^0\cdot5^2\cdot7^0$
*$3^1\cdot5^0\cdot7^1$
*$3^2\cdot5^1\cdot7^0$
*$3^3\cdot5^2\cdot7^1$
*$3^0\cdot5^0\cdot7^0$
*$3^1\cdot5^1\cdot7^1$
*$3^2\cdot5^2\cdot7^0$
*$3^3\cdot5^0\cdot7^1$
*$3^0\cdot5^1\cdot7^0$
*$3^1\cdot5^2\cdot7^1$
*$3^2\cdot5^0\cdot7^0$
*$3^3\cdot5^1\cdot7^1$
*$3^0\cdot5^2\cdot7^0$
*$3^1\cdot5^0\cdot7^1$
*$3^2\cdot5^1\cdot7^0$
*$3^3\cdot5^2\cdot7^1$
Each of those divisors should be multiplied by $2^1$ of course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is the sequence $\sqrt{1+\frac{1}{n^2}}$ increasing or decreasing? Is the sequence $\sqrt{1+\frac{1}{n^2}}$ increasing or decreasing?
I simplified it to $\frac{\sqrt{n^2+1}}{n}$, and I tried $a_{n+1}-a_n$ and $\frac{a_{n+1}}{a_n}$, but neither seem to work, how should I proceed?
| $\sqrt{1+\frac{1}{n^2}} > \sqrt{1+\frac{1}{(n+1)^2}}$
$\iff 1+\frac{1}{n^2} > 1+\frac{1}{(n+1)^2}$
$\iff \frac{1}{n^2} > \frac{1}{(n+1)^2}$
$\iff {n^2} < {(n+1)^2}$
$\iff {n^2} < n^2+2n+1$
$\iff 0< 2n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Prove that $a+\frac{1}{b}>2$ or $b+\frac{1}{a}>2$ for two strict positive numbers Another Olympiad Problem, let $x$ and $a$ and $b$ be strictly real positive numbers.
*
*Prove that $x$+$\frac{1}{x}$$>$$2$ (proven)
*Than conclude that $a$+$\frac{1}{b}$$>$$2$ or
$b$+$\frac{1}{a}$$>$$2$
For the second question, I know I am supposed to replace $x$ with a number in function of $a$ and $b$ , but I can't find it .
if we replace $x$ by $ab$ we still have a problem.
$ab$+$\frac{1}{ab}$$+2>$$4$ , we can't just conclude that $a$+$\frac{1}{b}$$>$$2$ or $b$+$\frac{1}{a}$$>$$2$
| Your first statement $x + \frac 1x > 2$ is not fully correct, you have
$$
x + \frac 1x \ge 2
$$
for $x > 0$, with equality exactly for $x = 1$.
And as the example $a=b=1$ shows, you can only conclude that
$$
a+\frac{1}{b} \ge 2 \quad \text{ or } \quad b+\frac{1}{a} \ge 2 \, .
$$
which is your claim with $\ge$ instead of $>$.
This follows from (similar to André's solution):
$$
\left(a+\frac{1}{b}\right) + \left(b+\frac{1}{a}\right)= a + \frac 1a + b + \frac 1b \ge 2 + 2 = 4
$$
(with equality exactly for $a=b=1$),
so at least one term on the left-hand side must be greater or equal
to $2$.
If $a$ and $b$ are not both equal to one then you can conclude that
$$
a+\frac{1}{b} > 2 \quad \text{ or } \quad b+\frac{1}{a} > 2 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
| Assume that,
$\sin x=a, \cos x=b $
Given that : $$\sin x+\cos x=\frac75$$
$$a+b=\frac75\tag 1$$
$$\sin^2 x+\cos^2 x=1$$
$$a^2 +b^2 =1$$
$$(a+b)^2-2ab =1$$
$$(7/5)^2-2ab =1$$
$$ab=12/25\tag 2$$
solving (1), (2), i get $a=3/5, b=4/5$
therefore,
$$\frac1{\sin x}+\frac1{\cos x}=\frac1a+\frac1b$$$$=\frac{1}{3/5}+\frac{1}{4/5}$$ $$=\frac{35}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 3
} |
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integer it has to be $k=101$ cause $gcd(k,k-101)=1\lor101$ and both $k,k-101$ can't be both squares of an integer,so $k=101t$ and $t(t-1)$ is never an square except for $t=1,0$ and $t=0$ is not possible hence $k=101$ is only possible integer solution
EDIT: So if $\gcd(k,k-101)=1$ then $k=h^2,k-101=(h-s)^2$ then $h(2h-s)=101$ which can be $s=1,h=51$ or $s=101,h=51$.$y=51^2+51\cdot50=51\cdot 101,x=51$
And since $x=2k-y$ is integer then $k=\frac{h}{2}$,if $h=2q$ then $k$ is integer otherwise if $h=2q+1$ then $$\frac{2q+1}{2}+\frac{1}{2}\sqrt{4q^2-400q-201}=\frac{2q+1}{2}+\frac{1}{2}\sqrt{(2q-10)^2-301}\\(2q-10)^2-301=r^2\\2q-10=z\\z^2-301=(z-c)^2\\c(2z-c)=301,c=1,z=151,c=7,z=25,c=43,z=25$$
The $z=151$ is impossible cause $2q-10$ is even,and $z=25$ is also impossible because $2q-10$ is even.Hence the only solutions are $(x,y)=(101,101),(51,5151),(5151,51)$ The last one clearly from symmetry
| $$\dfrac1x=\dfrac{2y-101}{101y}\iff x=\dfrac{101y}{2y-101}$$
If $d$ divides $2y-101,101y$
$d$ must divide $2(101y)-101(2y-101)=101^2$
So, $2y-101$ must divide $101^2$ to make $x$ an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$
$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$
$$\int \frac{1}{(1-\sqrt{u})}du$$
How do I continue from here? it seems that I have made it harder
| $x=\sin^2(\theta)$
$$
\begin{align}
\int\frac{\mathrm{d}x}{x\sqrt{1-x}}
&=\int\frac{2\,\mathrm{d}\theta}{\sin(\theta)}\\
&=-2\int\frac{\mathrm{d}\cos(\theta)}{1-\cos^2(\theta)}\\
&=-\int\left(\frac1{1-\cos(\theta)}+\frac1{1+\cos(\theta)}\right)\mathrm{d}\cos(\theta)\\
&=\log\left(\frac{1-\cos(\theta)}{1+\cos(\theta)}\right)+C\\
&=\log\left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)+C\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting the terms one place:
$$
\begin{align}
1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\
=1+3+5+7+\cdots & =2S
\end{align}
$$
This is the sum of the odd numbers
If we were to now sum the odd numbers and the even numbers like below:
$$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$
This supposedly tells us that:
$$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$
What is faulty with this proof.
| Interpreted literally (i.e., using the usual sense of limits of infinite series), the first line,
$$1 + 2 + 3 + \cdots = -\frac{1}{12} ,$$
is simply false, as the series on the l.h.s. diverges.
What's true, for example, is that there's a natural way to extend the function $$Z(s) := \sum_{k = 1}^{\infty} k^{-s} ,$$ which is defined on $(1, \infty)$ (and in particular not at $s = -1$), to a function $\zeta$ defined for most complex numbers, including $s = -1$, and this function satisfies $\zeta(-1) = -\tfrac{1}{12}$. The partial sums of the series, evaluated at $s = -1$, are $1 + \cdots + n$, but this is not the same as saying $1 + 2 + 3 + \cdots = -\tfrac{1}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Total number of $4$ digit numbers whose product of digits is $72$
Total number of $4$ digit numbers whose product of digits is $72$
$\bf{My\; Try::}$ Here the possible factor of $72 = 2^{3}\cdot 3^2$
Now here we divide $2^3\cdot 3^3$ into product of $4$ numbers
Like $1\cdot 1\cdot 8\cdot 9$
and $1\cdot 2\cdot 4\cdot 9$
and $1\cdot 2\cdot 6\cdot 6$
and $1\cdot 3\cdot 4\cdot 6$
and $1\cdot 3\cdot 3\cdot 8$
and $2\cdot 2\cdot 3\cdot 6$
and $2\cdot 2 \cdot 2\cdot 9$
So arrangement of these numbers is $$\frac{4!}{2!}+4!+\frac{4!}{2!}+4!+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{3!} = 12+12+12+24+24+12+4=100$$
But answer given is $=112$
So Where i have make mistake, Is there is any other better method,
Then plz explain here
Thanks
|
Where have I made a mistake?
*
*You forgot to count the number of permutations of $[2,3,3,4]$
*You miscounted the number of permutations of $[2,2,3,6]$ as $\frac{4!}{3!}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Another $1=2$ proof So a friend shows me this :
$x^4= x^2+x^2+ \cdots +x^2 $ ( i.e. $x^2$ added $x^2$ times)
Now take the derivative of both side;
$4x^3 = 2x + 2x + \cdots + 2x $;
So $4x^3 = 2x^3 \cdots $(1)
And so dividing by $x^3$ gives $2=1 \cdots $(2).
I know we can't divide by 0 so that makes (2) false, but to show that (1) is false too ?
| *
*First of all, the statement
$x^2$ added $x^2$ times
makes sense only if $x^2$ is a positive integer. Else if $x^2$ is not a positive integer, then the statement is meaningless.
*Moreover, from $(1)$, we have that $4x^3=2x^3 \Rightarrow 4x^3-2x^3=0 \Rightarrow 2x^3=0 \Rightarrow x^3=0$
And hence division by $x^3$ in the next step is meaningless.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$
I could not solve it further.
| Another way is, simply, to take the derivative of $$\frac{-x}{(x^2-1)^2}+c$$ and verify that this derivative is no other that $$\frac{3x^2+1}{(x^2-1)^3}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} = \frac{p}{q}.$$
By using Wolfram Alpha, I obtain the sum is $\frac{3}{20}$. But I have no idea on how to obtain the sum.
Any hint would be appreaciated.
[The question is taken from SMO 2015 Junior section]
| We know that
$$
S_n=\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+ \cdots+ \frac{1}{(n-1)n}=1-\frac{1}{n}
$$
Then yours sum is exactly $$S_{10}-S_4=(1-\frac{1}{10})-(1-\frac{1}{4})=\frac{1}{4}-\frac{1}{10}=\frac{5-2}{20}=\frac{3}{20}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How would you find the kinematic range using the position function? Knowing that the range is:
$$ R = \frac{v^2\sin2\theta}g $$
Taking the integral of the velocity function we have:
$$ R(T) = (V_i \cos\theta T + x_i)X +\left(-\frac{1}2gT^2+V_i\sin\theta T+y_i\right)Y$$
So, I know it hits the ground at:
$$ \text{time in flight} = \frac{V_i\sin\theta \pm \sqrt{ (V_i\sin\theta )^2 +2gy_i } }g $$
Knowing the time in flight, how would we derive the range?
Would I simply plug it back into the position function on the $x$-axis? (Assuming the object hits the ground when $y = 0$)
Edit:
After messing with it, I found out what the book wanted me to do:
Assuming $y_i = 0:$
$ timeInFlight = \frac{ V_isin\theta + \sqrt{ (V_isin\theta)^2 } }g = \frac{ 2V_isin\theta }g $
$ R_x(T)=V_icos\theta T+x_i $
$R_x(time in flight) = \frac{2V_i^2sin\theta cos\theta}g + x_i = \frac{V_i^2sin2\theta}g + x_i$
| \begin{align*}
x &= ut\cos \theta \\
y &= ut\sin \theta -\frac{gt^{2}}{2} \\
&= x\tan \theta -\frac{gx^{2}}{2u^{2}\cos^{2} \theta} \\
&= x\tan \theta \left( 1-\frac{gx}{2u^{2} \sin \theta \cos \theta} \right) \\
&= x\tan \theta \left( 1-\frac{gx}{u^{2} \sin 2\theta} \right) \\
R &= \frac{u^{2} \sin 2\theta}{g}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
determinant of sum of anticommuting matrices Let $A, B \in \mathcal{M}_n(\mathbb{C})$ such that $AB=-BA$ and $det(A)=0$ and $det(A+B)=0.$
Prove that $det(A^3+B^3)=0$.
I don't really know how to approach this question.
| You only need $AB=-BA$ and $\det(A+B)=0$, not $\det A=0$, to prove that $\det(A^3+B^3)=0$. The reason is that $A^3+B^3$ can be factored such that $A+B$ is a factor.
Using $BA=-AB$ and always normalising products to the form $A^pB^q$, i.e. all $A$ factors before all $B$ factors, the product
$$
(A+rB)(A+B)(A+sB)=A^3+(r+s-1)A^2B+(r+s-rs)AB^2+rsB^3
$$
which equals to $A^3+B^3$ if $rs=1$ and $r+s=1$. This happens if $(r,s)=(z,\bar z)$ where $z=e^{\pm\pi i/3}=\frac{1\pm\sqrt{3}i}{2}$, so you get
$$
A^3+B^3=(A+zB)(A+B)(A+\bar zB)
$$
and therefore
$$
\begin{align}
\det(A^3+B^3)&=\det[(A+zB)(A+B)(A+\bar zB)]\\
&=\det(A+zB)\,\det(A+B)\,\det(A+\bar zB)=0.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
| The integrand is
$$f(x):=\frac{\cos(x)}{2\cos(x)+\sin(x)}.$$
We can form a linear combination to let the derivative of the denominator appear at the numerator:
$$af(x)+b=\frac{a\cos(x)+b(2\cos(x)+\sin(x))}{2\cos(x)+\sin(x)}=\frac{-2\sin(x)+\cos(x)}{2\cos(x)+\sin(x)},$$ is obtained with
$$b=-2,a=5.$$
Then by integrating,
$$5F(x)-2x=\ln(|2\cos(x)+\sin(x)|),$$
from $0$ to $\dfrac\pi4$,
$$5I-\frac\pi2=\ln\left(\frac{\frac3{\sqrt2}}{2}\right),$$
we get
$$I=\frac{\ln(9)-\ln(8)+\pi}{10}\approx0.3259375689\cdots$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
|
Let $AF=x\ , BF=y\ , CF=z$.
$F-$ Fermat point of $\triangle ABC\ $
So inequality we can rewrite as : $x+y+z\le 3R$ , which is obviously true)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Generating function to find the number of ways to put marbles in a basket Write a generating function for the number of ways to make a basket of $n$
marbles, if you need to use at least one orange marble , an even number of
yellow marbles, at most 2 green marbles, and any number of red marbles.
Because the question is asking for a generating function, it seems to me like it may be possible to create a recurrence relation and then solve the relation to create the generating function. However, I'm not sure how to go about finding that relation. The fact that we need an even number of yellow marbles is giving me trouble.
For instance, if we ignore that we need an even number of yellow marbles, we obtain:
(1) $n_o+n_y+n_g+n_r=n$ where $n_0>0$ and $n_g<3$
so what follows is
(2) $n'_o+n_y+n_g+n_r=n-1$ where $n_g<3$. Then we can solve the number of solutions to this equation minus the number of solutions where the number of green marbles is greater than or equal to 3 and take the difference between the two.
Where the number of solutions to (2)(ignoring the bounds on green marbles) is $n-1+4-1 \choose 3$=$n+2 \choose 3$
Any ideas on how to calculate the number of ways, say $f_n$, through a recurrence relation?
| Completing Brian Scott's answer, you are after the coefficient of $z^n$ in the following. This yields to partial fractions and using geometric series and (generalized) binomial theorem:
$\begin{align}
[z^n] \left( \sum_{k \ge 1} z^k \right)
\cdot &\left( 1 + z + z^2 \right)
\cdot \left( \sum_{k \ge 0} z^{2 k} \right)
\cdot \left( \sum_{k \ge 0} z^k \right) \\
&= [z^n] \frac{z}{1 - z}
\cdot \left( 1 + z + z^2 \right)
\cdot \frac{1}{1 - z^2}
\cdot \frac{1}{1 - z} \\
&= [z^n] \frac{z (1 + z + z^2)}{(1 - z)^2 (1 - z^2)} \\
&= [z^n] \left(
- \frac{1}{8} \frac{1}{1 + z}
+ \frac{7}{8} \frac{1}{1 - z}
- \frac{9}{4} \frac{1}{(1 - z)^2}
+ \frac{3}{2} \frac{1}{(1 - z)^3}
\right) \\
&= -\frac{1}{8} (-1)^n
+ \frac{7}{8}
- \frac{9}{4} (-1)^n \binom{-2}{n}
+ \frac{3}{2} (-1)^n \binom{-3}{n} \\
&= - \frac{1}{8} (-1)^n
+ \frac{7}{8}
- \frac{9}{4} \binom{n + 2 - 1}{2 - 1}
+ \frac{3}{2} \binom{n + 3 - 1}{3 - 1} \\
&= - \frac{1}{8} (-1)^n
+ \frac{7}{8}
- \frac{9}{4} (n + 1)
+ \frac{3}{2} \frac{(n + 2) (n + 1)}{2} \\
&= \frac{18 n^2 + 3 - 3 \cdot (-1)^n}{24}
\end{align}$
Tanks to maxima for the heavy lifting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| L'Hôpital's rule is not necessary. Here are the steps
$$\lim\limits_{x\to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
$$=\lim\limits_{x\to 2} \frac{8-x^2\sqrt{x+2}}{(x-2)(x+2)}$$
Let $t=\sqrt{x+2}$, then
$$\lim\limits_{t\to 2} \frac{8-\left(t^2-2\right)^2 t}{\left(t^2-4\right)t^2}$$
$$=\lim\limits_{t\to 2} \frac{\left(-t^5+4t^3-4t+8\right)}{(t-2)(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-(t-2)\left(t^4+2t^3+4\right)}{(t-2)(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-\left(t^4+2t^3+4\right)}{(t+2)t^2}$$
$$=\lim\limits_{t\to 2} \frac{-\left(t^2+2t+\frac{4}{t^2}\right)}{t+2}$$
$$=-\frac{\left(4+4+1\right)}{2+2}$$
$$=-\frac{9}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How to prove that the roots of this equation are integers? Let there be an equation $a^2 + 4ab + b^2 - 121 = 0$ where I want to prove that a,b are integers. Then I want to find whether there are integer values of $b$ for which $a$ is also an integer. Let us consider the case for $a$ $\rightarrow$ $a = \dfrac{-4b \pm \sqrt{16b^2 - 4(b^2-121)}}{2} = -2b \pm \sqrt{3b^2 +121}$ Thereby the problem reduces to showing that $3b^2 +121$ is a perfect square for certain integer values of $b$. This is what I cannot do.
Note: I know I can give examples and all for $3b^2 + 121$ to be a perfect square, but I'm looking for something with a little more substance. Thanks for the help.
| From $a^2+b^2+4ab-121=0$ we get $(a+b)^2+2ab=121$Or $2=\frac{11+a+b}{a} \frac{11-a-b}{b}$Now case 1: when the factors on right hand side are integers and coprime. Since 2 has only 2 factors, one of these has to be 1 and the other 2 Taking $\frac{11+a+b}{a}=2$ and $\frac{11-a-b}{b}=1$ we get $a=11 , b=0$
Similarly $\frac{11+a+b}{a}=1$ and $\frac{11-a-b}{b}=2$ gives $a=44$ and $b=-11$
I can't do the case where those factors are not integers or coprime right now (note that's the case of $a=5,b=4$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| If you set $u=x^3+1$ (so that $du = 3x^2dx$), then $x^2dx = \frac{du}{3}$ and
$$
\int \frac{x^2 dx}{x^3 +1} = \int \frac{\frac{du}{3}}{u} =\int \frac{du}{3u}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make sense:
Let $n = 1.$ Then $1 + a + a^2 + \ldots +1 < \frac {1 + a + a^2 + \ldots +1 + a}{2} = \frac{2 + 2a + a^2+ \ldots}{2} = 1 + a + \frac{a^2}{2} + \ldots$
Then I did this below, but it's unclear if the difference is positive:
$\frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1} - \frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} = \frac {n + na +na^2+ \ldots + na^{n - 1} + na^n - n - 1 - an - a- na^2 - a^2 - \ldots -na^{n - 1} - a^{n - 1}}{n(n + 1)} = \frac{n(1 + a + a^2+ \ldots + a^{n - 1} + a^n - 1 -a - a^2 -a^{n - 1}) - 1 - a - a^2 - \ldots - a^{n - 1}}{n(n+1)} $
$ = \frac{- 1 + a(n(a^{n - 1}) - 1 - a - \ldots - a^{n - 2})}{n(n+1)}$
What can I try now?
| Multiply both sides on $n(n-1)$ then you can see:
$$na^{n} > \sum_{i=0}^{n}a^{i}$$
Right side is geometric progression:
$$na^{n}>\frac{a^{n}-1}{a-1}$$
Now you can continue by yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$ using generating function I'm preparing to an exam and trying to solve $a_n=2 a_{n-1} - a_{n-2} + 2^n$, where $a_0=0$ and $a_1=1$.
This is my approach:
Let $A(z)=\sum_{n \geq 0} a_{n+2} z^{n+2}$, then:
$$\sum a_{n+2} z^{n+2} = 2 \sum a_{n+1} z^{n+2} - \sum a_n z^{n+2} + \sum (2z)^{n+2}$$
$$A(z)-a_0-a_1 z = 2z(A(z)-a_0)-z^2 A(z) + z^2 \frac{1}{1-2z}$$
$$A(z)(1-2z+z^2)=z+\frac{z^2}{1-2z}$$
$$A(z)=\frac{z(1-z)}{(1-2z)(1-2z+z^2)}=\frac{z}{(1-2z)(1-z)}=\frac{1}{1-2z}-\frac{1}{1-z}=\sum (\underbrace{2^n - 1}_{=a_n}) z^n$$
According to Wolfram, the result is different. Can you tell me where did I make the mistake?
| You have to be more careful with your indexing. I prefer to always cast the equation so that it is defined for $n \geq 0$ for greater consistency, but of course it doesn't matter as long as you keep track of your indices.
Anyway, the correct equation would be
$$A(z)(z^2-2z+1) = z + z^2 \frac{4}{1-2z}$$
if you defined $A(z) = \sum_{n=0}a_nz^n$. If you do the algebra you'll get
$$A(z) = \frac{4}{1-2z}+\frac{z-4}{(z-1)^2} = \frac{4}{1-2z}-\frac{1}{1-z}-3\frac{d}{dz}\left(\frac{1}{1-z}\right)$$
and hence
$$\sum_{n=0}a_nz^n = 4\sum_{n=0}2^nz^n-\sum_{n=0}z^n-3\sum_{n=1}nz^{n-1}$$
or
$$\sum_{n=0}a_nz^n = 4\sum_{n=0}2^nz^n-\sum_{n=0}z^n-3\sum_{n=0}(n+1)z^{n}=\sum_{n=0}(-3n+2^{n+2}-4)z^n,$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right or is there a simple trick?
| $$
n^3+11n
= 12 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
= 6\left(2 \binom{n}{1} + \binom{n}{2} + \binom{n}{3}\right)
$$
This is obtained by using repeated differences and Newton's interpolation formula:
$$
\begin{array}{llll}
0 & 12 & 30 & 60 & 108 & \\
12 & 18 & 30 & 48 & \\
6 & 12 & 18 & \\
6 & 6 & \\
0 & \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| Evaluate the equation for different $x$:
$$\begin{align}\\
\frac\pi6&\to -a+b+2c&+d&=\frac1{64}\\
\frac\pi4&\to -a-b+\ c&+d&=\frac8{64}\\
\frac\pi3&\to +a-b&+d&=\frac{27}{64}\\
\frac\pi2&\to -a+b-\ c&+d&=\frac{64}{64}\\
\end{align}$$
(after normalization of the coefficient of $d$).
Elimination by the combination $(1)+3\times(3)+2\times(4)$ yields
$$6d=\frac{210}{64}.$$
The evaluation for $x=\dfrac\pi4$ was not even necessary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find the limit as $(x,y,z) \to (0,0,0)$ of a specific function $f(x,y,z)$
find $$\lim_{(x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}} $$ if exist
$\sin$ is a continuous function so $\lim_{(x,y,z) \to (0,0,0)} \sin(x^2+y^2+z^2)=0$.
$\sqrt{x}$ is not defined for $x<0$ over $\mathbb{R}$ I need to find $x,y,z$ function such that $\sqrt{x^2+y^2+z^2}<0$ to prove there is no limit?
| HINT:
$$\frac{\sin(x^2+y^2+z^2)}{\sqrt{x^2+y^2+z^2}}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\cdot\sqrt{x^2+y^2+z^2}\;.$$
What do you know about $\lim\limits_{x\to 0}\dfrac{\sin x}x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$ without L'Hospital I need to evaluate
$$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$
I solved this using L'Hospital's Theorem and I got 4/3
However, is there a way to do this without applying this theoerm?
| Since $$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}+\ldots=\sum_{k=0}^{\infty}\frac{(-1)^kt^{2k+1}}{(2k+1)!}$$
we have
\begin{align}
\frac{\sin(4x)}{\sin(3x)}&=\frac{4x-\frac{(4x)^3}{3!}+\frac{(4x)^5}{5!}+\ldots}{3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+\ldots}\\
&=\frac{4-\frac{4^3}{3!}x^2+\frac{4^5}{5!}x^4+\ldots}{3-\frac{3^3}{3!}x^2+\frac{3^5}{5!}x^4+\ldots}\\
\lim_{x\to 0}\frac{\sin(4x)}{\sin(3x)}&=\frac{4}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve the following trigonometric equation using product cancellation Say I want to solve this equation using product cancellation:
$$\cos(2x+\frac {\pi}{2})\tan 3x = 0$$
The solutions are
$$\cos(2x+\frac {\pi}{2}) = 0 \space \vee \tan 3x = 0$$
Therefore
$$\cos(2x+\frac {\pi}{2}) = 0 \space \rightarrow 2x + \frac {\pi}{2} = \frac {\pi}{2} + k\pi \rightarrow x = \frac {\pi}{2} k$$
and
$$\tan3x = 0 \rightarrow 3x = k\pi \rightarrow x = \frac {\pi}{3}k$$
The solutions should be
$$x = \frac {\pi}{2}k \vee x=\frac{\pi}{3}k$$
The solution suggested by my book is this one
$$x = \frac {\pi}{3}k$$
Why is my solution different? Any hints?
| The problem is that while $$\lim_{x \to \frac{\pi}{2}} \left[ \cos\left( 2x+\frac{\pi}{2} \right) \right]=0$$
is true, $$\lim_{x \to \frac{\pi}{2}} \left[ \tan\left(3x\right) \right]=\infty$$
so we cannot not just say that the product is equal to zero. We can take the limit of the product,
$$\lim_{x \to \frac{\pi}{2}} \left[ \cos\left( 2x+\frac{\pi}{2} \right) \tan \left(3x\right)\right]=-\frac{2}{3}\not=0$$
Thus only $\frac{\pi}{3}k$ works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.