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Divisibility of integers Let $n > 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?
In general, for a fixed positive integer $a > 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?
|
The idea used below is very close to the one used by @Zander. It will not be a surprise to those who have seen my other posts that the details take longer.
If $n$ is even, then $2^n-1$ is divisible by $3$, so $2^n-1$ cannot divide $3^n-1$ unless $n=0$.
So let $n>1$ be odd. Let $p$ be a prime that divides $2^n-1$. Then since $2^n \equiv 1 \pmod{p}$, the order of $2$ modulo $p$ is odd, so $2$ is a quadratic residue of $p$. If furthermore $2^n-1$ divides $3^n-1$, then $3^n \equiv 1 \pmod {p}$, and therefore $3$ is also a quadratic residue of $p$.
The number $2$ is a quadratic residue of the odd prime $p$ iff $p\equiv \pm 1 \pmod{8}$. So $p$ must be of the shape $24k+1$, $24k+7$, $24k+17$, or $24k+23$. But by Quadratic Reciprocity, $3$ is a non-residue of $p$ if $p$ is of the shape $24k+7$ or $24k+17$. Thus it is enough to show that if $n$ is odd, then $2^n-1$ has at least one prime factor of the shape $24k+7$ or $24k+17$.
To do this, we show that not all primes in the prime factorization of $2^n-1$ can be of shapes $24k+1$ and/or $24k+23$. Suppose to the contrary that they all are. We will obtain a contradiction.
Note that $24k+1$ is congruent to $1$ modulo both $3$ and $8$, while $24k+23$ is congruent to $-1$ modulo both $3$ and $8$.
Since $2^n-1$ has shape $8s-1$, its prime factorization must have an odd number of occurrences of (not necessarily distinct) primes of the form $24k+23$. But that implies that $2^n-1\equiv -1 \pmod 3$, which is not the case when $n$ is odd.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Absolute maximum value of $\sin^2(x)-\sin(x)$ in $[0,\frac{3\pi}{2}]$ I thought it does not have absolute maximum, but wanted to just check and see why
|
A continuous function on a closed interval always has both an absolute max and an absolute min. And, the absolute max and min can only occur at critical points or the end points of the interval (I know of at least one book that includes end points as critical points). Therefore, the strategy that calculus books give is to find all critical points (in your interval) and plug these, and the end points, in to the original. The highest value you get is your absolute max and the smallest value you get is your absolute min (and you get the $x$ values where the absolute max and min occur, as the $x$ you plugged in). Since they must occur, and since these are the only possible places where they can occur, this guarantees we will find the absolute max and min if we do this.
If $f(x) = \sin^2 x - \sin x$, then $f'(x) = 2\sin x \cos x - \cos x = \cos x(2\sin x - 1)$. This is always defined so the only critical points will be where this is 0. A product is 0 exactly when one of the two parts is 0, so this is 0 when $\cos x = 0$ or when $\sin x = \frac{1}{2}$. In $\left[0, \frac{3\pi}{2} \right]$, the solutions would be: $\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3\pi}{2}$, $\sin x = \frac{1}{2}$ when $x = \frac{\pi}{6}$ and $x = 5\pi / 6$.
Now, plugging in all these values, and the end points gives:
$\begin{align*}
f(0) &= 0 \\
f(\pi/6) &= - \frac{1}{4} \\
f(\pi/2) &= 0 \\
f(5\pi/6) &= - \frac{1}{4} \\
f(3\pi/2) &= 2
\end{align*}$
Therefore, the max is 2 at $3\pi/2$ and the min is $-\frac{1}{4}$ at both $\pi/6$ and $5\pi/6$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How I can find the value of $abc$ using the given equations? If I have been given the value of
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried by squaring the equations.
But could not get it.
Thanks in advance.
|
First $\displaystyle{(a+b+c)^2 = (a^2+b^2+c^2)+2(ab+bc+ca)}$, which implies $1 = 9+2X$ where $X=(ab+bc+ca) \implies (ab+bc+ca)=-4$
Using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, and substituting the known values $1-3Y = 9+4$ and solve for $Y=abc$
Note: You can always check with Wolfram Alpha if your answer is correct (not to solve your problem) Check http://tinyurl.com/7n6ey2t
|
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|
The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\ 2x-3 = -(2x-3)$ when $2x-3$ $\geqslant$ $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$ $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.
|
*
*case - when $|2x-3| = -2x+3$ so than there will be $-2x+3=-2x+3$ --- from what will result $0=0$ so for this case the answer will be E).
2.case - when $|2x-3| = 2x-3$ so than there will be $2x-3=-2x+3$ --- 4x=6 --- $x=6/4 =3/2$ --- so in this case $x=3/2$
Note that because in your exercise , in this equation there is sign of equality and not is inequality for this you not can getting interval solutions
|
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|
Express some equations as polynomial equations Given
$$\begin{align*}
x&=(2+\cos(2s))\cos(3s)\\
y&=(2+\cos(2s))\sin(3s)\\
z&=\sin(2s),\end{align*}$$
I was wondering how to express these equations as polynomial equations in $x$, $y$, $z$, $a=\cos(s)$, $b=\sin(s)$.
Thanks!
Edit: I expect that the polynomial equations can give the same surface in $\mathbb R^3.$
|
$$\begin{align*}
x^2&=(2+\cos(2s)^2\cos(3s)^2\\
y^2&=(2+\cos(2s)^2\sin(3s)^2\\
x^2+y^2&=(2+\cos(2s))^2(\cos(3s)^2+sin(3s)^2)=(2+\cos(2s))^2\\
x^2+y^2&=(2+\cos(2s))^2=4+4\cos(2s)+\cos(2s)^2=4+4\cos(2s)+(1-\sin(2s)^2)\\
x^2+y^2&=5+4\cos(2s)-z^2\\
x^2+y^2+z^2&=5+4(\cos(s)^2-\sin(s)^2)\\
x^2+y^2+z^2&=5+4a^2-4b^2\\
,\end{align*}$$
$$x^2+y^2+z^2-4a^2+4b^2-5=0$$
|
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|
Progessive terms adding and minusing What is the value of
$$1+2+3-4+5+6+7-8+9+10+11+12...+97+98+99-100 \ ?$$
Any help is appreciated, thank you!
I added the terms as an AP then subtracted 10 then all the numbers that were missed out, not sure if this is right though.
|
If you rearrange the terms, you can write it as:
$1+2+3+5+6+7+9+10+...+98+99-4-8-...-100=1+2+3+4+...+100-2(4+8+12+...+100)=
1+2+...+100-2\cdot 4(1+2+...+25)=\frac{100(100+1)}{2}-2\cdot4\frac{25(25+1)}{2}$.
|
{
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|
limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$ Help me with that problem, please.
$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$
|
Note that for $x>0$ near $0$ we have
$$
\frac{1}{x^2} - \cot x = \frac{1}{x^2}-\frac{\cos x}{\sin x} \geq \frac{1}{x^2}-\frac{1}{\sin x} = \frac{\sin x - x^2}{x^2 \sin x}.
$$
Then we have
$$
\lim_{x\to 0^+}\frac{\sin x - x^2}{x^2 \sin x}
\ \operatorname*{=}^{\small\mathrm{L'H}}\ \lim_{x\to 0^+} \frac{\cos x-2x}{2x\sin x + x^2\cos x } = \infty
$$
so it follows by the squeeze theorem that
$$
\lim_{x\to 0^+}\left(\frac{1}{x^2} - \cot x \right) = \infty.
$$
For $x<0$ near $0$ we have $\cot x < 0$ so $$\frac{1}{x^2}-\cot x \geq \frac{1}{x^2} \to \infty$$
and so
$$
\lim_{x\to 0}\left(\frac{1}{x^2} - \cot x \right) = \infty.
$$
|
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|
Factorial Divisors Q. For what maximum value of $n$ will the expression $\frac{10200!}{504^n}$ be an integer? I have the solution to this question and I would like you to please go through the solution below. My doubt follows the solution :)
The solution can be found by writing $504 = 2^3 \cdot 3^2 \cdot 7$ and then finding the number of $2^3$s, $3^2$s and $7$s in the numerator, which can be obtained by
Number of $2$s = $\left\lfloor\frac{10200}{2}\right\rfloor + \left\lfloor\frac{10200}{2^2}\right\rfloor + \left\lfloor\frac{10200}{2^3}\right\rfloor + \dots + \left\lfloor\frac{10200}{2^{13}}\right\rfloor= 10192$
where $\left\lfloor\dots\right\rfloor$ is the floor function.
Therefore, the number of $2^3\textrm{s} = \left\lfloor\frac{10192}{3}\right\rfloor = 3397$
$\begin{align}\textrm{Similarly, the number of }3^2\textrm{s} &= 2457\\
\textrm{and the number of }7\textrm{s} & = 1698\end{align}$
The number of factors of $2^3 \cdot 3^2 \cdot 7$ is clearly constrained by the number of $7$s, therefore $n = 1698$.
My question is, whether there is any way I can simply look at the prime factors of the divisor and know which prime factor is going to be the constraining factor? (as $7$ was, in this particular example)
|
In this answer, it is shown that the number of factors of $p$ in $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.
Factor
$$
504=2^3\cdot3^2\cdot7\tag{2}
$$
Write $10200$ in base-$2$, base-$3$, and base-$7$:
$$
\begin{array}{}10011111011000_2&111222210_3&41511_7\end{array}\tag{3}
$$
The number of factors of $2$ in $10200!$ is $\frac{10200-8}{2-1}=10192$.
The number of factors of $3$ in $10200!$ is $\frac{10200-12}{3-1}=5094$.
The number of factors of $7$ in $10200!$ is $\frac{10200-12}{7-1}=1698$.
Since $\left\lfloor\frac{10192}{3}\right\rfloor=3397$, $\left\lfloor\frac{5094}{2}\right\rfloor=2547$, and $\left\lfloor\frac{1698}{1}\right\rfloor=1698$, the maximum value of $n$ so that $\frac{10200!}{504^n}$ is an integer is $n=1698$.
To answer the question asked:
For large $n$, the sum of the digits of $n$ is small compared to $n$, so suppose
$$
d=p_1^{e_1}p_2^{e_2}\dots p_m^{e_m}\tag{4}
$$
Following the computations above, the greatest power of $d$ that divides $n!$ is
$$
\min_k \left\lfloor\frac{n-\sigma_{p_k}(n)}{e_k(p_k-1)}\right\rfloor\tag{5}
$$
Ignoring $\sigma_{p_k}(n)$ as negligible, the greatest of $e_k(p_k-1)$ is a strong indicator of which $p_k$ is the constraining factor.
In the current case, the greatest of $3(2-1)=3$, $2(3-1)=4$, and $1(7-1)=6$ hints strongly that $7$ is the constraining factor.
|
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|
Find the value of : $\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$
Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$
So, I have an intermediate form of $\infty - \infty$ and I tried multiplying by the conjugate; however, I seem to be left with another intermediate form of $\frac{\infty}{\infty}$ and wasn't sure what else to to do. Is there anything else I can do other than L'Hopital's rule?
|
$$\begin{align}
\sqrt{4x^2 + 4} - (2x+2) & = \frac{(\sqrt{4x^2 + 4} - (2x+2))(\sqrt{4x^2 + 4} + (2x+2))}{\sqrt{4x^2 + 4} + (2x+2)}\\
& = \frac{4x^2 +4 - 4(x+1)^2}{\sqrt{4x^2 + 4} + (2x+2)} = - \frac{8x}{\sqrt{4x^2 + 4} + (2x+2)}\\
& = - \frac{8}{\sqrt{4 + 4/x^2} + 2 + 2/x}
\end{align}
$$
Now take the limit as $x \rightarrow \infty$ to get the limit as $-2$.
|
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|
Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show me how to do this step by step.
The answer from the book is:
$$ A_2=\sqrt\frac{A_1^2q^2}{2A_1^2gh+q^2} $$
The closest I can get is the following:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
$$ \frac{q^2}{A_1^2} = \frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Invert:
$$ \frac{A_1^2}{q^2} = \frac{(\frac{A_1}{A_2})^2-1}{2gh} $$
Multiply both sides by $2gh$:
$$ 2gh\frac{A_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
$$ \frac{2ghA_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
Add 1 to both sides and re-arrange:
$$ \frac{A_1^2}{A_2^2} = \frac{2ghA_1^2}{q^2} +1 $$
Invert again:
$$ \frac{A_2^2}{A_1^2} = \frac{q^2}{2ghA_1^2} +1 $$
Multiply by $A_1^2$:
$$ A_2^2 = \frac{A_1^2q^2}{2ghA_1^2} +1 $$
Get the square root:
$$ A_2 = \sqrt{\frac{A_1^2q^2}{2ghA_1^2}+1} $$
I cannot see where the $q^2$ on the bottom of the textbook answer comes from.
|
$$
\begin{eqnarray*}
q &=& A_1\sqrt{\frac{2gh}{(\frac{A_1}{A_2})^2-1} }&\biggr| : A_1, (\;\;)^2\\
\left(\frac{q}{A_1}\right)^2 &=& \frac{2gh}{(\frac{A_1}{A_2})^2-1}&\biggr| (\;\;)^{-1},\cdot 2gh,+1 \\
2gh\left(\frac{A_1}{q}\right)^2+1 &=& (\frac{A_1}{A_2})^2&\biggr| (\;\;)^{-1/2},\cdot A_1\\
\pm\frac{A_1}{\sqrt{2gh\left(\frac{A_1}{q}\right)^2+1}} &=& A_2\\
\pm\sqrt{\frac{A_1^2q^2}{2ghA_1^2+q^2}} &=& \\
\end{eqnarray*}
$$
|
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|
Proving that: $\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$ Let $a$ and $b$ be positive reals. Show that
$$\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}$$
|
An elementary proof.
We use the Taylor series $e^x = 1 + x + O(x^2)$ and the fact that $\lim_{n\to\infty}(1+x/n)^n = e^x$.
If $a=b$ the identity is trivial.
Without loss of generality, assume $0<a<b$.
Then
$$\begin{eqnarray*}
\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n
&=& b \left(\frac{1+(\frac{a}{b})^{1/n}}{2}\right)^n \\
&=& b \left(\frac{1+e^{\frac{1}{n}\ln \frac{a}{b}}}{2}\right)^n \\
&=& b \left(1+\frac{1}{2}\frac{1}{n}\ln \frac{a}{b} + O(1/n^2)\right)^n \\
&=& b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n + O(1/n).
\end{eqnarray*}$$
Therefore,
$$\begin{eqnarray*}
\lim_{n\to\infty} \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n
&=& \lim_{n\to\infty} b \left(1+\frac{1}{n}\ln \sqrt{\frac{a}{b}}\right)^n \\
&=& b e^{\ln \sqrt{\frac{a}{b}}} \\
&=& \sqrt{a b}.
\end{eqnarray*}$$
|
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|
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
|
I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available.
Apply the double angle formula
$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$
to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$,
with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$
for convenience, recalling also that
$\cos(\pi\pm\theta)=-\cos\theta$:
$$
b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1
$$
$$
-a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1
$$
Next, subtracting the equations
$$
\matrix{
2a^2=1+b\\
2b^2=1-a}
$$
we get
$$
\eqalign{
2\left(a^2-b^2\right)&=b+a\\
2\left(a+b\right)\left(a-b\right)&=b+a\\
2\left(a-b\right)&=1\\
a-b&=\frac12\,.
}
$$
Furthermore, multiplying, we get
$$
4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab
$$
giving us the quadratic equation
$$4(ab)^2+(ab)-\tfrac12=0$$
$$8(ab)^2+2(ab)-1=0$$
$$\left(4ab-1\right)\left(2ab+1\right)=0$$
so that $ab=\frac14$ or $-\frac12$,
from which we can choose the former
since we know that $0<a<b$.
|
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|
How to transform the factored form of $\sin(x)$? We know $\sin(x)=0$ has solutions $0,\pm\pi,\pm2\pi,\pm3\pi,\dots$.
So $\sin(x)$, if interpreted as a polynomial, could be written as:
$a_0x^0+a_1x^1+a_2x^2+\cdots$ and we know this polynomial too:
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$
So, the question is, is it possible to transform the factored form of $\sin(x)$:
$$\sin(x)=a x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\dots$$
to
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\ ?$$
|
The proposal is:
$$\sin x = a(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots$$
The standard result, already posted by Sam, is in effect
$$
\sin x = \frac{x-\pi}{\pi}\cdot \frac{x+\pi}{\pi} \cdot \frac{x - 2\pi}{2\pi}\cdot \frac{x+2\pi}{2\pi}\cdot\frac{x-3\pi}{3\pi}\cdot\frac{x+3\pi}{3\pi} \cdots
$$
So the coefficient "$a$" in front of the whole thing is more . . . . interesting . . . than might be initally guessed. Might Euler have considered
$$
a = \frac{1}{\pi^2}\cdot\frac{1}{(2\pi)^2}\cdot\frac{1}{(3\pi)^2}\cdots
$$
to be some sort of "infinitely small number"? Might it actually be fruitful in some way to think of it that way?
|
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|
Finding the linear combinations of two vectors I am studying for my finals and I'm trying to answer the following question:
Consider the following two vectors in $\mathbb{R}^3$: $a=(1,2,3)$ and $b=(2,3,1)$. Decide
whether it is possible to express the vector $c=(2,4,5)$ as a linear
combination of $a$ and $b$.
I have used the following row operations to find what $C_1$,$C_2$ and $C_3$ are equal to:
\begin{align*}
\left[ \begin{array}{ccc|c} 1 & 2 &2 & 2\\
2 & 3 & 4 & 4 \\
3 & 1 & 5 & 5 \end{array}\right]
&\overset{R_2 = 2R_1 -R_1}{\Longrightarrow}
\left[ \begin{array}{ccc|c}
1 & 2 &2 &2 \\
0 & 1 & 0 & 0 \\
3 & 1 & 5 & 5
\end{array}\right]
\overset{R_3=3R_1-R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 &2 & 2 & 2 \\
0 &1 & 0 & 0 \\
0 & 5 &1 & 1
\end{array} \right]
\\
&\overset{R_3=5R_2+R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 & 2 & 2 & 2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1
\end{array}\right]
\end{align*}
Is the following argument correct?
By looking at the last row of the matrix, we see that it says $0 = 1$, which is impossible, and the system of equations therefore has no solutions. This means that we can not find the values for $C_1$ and $C_2$, and so $c$ can not be written as a linear combination of $a$ and $b$.
Thanks in advance!
|
$c=m\cdot a+n \cdot b$
Hence :
$(2,4,5)=m\cdot(1,2,3)+n\cdot(2,3,1)$
So we have following system of equations :
$\begin{cases}
m+2n=2 \\
2m+3n=4 \\
3m+n=5
\end{cases}$
which has no solution ,therefore you cannot express $c$ as linear combination of $a$ and $b$ .
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$ $$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$$
Tried substitution ($u = \cos\frac{x}{2}$), but I get $-\frac{\cos^3\frac{x}{2}}{3}$ ($-\frac{2}{3}$) instead of the correct answer, which is $1\frac{1}{3}$
|
You can successively reduce this to simple integrals using sum formulae,
$$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2} dx = \frac{1}{2}\int_0^{2\pi}\sin(x)\cos\frac{x}{2} dx = \frac{2}{4}\int_0^{\pi}\left(\sin\frac{3x}{2} + \sin\frac{x}{2}\right) dx$$
This will give you, $\frac{1}{2} \times \frac{8}{3} = \frac{4}{3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is this equation correct? There is this equation:
$$M = \sum\limits_{i = 1}^{\log n} {\frac{{in}}{{{2^i}}}} = n\sum\limits_{i = 1}^{\log n} {\frac{i}{{{2^i}}}} \leqslant n\sum\limits_{i = 1}^\infty {\frac{i}{{{2^i}}} = 2n} $$
And I don't understand why the rightmost summation can be simplified to 2n. Can you please explain it to me?
|
Knowing that:
$$ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1 $$
And hence by removing successive terms from the left:
$$ \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{2} $$
$$ \frac{1}{8} + \frac{1}{16} + \cdots = \frac{1}{4} $$
$$ \vdots $$
We have:
$$\begin{eqnarray*}
\sum_{i=1}^\infty \frac{i}{2^i} &=& \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots \\
&=& (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots) + (\frac{1}{8} + \frac{1}{16} + \cdots) + \cdots \\
&& \text{Substituting the identites from above:} \\
&=& 1 + \frac{1}{2} + \frac{1}{4} + \cdots \\
&& \text{And substituting the first identity again:} \\
&=& 1 + 1 \\
&=& 2
\end{eqnarray*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\int \sqrt{a^2-x^2} dx$, where $a$ is a constant How to find $$\int \sqrt{a^2-x^2} dx\;,$$ where $a$ is a constant?
It appears to be $$\frac{\pi a}{2}\;,$$ but how do I get there?
|
One way of finding the definite integral is to consider the shape of the graph. If $y = \sqrt{a^2-x^2}$ then $y^2=a^2-x^2$, so $x^2+y^2=a^2$. Apply the Pythagorean theorem: that's the equation of a circle of radius $a$. $y = \pm\sqrt{a^2-x^2}$ is the whole circle; $y = \sqrt{a^2-x^2}$ is the top half of the circle. If you know that the area of the whole circle is $\pi a^2$, then the area of the top half is
$$
\int_{-a}^a \sqrt{a^2-x^2} \, dx = \frac{\pi a^2}{2}.
$$
Generally, if you have
$$
a^2-x^2
$$
in an integral, you can use $x = a\sin\theta$ and $dx = a\cos\theta\,d\theta$, and then $a^2-x^2$ becomes $a^2\cos^2\theta$.
If you have
$$
a^2 + x^2
$$
then you can use $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$, $a^2+x^2=a^2\sec^2\theta$.
If you have
$$
x^2 - a^2
$$
then you can use $x=a\sec\theta$, $dx=a\sec\theta\tan\theta\,d\theta$, $x^2 - a^2 = a^2\tan^2\theta$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$
Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$
The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.
|
Let $m, n > 0$. Then observe that
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx$$
is the area of the region given by inequalities
$$ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq \sqrt[n]{1-x^m}.$$
But the last inequality is equivalent to $0 \leq x^m + y^n \leq 1$. Thus
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = [\text{Area given by} \ 0 \leq x^m + y^n \leq 1, \ 0 \leq x, y \leq 1 ]$$
Thus by interchanging the role of $x$ and $y$, we have
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = \int_{0}^{1} \sqrt[m]{1-x^n} \; dx.$$
Of course, we can give a purely analytic approach. Let $y = \sqrt[3]{1 - x^7}$. Then $x = \sqrt[7]{1 - y^3}$ and hence by integration by substitution,
$$\begin{align*}
\int_{0}^{1} \sqrt[3]{1 - x^7} \; dx
&= \int_{0}^{1} y(x) \; dx \\
&= \int_{1}^{0} y \; dx(y) \\
&= [y x(y)]_{1}^{0} - \int_{1}^{0} x(y) \; dy \\
&= \int_{0}^{1} \sqrt[7]{1 - y^3} \; dy.
\end{align*}$$
|
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|
Represent lengths rectangle using given terms In a rectangle, $GHIJ$, where $E$ is on $GH$ and $F$ is on $JI$ in such a way that $GEIF$ form a rhombus. Determine the following: $1)$ $x=FI$ in terms of $a=GH$ and $b=HI$ and $2)$calculate $y=EF$ in terms of $a$ and $b$.
|
We have $GHIJ$ is rectangle and $GEIF$ is a rhombus and also $GH = a = IJ$ and $HI = b=GJ$
We have to find $x=FI$ (side of rhombus) and $y=EF$ (one of the diagonals of rhombus)
(Here I have to draw picture of your problem. I know diagram. But i am not able to draw a picture in mac OS X. You can draw diagram easily)
$$x= FI = EI = GE = FG$$ (since sides of the rhombus are equal)
$$EH = GH-GE = a-x$$
$\triangle EHI$ is right angle triangle. That means
$$\begin{align*}(EI)^2 &= (EH)^2 + (IH)^2\\
X^2 &= (a-x)^2 + b^2\end{align*}$$
From above, we will get $x = \dfrac{a^2 + b^2}{2a}$
Draw rectangle $CEDF$ such that $GE\perp FC$ and $ED\perp FI$.
$EF$ is the diagonal of rectangle $CEDF$ and also $ED = HI = b$
$$FD = FI-DI = FI-(EH)$$ (since $DI = EH$)
$$ x - a + x = 2x - a = \frac{a^2 + b^2}{a} - a = \frac{b^2}{a}$$
$\triangle EDF$ is right angle triangle.
That means
$$\begin{align*}y^2 &= (EF)^2 = (ED)^2 + (FD)^2\\
&= b^2 + \left({\frac{b^2}{a^2}}\right)^2\\
&= \frac{b^2(a^2 + b^2)}{a^2}
\end{align*}$$
then you can easily get value of y.
|
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|
Finding the integral of $(1+2^{2x})/2^x$
Evaluate the integral:
$$\int\frac{1+2^{2x}}{2^x}\,dx = \int \frac{ 1 + (2^x)^2}{2^x}\,dx$$
Let $u = 2^x$. Then $du = 2^x\ln2\,dx$, which yields $\frac{du}{2^x\ln2} = dx$ so
$$ \int \frac{ 1 + (2^x)^2}{2^x}\,dx = \int \frac{1+u^2}{u}du =
\left( x+ \frac{u^3}{3} \right)\ln u+C$$
$$=\left(x+ \frac{(2^x)^3}{3} \right)\ln 2^x +C$$
I'm not sure if I integrated this correctly. Any help would be appreciated.
|
Your change of variable is fine; your substitution is not quite right and your integral is not quite right. If $u=2^x$, then $du = 2^x\ln(2)\,dx = u\ln(2)\,dx$, so $dx = \frac{du}{\ln(2)u}$. So we have:
$$\begin{align*}
\int \frac{1+2^{2x}}{2^x}\,dx &= \int\frac{1 + (2^x)^2}{2^x}\,dx\\
&= \int\left(\frac{1 + u^2}{u}\right)\frac{du}{\ln(2)u}\\
&= \frac{1}{\ln(2)}\int\frac{1+u^2}{u^2}\,du\\
&=\frac{1}{\ln(2)}\int\left(u^{-2} + 1\right)\,du\\
&=\frac{1}{\ln(2)}\left(-u^{-1} + u\right) + C\\
&= \frac{1}{\ln(2)}\left(2^x - \frac{1}{2^x}\right) + C
\end{align*}$$
|
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|
How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$ I need to calculate the length of a curve $y=2\sqrt{x}$ from $x=0$ to $x=1$.
So I started by taking $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$, and then doing substitution: $\left[u = 1+\frac{1}{x}, \text{d}u = \frac{-1}{x^2}\text{d}x \Rightarrow -\text{d}u = \frac{1}{x^2}\text{d}x \right]^1_0 = -\int\limits^1_0 \sqrt{u} \,\text{d}u$ but this obviously will not lead to the correct answer, since $\frac{1}{x^2}$ isn't in the original formula.
Wolfram Alpha is doing a lot of steps for this integration, but I don't think that many steps are needed.
How would I start with this integration?
|
Put $x= \tan^{2}\theta$, then you have the integral as
\begin{align*}
\int_{0}^{1} \sqrt{1+\frac{1}{x}} \ dx &= \int_{0}^{\pi/4} \sqrt{\frac{1+\tan^{2}\theta}{\tan^{2}\theta}} \cdot 2\tan\theta \cdot\sec^{2}\theta \ d\theta \\\ &= 2 \cdot\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta
\end{align*}
Now integrate this function by parts. Take $u = \sec\theta$ then $du = \sec\theta \cdot \tan\theta$ and $dv = \sec^{2}\theta$. Then you have $v = \tan\theta$, so
\begin{align*}
\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta &= (\sec\theta\cdot\tan\theta)\:\biggl|_{0}^{\pi/4} - \int_{0}^{\pi/4} \sec\theta \cdot \tan^{2}\theta \ d\theta \\\ &= \frac{1}{\sqrt{2}} -\int_{0}^{\pi/4} \sec^{3}\theta \ d\theta + \int_{0}^{\pi/4} \sec\theta \ d\theta \\\ &= \frac{1}{2} \cdot \biggl\{ \frac{1}{\sqrt{2}} + \int_{0}^{\pi/4} \sec\theta \ d\theta \:\biggr\} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \bigl(\:\log(\sec\theta +\tan\theta)\bigr)_{0}^{\pi/4} \\\ &= \frac{1}{2\sqrt{2}} + \frac{1}{2} \cdot \log(\sqrt{2}+1)
\end{align*}
|
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|
Summing $\sum_{n=1}^{\infty} \sin\frac{n!\pi}{120}$ How do I sum $$\sum_{n=1}^{\infty} \sin\frac{n!\pi}{120}$$
|
Note that $\sin \left(\dfrac{n! \pi}{120} \right) = 0$ for all $n \geq 5$. Hence, $$\sum_{n=1}^{\infty} \sin \left(\dfrac{n! \pi}{120} \right) = \sin \left(\dfrac{1! \pi}{120} \right) + \sin \left(\dfrac{2! \pi}{120} \right) + \sin \left(\dfrac{3! \pi}{120} \right) + \sin \left(\dfrac{4! \pi}{120} \right)$$
|
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|
How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$
|
Now with $n = 5$ and $k = 5$.
With $n = 5$ and $k = 5$ (missing a $9$ for now but I'll come back to it later).
$\dfrac{55}{5}-5-5=1$
$\dfrac{5+5}{5}-5+5=2$
$\dfrac{5+5}{5}+\frac{5}{5}=3$
$\dfrac{5+5+5+5}{5}=4$
$5 - 5 + 5 - 5 + 5 = 5$
$5 + \dfrac{5}{5} - 5 + 5 = 6$
$5 + \dfrac{5}{5}+\dfrac{5}{5} = 7$
$5 + 5 - \dfrac{5+5}{5} = 8$
$5 + \dfrac{5(5) - 5}{5}=9$
$\dfrac{55}{5} - \dfrac{5}{5} = 10$
$\dfrac{55}{5} - 5 + 5 = 11$
$\dfrac{5+5}{5} + 5 + 5 = 12$
Thanks to tzador for $9$.
|
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|
Summation of $ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$ till $n$ terms What is the pattern in the following?
*
*Sum to $n$ terms of the series: $$ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$$
|
So, what you want is basically $$\left(1 - \frac {1} {2}\right) + \left(1 - \frac {1} {4}\right) + \cdots + \left(1 - \frac {1} {2^n}\right) = n - \left(\frac {1} {2} + \frac {1} {4} + \cdots + \frac {1} {2^n}\right) = n - 1 + \frac {1} {2^n}.$$
|
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|
Integral of $\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$ I am trying to find $$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
$t = \sec \theta$ $dt = \sec \theta \tan\theta $
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \sqrt{\sec^2 \theta-1}}$$
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{\sec \theta \tan\theta}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{1}{\sec \theta}$$
$$\int_\sqrt{2}^2 \cos \theta$$
$$\sin \theta$$
Then I need to make it in terms of t.
$t = \sec \theta$
So I just use the arcsec which is
$\theta =\operatorname{arcsec} t$
$$\sin (\operatorname{arcsec} t)$$
This is wrong but I am not sure why.
|
$$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
Substitute :
$t = \sec \theta$
$dt = \sec \theta \tan\theta d\theta$
The limits will also change accordingly
When $t=2$ , $\theta = \ arc sec(2) = \frac{\pi}{3}$
When $t=\sqrt2$ , $\theta = \ arc sec(\sqrt2) = \frac{\pi}{4}$
$$=\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{\sec \theta \tan\theta d\theta}{\sec ^2 \theta \tan \theta}$$
$$=\int_\frac{\pi}{4}^\frac{\pi}{3} \cos \theta d \theta$$
$$=\sin \frac{\pi}{3} - \sin \frac{\pi}{4}$$
$$=\frac{\sqrt3 - \sqrt2}{2}$$
I think the answer you got is also correct
$$\sin (\operatorname{arcsec} (t))$$
by applying simple trigonometric rules
$$=\frac{\sqrt{t^2-1}}{t}$$
and then applying the limits we get the same answer
$$=\frac{\sqrt3 - \sqrt2}{2}$$
|
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|
Evaluating $\int_{0}^{1} \frac{x^{2} + 1}{x^{4} + 1 } \ dx$ How do I evaluate $$\int_{0}^{1} \frac{x^{2} + 1}{x^{4} + 1 } \ dx$$
I tried using substitution but I am getting stuck. If there was $x^3$ term in the numerator, then this would have been easy, but this one doesn't.
|
Another way, if you want to sweat harder instead of the elegant suggestion of Chandrasekhar:$$x^4+1=(x^2+\sqrt{2}\,x+1)(x^2-\sqrt{2}\,x+1)\Longrightarrow$$$$ \frac{x^2+1}{x^4+1}=1-\frac{\sqrt 2\,x}{x^2+\sqrt 2\,x+1}+1+\frac{\sqrt 2\,x}{x^2-\sqrt 2\,x+1}$$ so for example$$\int\frac{\sqrt 2\,x}{x^2+\sqrt 2\,x+1}dx=\frac{1}{\sqrt 2}\int\frac{2x+\sqrt 2}{x^2+\sqrt 2\,x+1}dx-\frac{1}{2\sqrt 2}\int\frac{\sqrt 2dx}{(\sqrt 2 x+1)^2+1}=$$$$=\frac{1}{\sqrt 2}\log|x^2+\sqrt 2\,x+1|-\frac{1}{2\sqrt 2}\arctan(\sqrt 2\,x+1)+C$$ and etc.
|
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|
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
|
Let us consider $$A=\iint_{[0,1]^2}\frac{x}{(1+xy)(1+x^2)}dx dy$$
By Fubini's theorem, we have : $$A=\int_0^1\left[\frac{1}{1+x^2}\int_0^1\frac{x\,dy}{1+xy}\right]dx=\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$ and$$A=\int_0^1\left[\int_0^1\frac{x}{(1+xy)(1+x^2)}dx\right]dy$$But$$\frac{x}{(1+xy)(1+x^2)}=\frac{1}{1+y^2}\left(\frac{-y}{1+xy}+\frac{x+y}{1+x^2}\right)$$and therefore$$A=\int_0^1\frac{1}{1+y^2}\left(-\ln(1+y)+\frac{\ln(2)}{2}+\frac{\pi y}{y}\right)dy=-A+\frac{\pi\ln(2)}{4}$$Finally :$$\boxed{\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln(2)}{8}}$$
|
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|
Evaluate the integral $H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$
*
*$y\geq0$ define $$H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$$ Show that $H$ is a continuous function of $y$ and show $\lim\limits_{y \to +\infty}H(y)=0$.
|
This is kind of a brute force method where we explicitly find the function $f(a)$.
$$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$
This can be obtained as shown below. We have that for $a>0$, $$\dfrac1{x^4+ax} = \dfrac1{x(x^3+a)} = \dfrac1{ax} - \dfrac{x^2}{a(a+x^3)}$$
Hence, $$f(a) = \int_1^{\infty} \dfrac{dx}{x^4+ax} = \int_1^{\infty} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx = \lim_{R \rightarrow \infty} \int_1^{R} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx$$
The first integral $$I_1 = \int_1^{R} \dfrac{dx}{ax} = \dfrac{\log(R)}a.$$
The second integral $$I_2 = \dfrac1{3a} \int_1^{R} \dfrac{3x^2dx}{(a+x^3)} = \left. \dfrac1{3a} \log(a+x^3) \right \rvert_{1}^{R} = \dfrac{\log(a+R^3) - \log(a+1)}{3a}$$
Putting these together, we get that
\begin{align}
f(a) & = I_1 - I_2\\
& = \lim_{R \rightarrow \infty} \left(\dfrac{\log(R)}a - \left( \dfrac{\log(a+R^3) - \log(a+1)}{3a}\right) \right)\\
& = \lim_{R \rightarrow \infty} \dfrac{\log(R^3)-\log(a+R^3) + \log(a+1)}{3a}\\
& = \lim_{R \rightarrow \infty} \dfrac{\log \left(\dfrac{R^3}{a+R^3} \right) + \log(a+1)}{3a}\\
& = \dfrac{\log (1) + \log(a+1)}{3a}\\
& = \dfrac{\log(a+1)}{3a}\\
\end{align}
If $a=0$, then $f(0) = \displaystyle \int_1^{\infty} \dfrac{dx}{x^4} = \dfrac13$. Hence, we have that $$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$
Clearly, $f$ is a continous function of $a$ for all $a \geq 0$ and $\lim_{a \rightarrow \infty} f(a) = 0$.
|
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|
Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form As the title says, I'm trying to transform $\displaystyle{n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ into a closed form. My work:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \displaystyle\sum_{p=0}^{n}\binom{n}{p}\exp\left(\frac{p\cdot2i\pi}{5} \right)$
$\displaystyle=\binom{n}{0} + \binom{n}{1}\exp\left(\frac{1\cdot2i\pi}{5} \right) + \binom{n}{2}\exp\left(\frac{2\cdot2i\pi}{5} \right) + \binom{n}{3}\exp\left(\frac{3\cdot2i\pi}{5} \right) + \binom{n}{4}\exp\left(\frac{4\cdot2i\pi}{5} \right) + \binom{n}{5} + \cdots = \left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] + \exp\left(\frac{2i\pi}{5} \right)\left[\binom{n}{1} + \binom{n}{6} + \binom{n}{11} \right ] + \exp\left(\frac{4i\pi}{5} \right)\left[\binom{n}{2} + \binom{n}{7} + \binom{n}{12} \right ] + \exp\left(\frac{6i\pi}{5} \right)\left[\binom{n}{3} + \binom{n}{8} + \binom{n}{13} \right ] + \exp\left(\frac{8i\pi}{5} \right)\left[\binom{n}{4} + \binom{n}{9} + \binom{n}{14} \right ] + \cdots$
I'll recall $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = k$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = u$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = v$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = w$ and $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = z$. Thus
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + u\cdot\exp\frac{2i\pi}{5} + v\cdot\exp\frac{4i\pi}{5} + w\cdot \exp\frac{6i\pi}{5} + z\cdot\exp\frac{8i\pi}{5} = k + u\cdot\left (\cos\frac{2\pi}{5} + i\cdot\sin\frac{2\pi}{5} \right ) + v\cdot\left (\cos\frac{4\pi}{5} + i.\sin\frac{4\pi}{5} \right ) + w\cdot\left (\cos\frac{6\pi}{5} + i.\sin\frac{6\pi}{5} \right ) + z\cdot\left (\cos\frac{8\pi}{5} + i.\sin\frac{8\pi}{5} \right )$
Noting that $\cos\frac{2\pi}{5} = \cos\frac{8\pi}{5}$, $\cos\frac{4\pi}{5} = \cos\frac{6\pi}{5}$, $\sin\frac{2\pi}{5} = -\sin\frac{8\pi}{5}$ and $\sin\frac{4\pi}{5} = -\sin\frac{6\pi}{5}$:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + \left(u + z\right)\cos\frac{2\pi}{5} + i\cdot\left(u - z \right)\sin\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} + i\cdot\left(v - w \right)\sin\frac{4\pi}{5} = \left(k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5}\right) + i\cdot\left(\left(u - z \right)\sin\frac{2\pi}{5} + \cdot\left(v - w \right)\sin\frac{4\pi}{5} \right)$
But $\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \left(2\cos\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{i\pi}{5}\right)\right)^n = \left(2^n\cos^n\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\cos\frac{n\pi}{5} + i.\sin\frac{n\pi}{5} \right ) = \left(2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5} \right) + i\cdot \left(2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5} \right)$
So,
$\displaystyle k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5}$
and
$\displaystyle\left(u - z \right)\sin\frac{2\pi}{5} + \left(v - w \right)\sin\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5}$
and I'm stuck here. I noted that $k + u + v + w + z = 2^n$ but I couldn't isolate $k$. So, any help finishing this result will be fully appreciated. Thanks.
|
$$
\color{red}{\mathbf 5}\cdot\sum_{k\geqslant0}{n\choose \color{red}{\mathbf 5}k}=\sum_{\ell=1}^{\color{red}{\mathbf 5}}\left(1+\mathrm e^{2\mathrm i\pi\ell/\color{red}{\mathbf 5}}\right)^n=\sum_{\zeta\in\mathbb C\,:\,\zeta^\color{red}{\mathbf 5}=1}\left(1+\zeta\right)^n
$$
|
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|
What other substitutions could I use to evaluate this integral? Consider the integral
$$ \int x^2\sqrt{2 + x} \, dx$$
I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover both terms in the integrand, as well as the composed function inside the root.
$\boxed{\text{Let }u = (2 + x)}$
$$u = 2 + x$$
$$du = 1 \space dx$$
$$dx = du$$
$$\int x^2 \sqrt{u} \space du$$
$\boxed{\text{Let } u = \sqrt{2+x}}$
$$ u = \sqrt{2 + x}$$
\begin{align*} du &= \frac{1}{2}(2 + x)^{-\frac{1}{2}} \space dx \\
&= \frac{1}{2\sqrt{2 + x}} \space dx
\end{align*}
$$ dx = 2\sqrt{2 + x} \space du$$
$$ 2\int ux^2\sqrt{2 + x} \space du $$
$\boxed{\text{Let }u = x^2}$
$$ u = x^2$$
$$ du = 2x \space dx$$
$$ dx = \frac{1}{2x} \space du$$
$$ \frac{1}{2} \int \frac{u\sqrt{2 + x}}{x} \space du$$
As you can see, none of these allow me to move forward into integrating with respect to $u$. What other substitutions can I use that would help me move on with this integral?
|
Your first substitution should work. That is let $u=2+x$, then $ x= u-2$ and thus $$x^2 = (u-2)^2.$$ You should get $$ \int \left(u^{5/2}-4u^{3/2}+4u^{1/2}\right)~du. $$
The second substitution also works. Let $u =\sqrt{2+x}$. So $u^2 = 2+x$. Then $x^2 = (u^2-2)^2$. Putting everything together gives you $$\int 2\left(u^6-4u^4+4u^2\right)~du.$$
|
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|
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
|
$AM-GM$:
$\frac{a^2}{b-1}+4(b-1)\geq 4a$
$\frac{b^2}{a-1}+4(a-1)\geq 4b$
$\Rightarrow \frac{a^2}{b-1}+4(b-1)+\frac{b^2}{a-1}+4(a-1)\geq 4a+4b$
$\Rightarrow \frac{a^2}{b-1}+\frac{b^2}{a-1}\geq 8$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Under what situations does $x+1$ divide $4n^2-x$? I am looking at the equation
$$\frac{4n^2-x}{x+1} = y$$
for even $x$ and $y$, both positive. Under what situations does $x+1$ divide $4n^2-x$?
|
I will throw another wooden nickel on the fire.
Since $\dfrac{4n^2-x}{x+1}+1=\dfrac{4n^2+1}{x+1}$, we get that the question is equivalent to asking when
$$
x+1\,|\,4n^2+1\tag{1}
$$
Trivially, $x=0$ and $x=4n^2$ satisfy $(1)$, and these are the only solutions when $4n^2+1$ is prime. Let's look for less trivial solutions. That is, let us look at when $4n^2+1$ is composite.
Suppose there is a prime $p$ so that
$$
p\,|\,4n^2+1\tag{2}
$$
$4n^2+1$ is odd, so $p$ must be odd. Furthermore, condition $(2)$ says that $-1\equiv(2n)^2\pmod{p}$; that is, $-1$ is the square of some number $\bmod{p}$ (usually said as "$-1$ is a quadratic residue $\bmod{p}$").
Intro to Quadratic Residues
Suppose that $p$ is an odd prime and $a\not\equiv0\pmod{p}$ is a quadratic residue $\bmod{p}$. There is an $x$ so that $x^2\equiv a\pmod{p}$. First, note that $x\not\equiv0\pmod{p}$. Therefore, by Fermat's Little Theorem,
$$
a^{\frac{p-1}{2}}\equiv x^{p-1}\equiv1\pmod{p}\tag{3}
$$
Consider the equation
$$
(x^{\frac{p-1}{2}}+1)(x^{\frac{p-1}{2}}-1)=x^{p-1}-1\equiv0\pmod{p}\tag{4}
$$
Since $\mathbb{Z}_p$ is a field, there are at most $\frac{p-1}{2}$ solutions to each of
$$
x^{\frac{p-1}{2}}+1\equiv0\pmod{p}\quad\text{and}\quad x^{\frac{p-1}{2}}-1\equiv0\pmod{p}\tag{5}
$$
However, Fermat's Little Theorem says there are $p-1$ solutions to $(4)$. Therefore, there are exactly $\frac{p-1}{2}$ solutions to each relation in $(5)$.
Consider the set of non-zero, quadratic residue classes, $Q$. For each $a\in Q$, there are exactly two solutions for $x^2=a$ ($x$ and $-x$). Thus, $|Q|=\frac{p-1}{2}$. For each $a\in Q$, we showed in $(3)$ that $a^{\frac{p-1}{2}}\equiv1$. The pigeonhole principle and $(5)$ say that for the $\frac{p-1}{2}$ non-zero, non-quadratic residue classes $\bmod{p}$, we have $a^{\frac{p-1}{2}}\equiv-1\pmod{p}$.
Thus, we have shown that for an odd prime $p$,
$$
a^{\frac{p-1}{2}}\equiv\left\{\begin{array}{}+1\pmod{p}&\text{if }a\text{ is a quadratic residue }\bmod{p}\\-1\pmod{p}&\text{if }a\text{ is not a quadratic residue }\bmod{p}\end{array}\right.\tag{6}
$$
For example, $-1$ is a quadratic residue $\bmod{p}$ if and only if $(-1)^{\frac{p-1}{2}}\equiv1\pmod{p}$; that is, if and only if $p\equiv1\pmod{4}$.
Therefore, if $x+1$ is divisible by any prime $p\not\equiv1\pmod{4}$, then $\dfrac{4n^2-x}{x+1}$ cannot be an integer.
If $4n^2+1$ is prime, there are only trivial solutions, and if $4n^2+1$ is composite, then $x+1$ can be any factor of $4n^2+1$.
|
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|
Two algebra questions I have two questions that I need help with:
1) How many single digit even natural number solutions are there for the equation $A+B+C+D = 24$ such that $A+B > C+D$
A)20 B)11 C)16 D)24
2) Three positive real numbers $x,y,z$ are such that $x+y+Z = 1$. which of the following inequalities best discribe the relation between $XY,YZ,ZX$.
A) $xy+yz+zx > 1/3$
B) $xy+yz+zx <1/3$
C) $xy+yz+zx \le 1/3$
D) $xy+yz+zx \le 2/3$
|
For the first question, you have on the one hand that $(A+B)+(C+D)=24$ and on the other hand that $A+B>C+D$. What are the possibilities for $A+B$ and $C+D$? Clearly you must have $A+B>12$ and $C+D<12$. If two single-digit natural numbers add up to $11$ or less, what are the possibilities? (Note that the answer will depend on whether $0$ counts as a natural number.)
The second question is harder. Because it’s multiple choice, however, you can solve it by elimination. First, if you set $x=y=z=\frac13$, you find that $xy+yz+zx=\frac13$; assuming that one of the answers actually does give the set of possible values, this rules out choices (A) and (B), since they both exclude $\frac13$ from the set of possible values. Next, note that
$$1=1^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\;,$$ so
$$xy+yz+zx=\frac12\Big(1-(x^2+y^2+z^2)\Big)\;.$$
Now $x^2+y^2+z^2$ is certainly positive, so $1-(x^2+y^2+z^2)<1$, and therefore
$$xy+yz+zx<\frac12\;.$$
Answer (D) therefore includes too much to be the exact set of possible values: it includes $\frac12$, which we now know isn’t possible. Thus, only (C) can be the exact set of possible values of $xy+yz+zx$.
In fact this can be proved by showing that $x^2+y^2+z^2\ge\frac13$ when $x+y+z=1$ and $x,y$, and $z$ are positive, but I don’t immediately see a way to do that without using a bit more than what I consider precalculus algebra.
|
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|
Maximum value of the modulus of a holomorphic function I'm looking for the maximum value of the modulus of a holomorphic function, and I am getting a bit stuck.
The function is $$(z-1)\left(z+\frac{1}{2}\right)$$ with domain $\,|z| \leq 1\,$
Now, I know by the maximum modulus principle the max value will occur on the boundary. So by multiplying the two expressions I get:
$$\left|z^2 - \frac{1}{2}z - \frac{1}{2}\right|$$
writing in complex polar form (and applying MMP, so $\,r = 1\,$) I then get:
$$\left|e^{2i\theta} -\frac{1}{2}\,e^{i\theta}-\frac{1}{2}\right|$$
And... this is where I am stuck. So any help would be greatly appreciated!
|
$$f(z) = (z-1)(z+1/2) = z^2 - z/2 - 1/2$$ Since you know that the maximum is hit on the boundary, $z = e^{i\theta}$, we get that $$F(\theta) = e^{2i \theta} - \dfrac{e^{i\theta}}2 - \dfrac12 = \left(\cos(2\theta) - \dfrac{\cos(\theta)}2 - \dfrac12 \right)+i \left(\sin(2 \theta) - \dfrac{\sin(\theta)}2\right)$$
Let $g(\theta) = \vert F(\theta) \vert^2$.
\begin{align}
g(\theta) & = \vert F(\theta) \vert^2 = \left(\cos(2\theta) - \dfrac{\cos(\theta)}2 - \dfrac12 \right)^2 + \left(\sin(2 \theta) - \dfrac{\sin(\theta)}2\right)^2 \\
& = \cos^2(2\theta) + \dfrac{\cos^2(\theta)}4 + \dfrac14 - \cos(2\theta) \cos(\theta) - \cos(2\theta) + \dfrac{\cos(\theta)}2\\
& + \sin^2(2\theta) + \dfrac{\sin^2(\theta)}4 - \sin(2\theta) \sin(\theta)\\
& = 1 + \dfrac14 + \dfrac14 - \dfrac{\cos(\theta)}2 - \cos(2\theta)\\
& = \dfrac32 - \dfrac{\cos(\theta)}2 - \cos(2 \theta)\\
& = \dfrac52 - \dfrac{\cos(\theta)}2 - 2 \cos^2(\theta)\\
& = \dfrac52 - 2 \left( \cos(\theta) + \dfrac18\right)^2 + 2 \left(\dfrac18 \right)^2\\
& = \dfrac52 + \dfrac1{32} - 2 \left( \cos(\theta) + \dfrac18\right)^2
\end{align}
The maximum is at $\cos(\theta) = -\dfrac18$ and the maximum value of $g(\theta) = \dfrac{81}{32}$. Hence, the maximum value is $$\max_{\vert z \vert \leq 1}\vert f(z) \vert = \sqrt{\dfrac{81}{32}} = \dfrac98 \sqrt{2}$$
|
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|
Evaluating $\int^1_0\frac{x^2}{4x+5}dx$ How can I integrate this function ?
$$\int^1_0\frac{x^2}{4x+5}dx$$
|
Divide the $x^2$ by $4x+5$, using ordinary division of polynomials. We get
$$\frac{x^2}{4x+5}=\frac{1}{4}x-\frac{5}{16}+\frac{25}{16}\frac{1}{4x+5}.$$
Now the integration should be straightforward.
Alternately, as a second choice, let $u=4x+5$. Then $du=4\,dx$, so $dx=\frac{1}{4}du$. Also, $x=\frac{1}{4}(u-5)$, so $x^2=\frac{1}{16}(u^2-10u+25)$. We end up with
$$\int_{u=5}^9 \frac{1}{64}\frac{u^2-10u+25}{u}\,du.$$
The integration is easy, because of the cancellations.
|
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|
Find $\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$ $$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$$
I have tried multiplying by $\frac{1}{\sqrt{x^2+4}}$ and it's reciprocal, but I cannot seem to find the solution. L'Hospital's doesn't seem to work either, as I keep getting rational square roots.
|
$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=$\lim\limits_{x \to \infty} \frac{\sqrt{x^2(1 + \frac{4}{x^2})}}{x+4}$=$\lim\limits_{x \to \infty} \frac{x\sqrt{1 + \frac{4}{x^2}}}{x+4}$.
When $ x\longrightarrow\propto $ $\Rightarrow$ $\frac{4}{x^2}$$\longrightarrow$ 0 .
The above integral then takes the form:
$\lim\limits_{x \to \infty} \frac{x}{x+4}$=$\lim\limits_{x \to \infty} \frac{x}{x(1+\frac{4}{x})}$.=$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$.
When $ x\longrightarrow\propto $ $\Rightarrow$ $\frac{4}{x}$$\longrightarrow$ 0 .
Now the above integral then takes the form:
$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$=$\lim\limits_{x \to \infty} \frac{1}{1+0}$=1.
$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=1
|
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|
$X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$ $X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way.
All I'm getting is $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$
|
Note that $XY + 5(X - Y) = (X - 5)(Y + 5) + 25$, so it suffices to find $(X - 5)(Y + 5)$.
$(X - 5) = \log_{12}(18) - 5 = \log_{12}{18 \over 12^5} = \log_{12}{3^{-3}2^{-9}} = -3\log_{12}(24)$.
$(Y + 5) = \log_{24}(54) + 5 = \log_{24}(54*24^5) = \log_{24}(2^{16}3^{8}) = 8\log_{24}(12)$.
Multiplying together gives $-24\log_{12}(24)\log_{24}(12) = -24\log_{12}(12) = -24$. Adding $25$ to this gives $1$, which is your answer.
|
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|
All integer solutions for $x^4-y^4=15$ I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?
Thanks!
|
First, assume $x$ and $y$ are strictly positive.
$x^4 - y^4 = (x-y)(x+y)(x^2+y^2)$. Of these three factors, only the first can equal 1. So they must be 1, 3, and 5. From 1 and 3 you already get $x=2, y=1$, and then you just have to check that this is a valid solution.
Now we allow the negative solutions, to get $x = \pm 2, y = \pm 1$.
|
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|
Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The book asks me to show that the sequence $\{C_n\}$, with base case $C_0 = 0$, also satisfies the recurrence relation $nC_n=(n+1)C_{n-1}+2n$ for $n=1,2,\cdots$.
I tried to solve it by induction. For this, I wrote the second recurrence relation for $n+1$:
$$(n+1)C_{n+1} = (n+2)C_{n} + 2n + 2$$
Then, assuming that the first recurrence relation holds for $n$, I tried to substitute $C_n = n + 1 + \frac{2}{n}\sum\limits_{k=0}^{n-1}C_k$ in the above equation, to see if I obtain $C_{n+1} = n + 2 + \frac{2}{n+1}\sum\limits_{k=0}^{n}C_k$.
$$\begin{align*}
(n+1)C_{n+1} &= (n+2)C_{n} + 2n + 2\\
(n+1)C_{n+1} &= (n+2)\left( n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k \right ) + 2n + 2\\
(n+1)C_{n+1} &= n(n+2) + n + 2 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k + 2n + 2\\
(n+1)C_{n+1} &= n^2 + 5n + 4 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
(n+1)C_{n+1} &= (n+1)(n+4) + \dfrac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
C_{n+1} &= n+4 + \dfrac{2(n+2)}{n(n+1)}\sum_{k=0}^{n-1}C_k
\end{align*}$$
From this point, I'm not sure how to proceed.
|
A simpler method :
$$n C_n = n(n + 1) + 2\sum_{k=0}^{n-1}C_k$$
$$(n-1) C_{n-1} = (n-1)n + 2\sum_{k=0}^{n-2}C_k$$
so that the difference is :
$$n C_n-(n-1) C_{n-1} = 2n+ 2C_{n-1}$$
that should give you the modified recurrence.
|
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|
Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the conversion from fractions to $\mathbb Z_7$ elements, like it's kind of obvious, it only reports $\frac{1}{3}=5$, $\frac{2}{3} = 10 = 3$, $\frac{32}{9}=2$. Will you please break it down for me?
|
Remember that $\frac{1}{x}$ means "the element which, when multiplied by $x$, gives $1$."
Alternatively, $\frac{a}{b}$ is just shorthand for "the solution to the equation $bx=a$."
So, in $\mathbb{Z}_7$, the fraction $\frac{1}{3}$ is shorthand for "the solution to the equation $3x=1$", which, when interpreted in $\mathbb{Z}_7$, is just "the solution to the integer congruence $3x\equiv 1\pmod{7}$."
Similarly, $\frac{2}{3}$ means "the solution to the integral congruences $3x\equiv 2\pmod{7}$", and $\frac{32}{9}$ means "the solution to the integral congruence $9x\equiv 32\pmod{7}$."
Since $\gcd(3,7)=\gcd(9,7)=1$, each of these congruences has a unique solution modulo $7$. $3(5)\equiv 1\pmod{7}$, so $5 = \frac{1}{3}$. Hence, $\frac{2}{3} = 2\frac{1}{3}=2(5) = 10 \equiv 3\pmod{7}$, so $\frac{2}{3}$ is $3$. And since $32\equiv 4\pmod{7}$, for $\frac{32}{9}$ you have $4(5)(5) = 100\equiv 2\pmod{7}$.
|
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|
Finding all primes $p$ such that $\frac{(11^{p-1}-1)}{p}$ is a perfect square How to Find all primes $p$ such that $\dfrac{(11^{p-1}-1)}{p}$ is a perfect square
|
If $p=2$, $\frac{11^{p-1}-1}{p}$ is not a perfect square. In the following, we assume that $p\neq 2$.
Suppose that $\frac{11^{p-1}-1}{p}$ is a perfect square and write $p=2k+1$.
$$11^{2k}-1 = (11^k-1)(11^k+1)= p n^2 $$
Now, as $gcd(11^k-1, 11^k+1)=2$, we have one of the following cases
$$ 11^k-1 = p a^2\text{ and }11^k +1 = b^2 $$
$$ 11^k-1 = a^2\text{ and }11^k +1 =p b^2 $$
$$ 11^k-1 = 2 p a^2\text{ and }11^k +1 = 2 b^2 $$
$$ 11^k-1 = 2 a^2\text{ and }11^k +1 =2 p b^2 $$
In the first case, $11^k=(b-1)(b+1)$ which is impossible.
In the second case, $11^k-1 = a^2$ so $a$ is even and $4$ divides $a^2$. But $a^2=pb^2-2$ so, $pb^2=2$ modulo $4$, and $b^2=2$ modulo $4$, which is impossible.
In the third case, $2 b^2 = 1$ modulo $11$ which is impossible, as it is easy to verify that the squares modulo $11$ are 0, 1, 4, 9, 5, 3.
The fourth case remains to be examined.
|
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|
Find the remainder for $(x^5-x^4+x^3+2x^2-x+4)/(x^3+1)$
Find the remainder for $\dfrac{x^5-x^4+x^3+2x^2-x+4}{x^3+1}$
I know exactly how to synthetically divide in the format of: $(x\pm a)$. But not $(x^n\pm a)$ (with an exponent). So if anyone can tell me if anything changes or if the steps are the exact same just with the remainder as $\dfrac{remainder}{x^n\pm a}$. And please do not solve the problem for me.
Edit:
I think I've reached an answer. Long division confused me so I used expanded synthetic division and got: $x^2-x+1+\left(\dfrac{x^2+3}{x^3+1}\right)$. Can anyone verify my answer please.
|
It works like any standard division you've done. You start by taking the highest power of $x$ possible to substract to the numerator. In your case, your first term to be $x^2$. You keep on going until you are done, like any usual division. so for the first step you get
$$
\frac{x^5-x^4+x^3+2x^2-x+4}{x^3+1}=x^2+\frac{-x^4+x^3+x^2-x+4}{x^3+1}
$$
|
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|
Evaluate :$\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx$ How to evaluate
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx
$$
I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in advance.
I tried this way,
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\frac{\pi}{2}}dx
$$
then I put the value $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ so
$$
\frac{2}{\pi}\int\left(\sin^{-1} \sqrt{x} -\left(\frac{\pi}{2}-\sin^{-1} \sqrt{x}\right)\right)dx
$$
Is this right?
after that I integrate by part and get,
$$ \int \frac{\sqrt{x}}{\sqrt{1-x}}$$ now,what can i do?
|
Let $$ I_0=\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$$$\Rightarrow I_0=\int\frac{\frac{\pi}{2}-2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}}dx$$$$\Rightarrow I_0=\int (1-\frac{4}{\pi}\cos^{-1}\sqrt{x})dx$$ $$\Rightarrow I_0=x-\frac{4}{\pi}\int\cos^{-1}\sqrt{x})dx$$ Now Consider $$I_1= \int\cos^{-1}\sqrt{x}dx$$ $$\Rightarrow I_1=\int 2z\cos^{-1} zdz$$ Where $$ x=z^2$$ Hence Integrating by parts we get $$ I_1 = 2zcos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz$$ $$I_1 = 2zcos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$$ $$\int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$$ $$\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z$$
|
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|
Find the integer solution of $ a^b = 2^{2 c + 1} + 2^c + 1 $ Find the possible number of integer solution for this equation, such that $ b>1$
$$ a^b = 2^{2 c + 1} + 2^c + 1 $$
From $1$ to $1000$, $ {a = 2, b = 2, c=0} $ and $ {a = 23, b = 2, c=4} $ computationally. Are there any other possible solutions? How to show analytically that above two are it's solutions?
|
Assume $c>2$ for simplicity's sake, so that $a$ is odd among other things.
Firstly, we note that $a^b\equiv1$ mod $2^c$. Since the order of the multiplicative group mod $2^c$ is $2^{c-1}$, we note that either $b$ is even or that $a\equiv1$ mod $2^c$, but we can rule out the latter fairly easily - $1+2*2^c$ is already too big for $b=2$, 1 is always too small and $1+2^c$ is too small for $b=2$ and too big for $b>2$.
Since $b$ is even, $a^b$ will be a square so all we need to do is show that $a=23, c=4$ is the only solution for $b=2$ and we will be done.
Rewrite the equation as:
$$(a-1)(a+1)=2^c(2^{c+1}+1)$$
$\gcd(a-1,a+1)=2$ since $a$ is odd, so it must be the case that $2^{c-1}$ divides $a-1$ or $a+1$. Let's assume $2^{c-1}|a+1$ for now. Then there exists odd numbers $u$ and $v$ such that
$$a-1=2u \\
a+1=2^{c-1}v\\
uv=2^{c+1}+1$$
Combining the first two we get $2^{c-2}v-u=1$, which we can plug into the third to obtain
$$(2^{c-2}v-1)v=2^{c+1}+1$$
We know $v$ is odd. If $v\geq5$, then the left hand side will certainly be bigger than the right hand side ($(2^{c-2}v-1)v>2^{c-3}*v^2\geq\frac{25}{16}2^{c+1}>2^{c+1}+1$) and if $v=1$, then the left hand side will definitely be smaller. So $v=3$ and
$$2^{c-2}9-3=8*2^{c-2}+1\\
2^{c-2}=4\\
c=4$$
which is exactly the solution you found. Working out the case $2^{c-1}|a-1$ is essentially the same except you find that there are no solutions then.
|
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|
What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
|
As I commented, you can exploit the symmetry.
$$\eqalign{
& f(1 - x) + {x^2}f(x) = 2x - {x^4} \cr
& f(x) + {\left( {1 - x} \right)^2}f(1 - x) = 2\left( {1 - x} \right) - {\left( {1 - x} \right)^4} \cr} $$
Now you get
$$\eqalign{
& f(1 - x) + {x^2}f(x) = 2x - {x^4} \cr
& f(1 - x) = \frac{{2\left( {1 - x} \right) - {{\left( {1 - x} \right)}^4} - f\left( x \right)}}{{{{\left( {1 - x} \right)}^2}}} \cr} $$
So all you have to do is solve for $f(x)$ in
$$2x - {x^4} - {x^2}f(x) = \frac{{2\left( {1 - x} \right) - {{\left( {1 - x} \right)}^4} - f\left( x \right)}}{{{{\left( {1 - x} \right)}^2}}}$$
|
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|
Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$ Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!
|
Notice that for $\theta = 20, 40,$ and $80$ degrees you have $\tan^2(3\theta) = 3$. The tangent triple angle formula, which you can get from the tangent angle addition formula, says that
$$\tan(3\theta) = {3\tan(\theta) - \tan^3(\theta) \over 1 - 3 \tan^2(\theta)}$$
So the equation $\tan^2(3\theta) = 3$ can be expressed as
$$(3\tan(\theta) - \tan^3(\theta))^2 = 3(1 - 3 \tan^2(\theta))^2$$
After a little algebra, this becomes the following, where $x = \tan(\theta)$.
$$x^6 - 33x^4 + 27x^2 - 3 = 0$$
By the above, this has roots $x = \tan(20^\circ), \tan(40^\circ),$ and $\tan(80^\circ)$. Since $x$ only appears to even powers here, the other roots must be $x = -\tan(20^\circ), -\tan(40^\circ),$ and $-\tan(80^\circ)$.
The sum of the squares of all six roots is thus given by $2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ))$. However, if we write these roots as $r_1,...,r_6$, then we also have
$$\sum_i r_i^2 = \left(\sum_i r_i\right)^2 - 2\sum_{i < j} r_ir_j$$
But $\sum_i r_i$ is the coefficient of $x^5$ in the above equation, namely zero, and
$\sum_{i < j} r_ir_j$ is the coefficient of $x^4$, namely $-33$. So you get
$$2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ)) = -2\times-33$$
So we conclude that
$$\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$$
|
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|
Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove
$$
\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)
$$
I am not sure what trig identity I should use to begin this problem.
|
$\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$
either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$
or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$
as $\sin(-x)=-\sin(x)$
Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$
So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$
$=\sin{A}(\sin{A}+\sin(B-C))$
$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$
$=\sin{A}(2\sin{B}\cos{C})$
$=2\sin{A}\sin{B}\cos{C}$
|
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|
Prime numbers $(x,c,p)$ such that $x^3-p x^2-cx-5c=0$ How should I proceed to find all prime numbers $x,c,p$ such that
$$x^3-px^2-cx-5c=0$$
|
If $x^3-px^2-cx-5c=0$, then $x$ divides $5c$. Since $x$ and $c$ are prime, we have the two possibilities $x=c$ and $x=5$.
Suppose $x=c$. Substitute. We get $c^3-(p+1)c^2-5c=0$, so $c^2-(p+1)c-5=0$, so $c$ divides $5$, so $c=5$. Therefore definitely $x=5$.
Substitute. We get $125-25p-10c=0$. Since $25$ divides $10c$, it follows that $c=5$. Thus $p=3$.
|
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|
Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$? Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?
|
$\sqrt{a} = x, b=y^2, c=z^2 => x^2+y^2+z^2=1$
We have to prove $$\frac{x}{y^{2}+z^{2}}+\frac{y}{x^{2}+z^{2}}+\frac{z}{x^{2}+y^{2}}\geq \frac{3\sqrt{3}}{2}$$:
$$\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )\leq \left ( x^{2}+\frac{1}{3} \right )\left ( y^{2}+z^{2} \right )\leq \frac{\left ( x^{2}+y^{2}+z^{2}+\frac{1}{3} \right )^{2}}{4}=\frac{4}{9}$$
Do it the same for $\frac{2\sqrt{3}}{3}y\left ( z^{2}+x^{2} \right ), \frac{2\sqrt{3}}{3}z\left ( y^{2}+x^{2} \right )$
So: $$\frac{leftside}{\frac{2\sqrt{3}}{3}}=\sum \frac{x^{2}}{\frac{2\sqrt{3}}{3}x\left ( y^{2}+z^{2} \right )}\geq \frac{\sum x^{2}}{\frac{4}{9}}=\frac{9}{4}$$
$$\Rightarrow leftside=\sum \frac{x}{y^{2}+z^{2}}\geq \frac{3\sqrt{3}}{2}$$
|
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|
Prove that $ \frac{n}{\phi(n)} = \sum\limits_{d \mid n} \frac{\mu^2(d)}{\phi(d)} $ I am trying to show that:
\begin{equation}
\frac{n}{\phi(n)} = \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)}
\end{equation}
where $\phi(n)$ is Euler's totient function and $\mu$ is the Möbius function.
The identity clearly holds for $n=1$, so if we write $n=p_1^{a_1} \cdots p_k^{a_k}$ for the prime decomposition of $n$, the left-hand side becomes:
\begin{eqnarray*}
\frac{n}{\phi(n)} & = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{\phi(p_1^{a_1}) \cdots \phi(p_k^{a_k})} \\
& = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{(p_1^{a_1-1})(p_1-1) \cdots (p_k^{a_k})(p_k-1)} \\
& = & \frac{p_1 \cdots p_k}{(p_1-1)\cdots (p_k-1)}
\end{eqnarray*}
Whereas the right-hand side becomes:
\begin{eqnarray*}
\sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} & = & \sum_{\substack{d \mid n \\ p^2 \nmid d}} \frac{1}{\phi(d)} \\
& = & 1 + \frac{1}{\phi(p_1)} + \dots + \frac{1}{\phi(p_k)} + \frac{1}{\phi(p_1 p_2)} + \dots + \frac{1}{\phi(p_{k-1}p_k)} \\
& + & \dots + \frac{1}{\phi(p_1 \cdots p_k)} \\
& = & 1 + \frac{1}{p_1 - 1} + \dots + \frac{1}{p_k - 1} + \frac{1}{(p_1 - 1)(p_2 - 1)} + \dots + \frac{1}{(p_{k-1}-1)(p_k-1)} \\
& + & \dots + \frac{1}{(p_1 - 1) \cdots (p_k - 1)}
\end{eqnarray*}
I am unsure of how to proceed from this point however. Any help would be much appreciated.
|
$$\sum_{d|n}\frac{\mu^2(d)}{\phi(d)}=1+\frac{1}{p_1-1}+\frac{1}{p_2-1}+\cdots+\frac{1}{(p_1-1)(p_2-1)+\cdots}$$
$$=(1+\frac{1}{p_1-1})(1+\frac{1}{p_2-1})\cdots=\prod_{p_i|n}(1+\frac{1}{p_i-1})=\prod_{p_i|n}(\frac{1}{1-1/p_i})
$$
$$
=\frac{n}{\phi(n)}
$$
|
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How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$? How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $\frac{a}{c} = \frac{2}{5}$ but that is not a correct answer.
|
Divide a:b with c:d
i.e. $$\frac{(\frac{a}{b})}{(\frac{c}{d})} = \frac{(\frac{2}{5})}{\frac{5}{2}}$$
$\implies$ $\frac{a}{c}*\frac{d}{b} = \frac{4}{25}$
$\implies$ $\frac{a}{c}*\frac{3}{2} = \frac{4}{25}$
$\implies$ $\frac{a}{c}= \frac{8}{75}$
Now talking about the approach, So you must try to figure the ratio whose value is required from all the ratios given to you. Like in this question, we just divided the first 2 ratios and after putting the values we got the answer.
|
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|
is there a method to find all the solutions to, $y^2-6y+\sqrt{y}+4=0$ is there a method to find all the solutions to the following set of irrational equations,
$\sqrt{x}+y=3$
$x+\sqrt{y}=5$
NOTE: $(4-1)=(2-1)(2+1)=3$ and $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$
|
If you make the change of variables
$$\begin{equation*}
u=\sqrt{x}\geq 0,\qquad v=\sqrt{y}\geq 0,
\end{equation*}$$
then you need to solve
$$\begin{equation*}
\left\{
\begin{array}{c}
u+v^{2}=3 \\
u^{2}+v=5
\end{array}
\right.
\end{equation*}$$
or
$$\begin{eqnarray*}
&&\left\{
\begin{array}{c}
u=3-v^{2} \\
v^{4}-6v^{2}+v+4=0
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
u=3-v^{2} \\
\left( v-1\right) \left( v^{3}+v^{2}-5v-4\right) =0.
\end{array}
\right.
\end{eqnarray*}$$
So $u=2,v=1$ is a solution, i.e. $x=4,y=1$ in the original variables. And
since $v^{3}+v^{2}-5v-4=0$ has 3 real solutions, one positive $v_{1}\approx
2.164\,2\gt \sqrt{3}$ and two negative $v_{2}\approx -2.391\,4,v_{3}\approx
-0.772\,87$, there are no other solutions because $u=3-v_{1}^{2}<0$.
|
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|
Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written using pluses only, minuses are allowed as well.
For example, $\frac{1}{2}$ is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$
So, what about $\frac{1}{3}$?
|
There is no way to write it as a finite sum. For if you bring such a sum to a common denominator, that denominator will be a power of $2$. Minus signs won't help.
It can be expressed as the infinite "sum"
$$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots.$$
For note that if $|r|\lt 1$ the geometric series $a+ar+ar^2+ar^3+\cdots$ has sum $\frac{a}{1-r}$. Put $a=\frac{1}{4}$ and $r=\frac{1}{4}$, and simplify.
Another interesting representation of $\frac{1}{3}$ is $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots$.
|
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|
how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
|
Well, $x = (2 + x)^.5 $ from the expression of x. Implying $x^2 = 2 + x$. And now you can solve the quadratic for x, giving x = -1 and x = 2. X can't be negative, so x = 2
|
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|
A cube is divided into two cuboids A cube is divided into two cuboids. The surfaces of those cuboids are in the ratio $7: 5$. Calculate the ratio of the volumes.
How can I calculate this?
|
Without loss of generality we can let the sides of the original cube be $1$.
The larger of the two cuboids has four of its sides equal to say $x$. (The other $8$ sides are $1$.) Then the smaller cuboid has four of its sides equal to $1-x$.
The surface area of the larger cuboid is $2+4x$. (Two $1\times 1$ faces, and four $x\times 1$ faces). Similarly, the surface area of the smaller cuboid is $2+4(1-x)=6-4x$.
We are told that
$$\frac{2+4x}{6-4x}=\frac{7}{5}.$$
Solve. We get $x=\frac{2}{3}$, making $1-x=\frac{1}{3}$. So the volumes are in the ratio $2:1$.
|
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|
Permutation across matrices. Matrices may be used to permute the order of elements in a set. For example:
$$
\begin{bmatrix}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}
=
\begin{bmatrix}
w \\
x \\
y \\
z
\end{bmatrix}
$$
My problem is that I need to permute values in different matrices:
$$
A = \begin{bmatrix}
0 & 0 & x & 0 & 0 \\
x & x & x & x & x \\
x & 0 & 0 & 0 & x \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
$$
B = \begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & y & 0 & 0 \\
y & y & y & y & y \\
y & 0 & 0 & 0 & y
\end{bmatrix}
$$
$$
C = \begin{bmatrix}
z & 0 & 0 & 0 & z \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & z & 0 & 0 \\
z & z & z & z & z
\end{bmatrix}
$$
$$
D = \begin{bmatrix}
w & w & w & w & w \\
w & 0 & 0 & 0 & w \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & w & 0 & 0
\end{bmatrix}
$$
Thus given the permutation matrix (from above) and the following pattern matrix how would you change all the $x$ in $A$ to $w$, all the $y$ in $B$ to $x$, all the $z$ in $C$ to $y$, and all the $w$ in $D$ to $z$:
$$
\begin{bmatrix}
0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
|
There's a pretty good chance I don't know what you're asking, but maybe the matrix you're looking for is $$M=\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr1&0&0&0\cr}$$ You get $$MA=\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr1&0&0&0\cr}\pmatrix{0&0&x&0&0\cr x&x&x&x&x\cr x&0&0&0&x\cr0&0&0&0&0\cr}=\pmatrix{x&x&x&x&x\cr x&0&0&0&x\cr0&0&0&0&0\cr0&0&x&0&0\cr}$$ which is exactly the pattern of $D$, but with $x$ instead of $w$, and you get similar things for $MB,MC,MD$.
|
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|
All pairs (x,y) that satisfy the equation $xy+(x^3+y^3)/3=2007$ How we can find the all pairs $(x,y)$ from the integers numbers ,that satisfy the equation :
$$xy+\frac{x^3+y^3}{3} =2007$$
|
Note that the given equation is $$x^3+y^3+3xy=6021$$ or $$x^3+y^3-1+3xy=6020$$
Factoring it we get , $$(x+y-1)(x^2+y^2+1+x+y-xy)=2^2.5.7.43$$
Obviously check $\equiv 3$ and see $$x+y-1\equiv 2 \mod 3$$
Also, $$x+y-1 < x^2+y^2+1+x+y-xy$$
This means, $$x+y-1 \rightarrow 20,5,2,35$$
so now it's easy to see that $$x+y-1=20$$ and so $(x, y)=(18,3)$ or $(3,18)$
|
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|
Small math help with polynomials If one solution of the equation $3x^2 = 8x + 2k + 1$ is $7$ times the other. Find the solutions and the value of $K$.
Note: This isn't a homework question. I'm skipping ahead in my textbook.
Thank you for the help.
Also please help me with this question no.2 also.
Find the other zeroes of $2x^4 - 3x^3 -3x^2 + 6x - 2$ if $-\sqrt{2}$ and $\sqrt{2}$ are two zeroes of the given polynomial
|
$\bigstar$ First question:
First we form the standard notation:$3x^2-8x-(2k+1)=0$
We know that the sum of the roots of a quadratic equation of the form $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $x_1+x_2 =\frac{8}{3}$
On the other hand: $x_1=7\times x_2$
Solving this two equations two unknowns yields:$x_2 = \frac13$ and $x_1=\frac73$
We also know that the product of the roots of a quadratic equation of the form $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$x_1 \times x_2 = \frac79 = -\frac{(2k+1)}{3} \implies k=-\frac 53 $$
$\bigstar$ Second question:
We can write a polynomial as: $P(x)=(x-x_1)(x-x_2)...(x-x_n)$ where $x_1$ to $x_n $ are the zeros of $P(x)$
So given two roots we have:
$$\begin{align}
P(x)&=2x^4 - 3x^3 -3x^2 + 6x - 2\\
&=(x-\sqrt2)(x+\sqrt2)(x-x_1)(x-x_2)\\
&=(x^2-2)(x-x_1)(x-x_2)
\end{align}$$
Dividing P by $x^2-2$ yields:
$$\begin{align}
P(x)&=(x^2-2)(2x^2-3x+1)\\
&=(x^2-2)(x-1)(2x-1)
\end{align}$$
So:
$$x_1=1 , x_2=0.5$$
|
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|
Inequality. $ab^2+bc^2+ca^2 \geq a+b+c.$ Using rearrangement inequalities prove the following inequality:
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that
$$ab^2+bc^2+ca^2 \geq a+b+c.$$
Thanks :)
|
I think I have the solution using arrangements inequalities.(source: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=497213)
We make the substitution $\displaystyle a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$. We have now:
$$\frac{z^2}{xy}+\frac{x^2}{yz}+\frac{y^2}{xz} \geq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}.$$
So:
$$z^3+x^3+y^3\geq y^2z+z^2x+x^2y.$$ And this inequality can be solved using rearrangements inequality.
Let $x \geq y \geq z$. Using rearrangement inequality for $(x^2,y^2,z^2)$ and $(x,y,z)$ we conclude that
$$x^2 \cdot x+ y^2 \cdot y + z^2 \cdot z \geq x^2 \cdot y+y^2 \cdot z+ z^2 \cdot x.$$
|
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|
Simple partial differentiation $x = r\cos\theta$ and $y = r\sin\theta$ If
\begin{align}
x &= r\cos\theta,\\
y &= r\sin\theta,
\end{align}
find
$$\dfrac{\partial^2\theta}{\partial{x}\partial{y}}.$$
How can I find this partial derivative?
I need to prove that
$$
\frac{\partial^2\theta}{\partial{x}\partial{y}} = -\frac{\cos2\theta}{r^2}.$$
|
Using $\theta = \arcsin\left(\frac{y}{r}\right)$ and $\arcsin'(t) = \frac{1}{\sqrt{1-t^2}}$ we get
$$ \frac{\partial \theta}{\partial y} = \frac{1}{r} \frac{1}{\sqrt{1- \left(\frac{y}{r}\right)^2}} = \frac{1}{x}$$
using $r^2 = x^2 + y^2$. Hence
$$\frac{ \partial \theta}{\partial x\partial y} = -\frac{1}{x^2} = -\frac{\cos^2(\theta)}{r^2} $$
|
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|
The value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ There was a multiple choices saying:
Find the value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ knowing that $x^2+x<0$
The right answer is $-2$.
For solving this Maths test, I solved $x^2+x<0$ which is $(-1,0)$ and then find the value of each term of the sum.
My question is: Is there any approach more formal than I did? Thank you for your time.
*
*$\lfloor x\rfloor$ is floor$(x)$
|
As $x^2+x<0, (x-0)(x-(-1))<0$
Now the product two terms is negative, so
either ($x-0>0$ and $x-(-1)<0$) or ($x-0<0$ and $x-(-1)>0$).
If $x>0$ and $x<-1\implies -1>x>0\implies -1>0$, which is clearly impossible.
If $x<0$ and $x>-1$, $-1<x<0$
$\implies -1<x^{2m+1}<0\implies \bigl\lfloor x^{2m+1}\bigr\rfloor=-1$
$\implies 0<x^{2m}<1\implies \bigl\lfloor x^{2m}\bigr\rfloor=0$.
So, $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor=-1+0-1+0=-2$
In general, if $(x-a)(x-b)<0$ where $a<b$,
either $x<a$ and $x>b\implies a>x>b\implies a>b$, but $a<b$(given)
or $x>a$ and $x<b\implies a<x<b$ Here $a=-1,b=0$
|
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|
Accuracy from approximating $\zeta(2)$ with a partial sum This is for an introductory numerical analysis class. The answer shouldn't be too complicated, but if you have one, feel free to post it.
Figure out what $n$ should be such that
$$\sum_{k=n+1}^\infty {1\over k^2}<10^{-8}.$$
My Simple Algebraic Attempt
We know that
$$\sum_{k=1}^\infty {1\over k^2} = \sum_{k=1}^n {1\over k^2} + \sum_{k=n+1}^\infty {1 \over k^2} \implies \sum_{k=n+1}^\infty {1 \over k^2} = \sum_{k=1}^\infty {1\over k^2} - \sum_{k=1}^n {1\over k^2}$$
And
$$\sum_{k=1}^\infty {1\over k^2} = \zeta(2)={\pi^2 \over6}$$
So,
$$\sum_{k=n+1}^\infty {1 \over k^2} = {\pi^2 \over6} - \sum_{k=1}^n {1\over k^2}$$
Then,
$$\sum_{k=n+1}^\infty {1 \over k^2} < 10^{-8} \implies {\pi^2 \over6} - \sum_{k=1}^n {1\over k^2} < 10^{-8} \implies \sum_{k=1}^n {1\over k^2} > {\pi^2 \over6} - 10^{-8}$$
I'm currently trying to brute force an answer to the last expression, but is there a better way to do this?
|
Since $\displaystyle \frac{1}{k^2} < \frac{1}{(k-1)k } = \frac{1}{k-1} - \frac{1}{k}$ we can bound your term by a telescoping sum: $$\sum_{k=n+1}^{\infty} \frac{1}{k^2} < \left(\frac{1}{n} - \frac{1}{n+1} \right)+\left(\frac{1}{n+1} - \frac{1}{n+2} \right)+\left(\frac{1}{n+2} - \frac{1}{n+3} \right) + \cdots = \frac{1}{n}$$ so $n=10^8$ works. The estimate we used isn't too weak, and this $n$ shouldn't be much worse than the minimal $n.$
|
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|
Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)
Thanks in advance =]
|
The anonymous commenter has already answered your question, but in case you have any remaining doubts, I will provide a detailed answer.
For starters, do note that
$A \, \sin (x + c) = \left(A \, \cos(c) \right) \, \sin(x) + \left( A \, \sin(c)\right) \, \cos(x)$
Since you have $\sin (x) + \sqrt{3} \, \cos (x)$, it follows that $A \, \cos(c) = 1$ and $A \, \sin(c) = \sqrt{3}$. Therefore, since $\sin^2 (x) + \cos^2 (x) = 1$, we have that $A^2 = 4$, which yields $A = 2$, and $2 \cos (c) = 1$, which yields $c = \pi / 3$. Finally, we conclude that
$\sin (x) + \sqrt{3} \, \cos (x) = 2 \sin (x + \pi / 3)$.
|
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|
Prove $ 1 + 2 + 4 + 8 + \dots = -1$
Possible Duplicate:
Infinity = -1 paradox
I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:
Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.
Therefore, $1+2+4+8+\dots = -1$.
I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?
|
It is similar to "proving" that 1 = 2 by saying that 1+infinity = 2+infinity.
|
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|
Least Common Multiple of 3 modulo equations i%7=1 -> i=1+7n
i%9=2 -> i=2+9n
i%11=3 -> i=3+11n
Besides writing out everything:
1,8,15,22...
2,11,20,29...
3,14,25,36...
to find the LCM. Is there a better way to do this?
|
In general, you should use the Chinese Remainder Theorem.
However, for these particular numbers, there is a trick! Note that the congruences are equivalent to $2x+5\equiv 0\pmod{7}$, $2x+5\equiv 0\pmod{9}$, and $2x+5\equiv 0\pmod{11}$. So we want $2x+5$ to be divisible by $7$, $9$, and $11$, and therefore by $(7)(9)(11)$, which is $693$. So $2x+5=693$ will work. That gives $x=344$.
Remark: If you are not familiar with congruence notation, look at the sequences of numbers you produced. We will concentrate on the first one, which was $1,8,15,22,\dots$ (there was a mild typo).
Double these, and add $5$. We get $7,21, 35, 49,\dots$, the odd multiples of $7$. Similarly, when you double the numbers in your second sequence, and add $5$, you get the odd multiples of $9$. Similar treatment to the third sequence gets you the odd multiples of $11$. So we want $2x+5$ to be an odd multiple of $(7)(9)(11)$. The smallest positive one is $693$. So now solve $2x+5=693$.
You can also carry through the same argument using the language of remainders. We want $(2x+5)\%7=0$, $(2x+5)\%9=0$, and $(2x+5)\%11=0$.
|
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|
List the primes for which the following system of linear equations DOES NOT have a solution in $\mathbb{Z}_p$ Let $p$ be a prime and consider the field $\Bbb Z_p$. List the primes for which the
following system of linear equations does not have a solution in $\Bbb Z_p$:
$$
\begin{align}
5x + 3y &= 4 \tag{1}\\
3x + 6y & = 1\tag{2}
\end{align}$$
My try is as follows:
Determinant of the coefficient matrix is $21$. $21$ will be 0 if $p=3$ or $7$. So the answer will be $3$ and $7$. is it correct?
|
Form the extended coefficients matrix and apply Gauss Reduction as much as possible.
To begin with, if $\,p=5\,$ then the first equation is $\,3y=4\Longrightarrow y=4\cdot 3^{-1}=4\cdot 2=3\,$ , and substituing in eq. 2 we get $\,3x+3=1\Longrightarrow 3x=-2=3\Longrightarrow x=1\,$ , so there's a unique solution: $\,(1,3)\,$
If $\,p\neq5\,$ we get
$$\begin{pmatrix}5&3&4\\3&6&1\end{pmatrix}\stackrel{R_1/5}\longrightarrow \begin{pmatrix}1&3/5&4/5\\3&6&1\end{pmatrix}\stackrel{R_2-3R_1}\longrightarrow \begin{pmatrix}1&3/5&4/5\\0&21/5&-7/5\end{pmatrix}$$
From here we get
$$R_2\Longrightarrow \frac{21}{5}y=-\frac{7}{5}\Longrightarrow 21y=-7\stackrel{\text{if}\,\,p\neq 3}\Longrightarrow y=-\frac{7}{21}=-\frac{1}{3}$$
and then
$$R_1\Longrightarrow x+\frac{3}{5}y=\frac{4}{5}\Longrightarrow x=\frac{4}{5}+\frac{3}{5}\frac{1}{3}=1$$
and the solution for $\,p\neq 3,5\,$ is $\,\displaystyle{\left(1\,,\,-\frac{1}{3}\right)}\,$
When $\,p=3\,$ the system is clearly inconsistent (watch thesecond equation!), whereas for $\,p=7\,$ , as the reduction process we carried on above shows at the end in the second row $\,(0\,,\,0\,,\,0)\,$ , we get one single (linearly independent) equation in two unknowns and thus there are several solutions: each pair of the form
$$\left(x\,,\,y\right)\,\,\,,\,\,s.t.\,\,\,5x+3y=4\pmod 7$$
|
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|
$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.
Then I remember that one can write any complex number $a+bi$ like:
$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$
and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$ where $\phi$ is $\arctan{\frac{b}{a}} $
So it becomes,
$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre
$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$
Substituting my $a=1$ and $b=1$
$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$
$\phi$ is 45 degrees hence $\frac{\pi}{4}$
But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.
Best regards
|
Use: $$\cos(n\theta)=\cos^n(\theta)-\frac{n(n-1)}{2!}\cos^{n-2}(\theta)\sin^2(\theta)+\frac{n(n-1)(n-2)(n-3)}{4!}\cos^{n-4}(\theta)\sin^4(\theta)-...$$ and $$\sin(n\theta)=n\cos^{n-1}(\theta)\sin(\theta)-\frac{n(n-1)(n-2)}{3!}\cos^{n-3}(\theta)\sin^3(\theta)-...$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Erased number? A set of consecutive positive integers starting with 1 is written on the board. A student came along and erased one number. Average of remaining numbers is 61 15/20 . What was the number erased
|
Let $n$ be the last number written. Lets say that $m$ is the erased number.
Then the sum of the numbers on the board is $\frac{n(n+1)}{2}-m$. Their average then is
$$\frac{\frac{n(n+1)}{2}-m}{n-1}=61 \frac{15}{20}$$
Multiplying by 2 you get
$$\frac{n(n+1)-2m}{n-1}=122\frac{3}{2}$$
$$\frac{n^2+n-2}{n-1}+\frac{2}{n-1}-\frac{2m}{n-1}=123\frac{1}{2}$$
$$n+2+\frac{2-2m}{n-1}=123 \frac{1}{2}.\tag{$*$}$$
Now, since $1 \leq m \leq n$ we have
$$-2 \leq \frac{2-2m}{n-1} \leq 0 \,.$$
Using the fact that $n+2$ is an integer and $-2 \leq \frac{2-2m}{n-1} \leq 0 \,,$ in $(*)$, you see immediately that there are only two possibilities:
Case 1:
$n+2=124$ and $\frac{2-2m}{n-1}=-\frac{1}{2}$
Case 2:
$n+2=125$ and $\frac{2-2m}{n-1}=-\frac{3}{2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Multiplicative inverse of a quadratic algebraic number $\,a+b\sqrt 2$ Find the multiplicative inverse of $1+ 3\sqrt{2}$ in the ring $\mathbb{Q}(\sqrt{2})$ and use it to solve the equation $(1+3\sqrt{2})x=1-5\sqrt{2}$.
I think that the inverse is the conjugate, so it would be $1-3\sqrt{2}$, but then I don't know where to use in the equation that needs to be solved.
|
Instead of finding inverse, you can directly find(less computation) $x$
Let $x=a+b\sqrt 2$
Then, $(1+3\sqrt 2)(a+b\sqrt 2)=1-5\sqrt 2\implies (a+6b)+(3a+b)\sqrt 2=1-5\sqrt 2$
$\implies a+6b=1$ and $3a+b=-5$.
Solving these equations, we get
$a=-\frac{31}{17}$ and $b=\frac{8}{17}\implies x=-\frac{31}{17}+\frac{8}{17}\sqrt 2$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
the least possible value for :$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $ If we know that for every $a,b,c>0$ ,how we can find the least possible value for :
$$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $$
|
Put $(a,b,c)=(3,4,4)$ to get 4. I will show this is optimal.
Assume without loss of generality that $a \leq b \leq c$. Two cases:
If $c \geq a+b$ then $\lfloor\frac{c+a}{b}\rfloor \geq 1$ and $\lfloor\frac{c+b}{a} \rfloor \geq \lfloor\frac{a+b+b}{a} \rfloor \geq 3$ so the sum is at least 4.
If $c \leq a+b$ then $\lfloor\frac{a+b}{c}\rfloor \geq 1$, $\lfloor\frac{b+c}{a}\rfloor \geq 2$, $\lfloor\frac{c+a}{b}\rfloor \geq 1$ so the sum is at least 4.
|
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|
How do we solve the equation? How do we solve the following equation in the set of real numbers?
$$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$
I wrote the given equation has the form
\begin{equation*}
(x+1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4)
\end{equation*}
This equation is equivalent to
\begin{equation*}
(x-2)\left(\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4\right) = 0.
\end{equation*}
But I can not prove that the equation
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 = 0
\end{equation*}
has no solution. Detail
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 <0, \forall x \geqslant -2.
\end{equation*}
|
Not much simpler, but you can also try something like $a:=\sqrt{x+2}$ then $\sqrt{x+7}=\sqrt{a^2+5}$, then it will have only one sqrt in the equation, put that on one side and the rest on the other side..
$$(a^2-1)a+(a^2+4)\cdot\sqrt{a^2+5} = (a^2+1)(a^2+3)$$
|
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|
Chebyshev Diff EQ Find a power series solution about $x_0=0$ for the Chebyshev differential equation
$$(1-x^2)y''-xy'+n^2 y=0,$$
as a function of of the integer $n$. Show that the solutions form a terminating expansion for each value of $n$. What is the orthogonality relationship for these polynomials?
|
The power series around zero is
$$y(x) = \sum_{k=0}^\infty a_k \,x^k.$$
Therefore
$$
y' = \sum_{k=0}^\infty k \,a_{k} \,x^{k-1} =\sum_{k=1}^\infty k \,a_{k} \,x^{k-1}=\sum_{k=0}^\infty \,(k+1) \,a_{k+1} \,x^{k},
$$
and
$$
y'' = \sum_{k=0}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=1}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=0}^\infty \, (k+1)\,(k+2) \, a_{k+2} \,x^{k}.
$$
Substituting these series into differential equation, we get
$$
\begin{aligned}
0 & = \left(1-x^2\right)y''-xy'+n^2 y=
\\
&=
\left(1-x^2\right)\sum_{k=0}^\infty \left( (k+1)\,(k+2) \, a_{k+2} \,x^{k}\right) -x\sum_{k=0}^\infty \left((k+1) \,a_{k+1} \,x^{k}\right)+n^2 \sum_{k=0}^\infty a_k \,x^k =
\\
&=
\sum_{k=0}^\infty \Big( (k+1)\,(k+2) \, a_{k+2} \left(1-x^2\right)x^{k} (k+1) -\,a_{k+1} \,x^{k+1} +n^2 a_{k} \,x^k\Big) =
\\
& =
\big(2 a_2 + n^2 a_0 \big) + \Big(\big(n^2 -1\big)a_1 + 6a_3 \Big)\, x +
\\
& \phantom{=\big(}
\sum_{k=2}^\infty \Big( (k+1)\,(k+2) \, a_{k+2} + \left(n^2 - k^2\right)a_k \Big) \,x^k
=0
\end{aligned}
$$
Thus,
$$
\begin{aligned}
2 a_2 + n^2 a_0 =0 , \\
\big(n^2 -1\big)a_1 + 6a_3 = 0.
\end{aligned}\label{*}\tag{*}
$$
By induction, for integer $k \geq 2$
$$
a_{k+2} = \frac{(k-n)\,(k+n)}{(k+1)\,(k+2)} a_{n}
$$
Determining initial coefficients for odd $k$ and for even $k$ from the system $\eqref{*}$, you will be able to get explicit formula for both even and odd part of the series.
Note that the series terminates at $k=n$.
Orthogonality for a proper weighted inner product is discussed here.
|
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|
Ideas on the ways to integrate $\int \tan^2( x)\sec^3( x) dx$ I would proceed by thus , let $y = [\sec (x)]^2 $
then
$$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $$
so,
$$
2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy
$$
since $$\sec(x) = y^{\frac{1}{2}}$$ and by considering
positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields
$$
2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy
$$
and this later form can be reduced to the standard form $\int(z^2 - a^2)^{1/2} dz$ since
$$
y(y-1)=(y-(1/2))^2 - (1/2)^2 .
$$
What are the other ways to integrate this expression, except for the substitution
$\tan^2(x)^2=\sec (x)^2 -1$ which gives
$$
\sec^5(x) - \sec^3 (x)
$$
in the integrand which I quite don't like.
|
We can do this by intgration by parts
$ I=\int tan^2 x \cdot sec^3x \space dx$
$=\int (sec^2 x-1)\cdot sec^3 x \space dx$
$=\int sec^5 x \space dx-\int sec^3 x \space dx$
$=sec^3 x \cdot tan x-\int 3sec^2 x \cdot sec x tan x \cdot tan x \space dx-\int sec^3 x \space dx $
$= sec^3x \cdot tanx-3I-I_1$
$ or, 4I= sec^3x \cdot tanx-I_1$
We can now find out the second integral
$I_1=\int sec^3 x ~dx=\int sec x ~sec^2x~dx=sec x~tan x-\int sec x~(sec^2 x-1)dx=sec x~tanx-I_1+\int sec x~dx$
$or, 2I_1=secx~tan x+ln|sec x+tan x|$
Now back to original problem:
$I={1 \over 4}sec^3x\cdot tanx-{1 \over 8}(sec x~tan x+ln|sec x+tan x|)+c$
|
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|
Evaluate $\sum_{n=1}^{1024}\left \lfloor \log_2n \right \rfloor$.
Evaluate $\sum_{n=1}^{1024}\left \lfloor \log_2n \right \rfloor$.
I thought the answer is $1+1*2+2*2^2+3*2^3+4*2^4+5*2^5+6*2^6+7*2^7+8*2^8+9*2^9+2^{10}=9219$, but the answer should be 8204. What mistake have I made?
|
We have $\lfloor \log_2 n \rfloor = k$ iff $2^k \le n < 2^{k+1}$, so for $0 \le k \le 9$, $k$ appears in the above sum exactly $2^{k+1} - 2^k = 2^k$ times. As $10$ appears exactly once (for $n =1024$), we have
\[
\sum_{n=1}^{1024} \lfloor \log_2 n \rfloor = 10 + \sum_{k=0}^9 k2^k
= 8204.
\]
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving recurrence relation in 2 variables We already know how to solve a homogeneous recurrence relation in one variable using characteristic equation. Does a similar technique exists for solving a homogeneous recurrence relation in 2 variables. More formally, How can we solve a homogeneous recurrence relation in 2 variables? For example,
F(n,m) = F(n-1,m) + F(n,m-1)
Given some initial conditions, how can we solve the above recurrence relation?
|
Use generating functions like the one variable case, but with a bit of extra care. Define:
$$
G(x, y) = \sum_{r, s \ge 0} F(r, s) x^r y^s
$$
Write your recurrence so there aren't subtractions in indices:
$$
F(r + 1, s + 1) = F(r + 1, s) + F(r, s + 1)
$$
Multiply by $x^r y^s$, sum over $r \ge 0$ and $s \ge 0$. Recognize e.g.:
\begin{align}
\sum_{r, s \ge 0} F(r + 1, s) x^r y^s
&= \frac{1}{x} \left( G(x, y) - \sum_{s \ge 0} F(0, s) y^s \right) \\
&= \frac{G(x, y) - G(0, y)}{x} \\
\sum_{r, s \ge 0} F(r + 1, s + 1) x^r y^s
&= \frac{1}{x} \left(
G(x, y)
- \sum_{s \ge 0} F(0, s) y^s
- \sum_{r \ge 0} F(r, 0) x^s
+ F(0, 0)
\right) \\
&= \frac{G(x, y) - G(0, y) - G(x, 0) + F(0, 0)}{x y}
\end{align}
Here $G(0, y)$ and $G(x, 0)$ are boundary conditions. If you are lucky, the resulting equation can be solved for $G(x, y)$.
In the specific case of binomial coefficients, you have $F(r, 0) = F(0, r) = 1$, so that $G(x, 0) = \frac{1}{1 - x}$ and $G(0, y) = \frac{1}{1 - y}$:
$$
\frac{G(x, y) - 1 / (1 - y) - 1 / (1 - x) + 1}{x y}
= \frac{G(x, y) - 1 / (1 - y)}{x} + \frac{G(x, y) - 1 / (1 - x)}{y}
$$
The result is:
\begin{align}
G(x, y) &= \frac{1}{1 - x - y} \\
&= \sum_{n \ge 0} (x + y)^n
\end{align}
This is:
$$
[x^r y^s] G(x, y) = \binom{r + s}{r} = \binom{r + s}{s}
$$
as expected.
|
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|
What is the pattern or relation in this table? Here is the table:
$$\begin{array}{c}
0\\
1\\
1& 1\\
3& 2& 3\\
5& 3& 3& 5\\
11 &8 &10 &8 &11\\
21 &13 &14 &14 &13 &21\\
43 &30 &37 &36 &37 &30 &43\\
85 &55 &61 &55 &55 &61 &55 &85\\
171 &116 &140 &140 &146 &140 &140 &116 &171
\end{array}$$
In this triangle like table, I am unable to find the relation between columnwise.
Say for 1st column, I have found the relation $A(n)= A(n-1)+2A(n-2)$.
For the 2nd column, I have the relation $A(n)= A(n-1)+2A(n-2) + x$ [ where $x$ is $-1$ for even row number and $+1$ for odd].
But for other columns, I have not found the relations yet. Can anyone help me find a generalized equation to generate the table?
|
Updated:
let $A(n,k)$ be the number in the table with row $n$ and column $k$, where $0\le n$ and $0\le k\le n$.
$$A(n,k)=A(n,n-k)$$
$$A(n,0)=A(n,n)=0$$
The table would look like this: (top left is row $0$ and column $0$)
$$\begin{array}{c}
0\\
0& 0\\
0& 1& 0\\
0& 1& 1& 0\\
0& 3& 2& 3& 0\\
0& 5& 3& 3& 5& 0\\
0& 11 &8 &10 &8 &11& 0\\
0& 21 &13 &14 &14 &13 &21& 0\\
0& 43 &30 &37 &36 &37 &30 &43& 0\\
0& 85 &55 &61 &55 &55 &61 &55 &85& 0\\
0& 171 &116 &140 &140 &146 &140 &140 &116 &171 & 0
\end{array}$$
Then, we have: $$A(n+1,k)+A(n,k)+A(n,k-1)=2^{n-1}$$
Example
$$A(8,3)+A(7,3)+A(7,2)=37+14+13=64=2^6$$
|
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|
Find the equation of the plane that contains Find the equation of the plane that contains,
the lines: $$
\frac{x-2}{2} = \frac{y+4}{3} = \frac{2 - z}{5}
$$
and
$$\begin{align}
x &= 3 + 4t \\
y &= -4 +6t \\
z &= 5 -10t.
\end{align}
$$
I'm not exactly sure on how to tackle this problem.
|
The line $\frac{x-2}{2} = \frac{y+4}{3} = \frac{2-z}{5}\,$ can be written as
$$x=2+2t\,,\quad y=-4+3t \,,\quad z=2-5t \,, $$
and the line
$$x = 3 + 4t \,,y = -4 + 6t \,,z = 5 - 10t \,.$$
The two lines have the same direction, since $ v_1=(2,3,-5) $ and $ v_2 = (4, 6,-10)\,, $ where $v_1$ and $v_2$ are the direction vectors of the two lines.
One can get two points lie in the plane. Putting $t=0$ in the equations of the lines gives $p_1=(2,-4,2)$ and $p_2=(3,-4,5)\,,$ which lie in the plane.
Constructing the vector $ v_3=p_2-p_1$ gives $v_3=(1,0,3)\,.$ Now, we can find the normal to the plane by taking the cross product of $v_3$ and $v_1$ or $v_2$.
$$ n = v_3 \times v_2 \,. $$
Once that done, the equation of the plane is given by
$$ n.(X-p_1)=0 \Rightarrow n.( x-2,y-4,z-2 )=0 \,.$$
|
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|
Transformation matrix to go from one vector to another I've two vectors $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$. How to find transformation matrix for transform from a to b?
|
Try using the dyadic product, the definition is
$$
\mathbf{a b} \equiv \mathbf{a}\otimes\mathbf{b} \equiv \mathbf{a b}^\mathrm{T} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix} = \begin{pmatrix} a_1b_1 & a_1b_2 & a_1b_3 \\ a_2b_1 & a_2b_2 & a_2b_3 \\ a_3b_1 & a_3b_2 & a_3b_3 \end{pmatrix}.
$$
You can construct the rotation matrix $\mathbf{R}$ you're looking for as following
$$
\mathbf{b} = \mathbf{R} \mathbf{a} \\ \\
\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} =
\frac{1}{3}
\begin{pmatrix}
\frac{1}{a_1}b_1 & \frac{1}{a_2}b_1 & \frac{1}{a_3}b_1 \\
\frac{1}{a_1}b_2 & \frac{1}{a_2}b_2 & \frac{1}{a_3}b_2 \\
\frac{1}{a_1}b_3 & \frac{1}{a_2}b_3 & \frac{1}{a_3}b_3
\end{pmatrix}
\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}
\\
\mathbf{R} =
\frac{1}{3}
\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}
\begin{pmatrix} \frac{1}{a_1} & \frac{1}{a_2} & \frac{1}{a_3} \end{pmatrix}
$$
|
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|
Generating Functions help Using generating functions, find the number of ways to make change for a $\$100$
bill using only dollar coins and $\$1,\$2$, and $\$5$ bills.
Thanks for the help.
|
Your generating function will have the form $$f(x)=\sum_{n\ge 0}a_nx^n\;,$$ where $a_n$ is the number of ways to make a total of $n$ dollars using the prescribed coin and bills. Each $\$1$ coin must therefore add $1$ to the exponent, as must each $\$1$ bill; each $\$2$ bill must add $2$ to the exponent, and each $\$5$ bill must add $5$. Thus,
$$f(x)=\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ coins}}\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ bills}}\underbrace{(1+x^2+x^4+\ldots)}_{\$2\text{ bills}}\underbrace{(1+x^5+x^{10}+\ldots)}_{\$5\text{ bills}}\;,$$
or more compactly,
$$f(x)=\left(\sum_{n\ge 0}x^n\right)^2\left(\sum_{n\ge 0}x^{2n}\right)\left(\sum_{n\ge 0}x^{5n}\right)\;.$$
Now use the basic geometric generating function to rewrite this as
$$f(x)=\left(\frac1{1-x}\right)^2\left(\frac1{1-x^2}\right)\left(\frac1{1-x^5}\right)=\frac1{(1-x)^2(1-x^2)(1-x^5)}\;,$$
which can be further simplified to $$f(x)=\frac1{(1-x)^3(1+x)(1-x^5)}\;,$$ if you wish.
The answer to the question is now the coefficient of $x^{100}$ in $f(x)$.
Added: Suppose that you want the number of $\$5$ bills to be at least $2$ and at most $6$. Then the factor that accounts for the $\$5$ bills would be
$$\begin{align*}
x^{2\cdot5}+x^{3\cdot5}+x^{4\cdot5}+x^{5\cdot5}+x^{6\cdot5}&=x^{10}+x^{15}+x^{20}+x^{25}+x^{30}\\
&=x^{10}\left(1+x^5+x^{10}+x^{15}+x^{20}\right)\\
&=\frac{x^{10}\left(1-x^{25}\right)}{1-x^5}\;.
\end{align*}$$
|
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|
Solve for $x$: $\big(x^3+\frac{1}{x^3}+1\big)^4=3\big(x^4+\frac{1}{x^4}+1\big)^3$ Solve for $x$
$$\big(x^3+\frac{1}{x^3}+1\big)^4=3\big(x^4+\frac{1}{x^4}+1\big)^3$$
let $x+\frac{1}{x}=t$ the equation equivalent to $(t^3-3t+1)^4=3(t^4-4t^2+3)^3$
but it's very complicated. Thanks.
|
Let $u=t^4-4t^2+3$ and $v=t^3-3t+1$. To finish off the problem (over real numbers), it suffices to prove that the equation $3u^3=v^4$ has no solutions with $|t|\ge 2$ other than $t=2$. Differentiate the ratio $f(t)=3u^3 v^{-4}$:
$$ f'(t)=\frac{d}{dt}(3u^3v^{-4})= 3u^2v^{-5}(3u'v-4uv') \tag1$$
The sign of $v$ is the same as the sign of $t$ when $|t|\ge 2$. It remains to find the sign of $3u'v-4uv'$. Direct computation shows
$$3u'v-4uv' = 12(t^3-t^2-2t+3) \tag2$$
which is not as bad as one might have expected from subtracting two polynomials of degree $6$.
*
*When $t \ge 2$ we have $t^3\ge 2t^2$, hence (2) is positive.
*When $t\le -2$ we have $t^3\le -2t^2$, hence (2) is negative.
Either way, (1) is positive when $|t|>2$.
*
*Since $f(2)=1$, it follows that $f(t)>1$ for $ t >2$.
*Since $\lim_{t\to-\infty}f(t)=3$, it follows that $f(t)>3$ for $ t <-2$.
Thus, the only root of $f(t)=1$ outside of $(-2,2)$ is $t=2$.
|
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|
Nonlinear equations and unique solution How to show that the following system of equations has a unique solution $(x,y)$? $x,y$ are scalars.
$x+\frac{3}{4}y+\frac{1}{20}\sin x=0$
$-\frac{37}{40}x+y+\frac{1}{10}\sin y=0$
I tried contraction mapping, but it didn't work.
|
I solved the top equation for $y$, substituted it into the second equation, and got
$$\frac{1}{10}\sin(\frac{1}{15}\sin x + \frac{4}{3}x) + \frac{271}{120}x + \frac{1}{15}\sin x = 0$$. One solution is $x = 0$, as you found. The derivative of the left-hand side is always positive:
$$\frac{1}{10}\cos(\frac{1}{15}\sin x + \frac{4}{3}x)(\frac{1}{15}\cos x+\frac{4}{3})+\frac{271}{120} +\frac{1}{15}\cos x \geq -\frac{1}{10}(\frac{1}{15} + \frac{4}{3}) + \frac{271}{120} - \frac{1}{15} > 0$$.
So $0$ is the only solution for $x$, and $y$ has to be $0$ too.
It is late and I don't guarantee against errors in the fractions, but I think the procedure works.
|
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|
Proving a sum using induction I am having a problem with this question.
I need to prove by induction that: $$\sum_{k=1}^n \sin(kx)=\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$
The relation is obvious for n=1
Now I suppose that the relation is true for a natural number n and I want to show that $$\sum_{k=1}^{n+1} \sin(kx)=\frac{\sin(\frac{n+2}{2}x)\sin(\frac{n+1}{2}x)}{\sin(\frac{x}{2})}$$
We have $$\sum_{k=1}^{n+1} \sin(kx)=\sin[(n+1)x]+\sum_{k=1}^{n} \sin(kx)
=\sin[(n+1)x]+\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}= \frac{\sin[(n+1)x]\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$
I am unable to simplify the expression using the trigonometric identities. I keep turning around in circles.
Can somebody help me please.
Thank you in advance
|
You can actually use telescopy, which is just induction in disguise.
We have that
$$\cos b - \cos a = 2\sin \frac{{a + b}}{2}\sin \frac{{a - b}}{2}$$
Now let $$b=\left(k+\frac 1 2 \right)x$$
$$b=\left(k-\frac 1 2 \right)x$$
Then
$$\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x = 2\sin kx\sin \frac{x}{2}$$
Now sum through $k=1,\dots,n$, to get
$$\sum\limits_{k = 1}^n {\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x} = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$
$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$
whence
$$\frac{{\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$
But using our first formula once more, we have
$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}$$
so finally
$$\frac{{\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}}}{{\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$
as desired.
|
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|
As shown in the figure: Prove that $a^2+b^2=c^2$ Geometry: Buildings in the triangle
Other triangles with the same property:
$1.$ 12 18 6 12 30 102
$2.$ 15 30 15 15 15 90
$3.$ 24 30 54 24 6 42
$4.$ 30 10 40 30 20 50 (proposed problem in sense of clockwise)
$5.$ 36 12 6 12 18 96
$6.$ 36 18 6 36 6 78
$7.$ 42 6 36 42 12 42
$8.$ 60 6 57 30 3 24
$9.$ 60 24 12 12 6 66
Using matlab we can find all triangles (integer solutions) with this property sums of squares:
|
I add the letters to your points
Using Theorem of Sine, we get
$$\frac{a}{\sin 30^\circ}=\frac{BD}{\sin(10^\circ+40^\circ)}=\frac{BD}{\sin 50^\circ}$$
$$\frac{BD}{\sin 40^\circ}=\frac{BA}{\sin(30^\circ+20^\circ+50^\circ)}=\frac{BA}{\sin 100^\circ}$$
so we get
$$a=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\sin 40^\circ$$
Also we have
$$\frac{b}{\sin 30^\circ}=\frac{BE}{\sin(10^\circ+40^\circ+30^\circ)}=\frac{BE}{\sin 80^\circ}$$
because $\angle BAE=\angle BEA=70^\circ$, we have
$$BE=BA$$
so we get
$$b=\frac{BA\cdot\sin 30^\circ}{\sin 80^\circ}=\frac{BA\cdot\sin 30^\circ\cdot\sin 50^\circ}{\sin 100^\circ\cdot\sin 50^\circ}=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\cos 40^\circ$$
Finally, we have
$$\frac{c}{\sin 30^\circ}=\frac{BC}{\sin(10^\circ+40^\circ+30^\circ+20^\circ)}=\frac{BC}{\sin 100^\circ}$$
$$BC=\frac{BA}{\sin 50^\circ}$$
So, we have
$$c=\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}$$
Since
$$\sin^2 40^\circ+\cos^2 40^\circ=1$$
So we have
$$a^2+b^2=c^2$$
|
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|
How do I solve this equation How do I solve this equation, I keep getting the wrong answer.
$$ \frac{-2x^2+50}{\sqrt{50-x^2}} = 1 $$
Thank you
|
We have:
$$\sqrt{50 - x^2} = 50 - 2x^2$$
Eliminating the square root and opening the brackets:
$$50 - x^2 = 4x^4 - 200x^2 + 2500$$
Make it more simple:
$$-4x^4 + 199x^2 - 2450 = 0$$
Substitute $y = x^2$, simplifying and find full square:
$$\left(y - \dfrac{199}{8}\right)^2 = \dfrac{401}{64}$$
So, there is two solutions:
$$y = \dfrac{199}{8} \pm \dfrac{\sqrt{401}}{8}$$
So, there is 4 solutions for $x$:
$$x = \pm\sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\tag{1}$$
$$x = \pm\sqrt{\dfrac{199}{8} + \dfrac{\sqrt{401}}{8}}\tag{2}$$
Substitution the solution for $x$ into original equations shows the correcteness of $(1)$ solutions and incorrectness of solutions $(2)$.
Finally:
$$x_1 = \sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\approx 4.7299\qquad x_2 = -\sqrt{\dfrac{199}{8} - \dfrac{\sqrt{401}}{8}}\approx -4.7299$$
|
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|
Prove the following relation: I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$
I got this far before I got stuck:
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}+\binom{2n+1}{n}\frac1{2^k}.
\end{eqnarray*}$
If I can combine the second and third terms and get something same as first term, I am done but I could not do that.
|
You correctly used Pascal's identity, but then you goofed going to the next line. (Should have an $n$ in that last exponent of $2$, not a $k$.) I recommend going a different way, though.
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^{n+1}\binom{n+k}{k-1}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=1}^{n+1}\binom{n+k}{k-1}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k+1}{k}\frac1{2^{k+1}}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+2}}+\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^{k+1}}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+2}}+\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\frac12\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}.
\end{eqnarray*}$
You see how we have half the original sum on the right-hand side now? If we subtract that and then multiply by $2$, we have
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & -\binom{2n+2}{n+1}\frac1{2^{n+1}}+2\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+1}}+2\binom{2n+1}{n+1}\frac1{2^{n+1}}+2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.
\end{eqnarray*}$
Finally, applying Pascal's identity to $\binom{2n+2}{n+1}$, and using the fact that $\binom{2n+1}{n}=\binom{2n+1}{n+1},$ the extraneous binomial coefficients cancel out, and we're left with $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k},$$ as desired.
|
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|
Probability that I will get a $4$ consecutive numbers I have five dice to roll. I roll them. What is the probability that I will get a straight with exactly four consecutive numbers and not $5$?
There are three options: $1,2,3,4$ or $2,3,4,5$ or $3,4,5,6$.
I have $1,2,3,4,*$ where $*$ can be either $1/2/3/4/6$. It cannot be five. Now we do $4!*5$.
Next, I take $2,3,4,5,*$ where $*= 2,3,4,5$. We get $4!*4$ and $4!*5$ for $3/4/5/6$.
I get the probability $(4!*14)/6^5$ but there are some complexities involved. Can anyone explain this clearly? I tried $4!*5*5$ since the position of * is variable. Now, I get $(4!*70)/6^5$ which is not correct either.
Can someone explain it systematically?
|
As you have noticed, there are $14$ possible multisets of $5$ numbers that are "good" in your sense, of containing four consecutive numbers but no five consecutive numbers. These $14$ multisets can be divided into two kinds:
*
*$12$ of them with some number repeated, namely: $\{1, 2, 3, 4, 1\}, \{1, 2, 3, 4, 2\}, \{1, 2, 3, 4, 3\}, \{1, 2, 3, 4, 4\}, \{2, 3, 4, 5, 2\}, \{2, 3, 4, 5, 3\}, \{2, 3, 4, 5, 4\}, \{2, 3, 4, 5, 5\}, \{3, 4, 5, 6, 3\}, \{3, 4, 5, 6, 4\}, \{3, 4, 5, 6, 5\}, \{3, 4, 5, 6, 6\}$
*$2$ of them with all five numbers distinct, namely: $\{1, 2, 3, 4, 6\}$ and $\{3, 4, 5, 6, 1\}$.
For each of the latter two multisets, it can be ordered in $5!$ ways, so the probability that you get exactly that multiset is: $$\frac{5!}{6^5}$$
For each of the former twelve multisets, it can be ordered in $\frac{5!}{2!}$ ways, so the probability that you get exactly that multiset is:
$$\frac{5!}{2! 6^5}$$
So the answer you want is:
$$ 2\frac{5!}{6^5} + 12\frac{5!}{2! 6^5} = \frac{2(5!) + 6(5!)}{6^5} = \frac{960}{6^5} = \frac{960}{7776} = \frac{10}{81}$$
|
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|
Trig Identities : $\frac{\sin (4x)}{1-\cos(4x)} \frac{1-\cos(2x)}{\cos(2x)} = \tan(x)$ I want to prove that $$\frac{\sin (4x)}{1-\cos(4x)} \frac{1-\cos(2x)}{\cos(2x)} = \tan(x)$$
\begin{align}
\text{Left hand side} : & = \sin(2x+2x)/(1-\cos(2x+2x)) \times ((1-\cos^2x+\sin^2x)/(\cos^2x-\sin^2x))\\
& = ((2\sin^2x)(\cos^2x))/(2\sin^2(2x)) \times (2\sin^2x/(2\cos^2x -1))\\
& = 2\sin^2(x)\cos^2(x)/\sin^2(2x)
\end{align}
Not sure where to go from here...
|
Recall the following identities:
$$\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$$
$$1-\cos(2 \phi) = 2 \sin^2(\phi)$$
Make use of the above identities and you will get your solution.
Move your cursor over the gray area for complete solution.
\begin{align}\dfrac{\sin(4x)}{1-\cos(4x)} \dfrac{1 - \cos(2x)}{\cos(2x)} & = \dfrac{\sin(2(2x))}{1-\cos(2(2x))} \dfrac{1 - \cos(2x)}{\cos(2x)}\\ & = \dfrac{2 \sin(2x) \cos(2x)}{2 \sin^2(2x)} \dfrac{1-\cos(2x)}{\cos(2x)}\\ & = \dfrac{1-\cos(2x)}{\sin(2x)} ( \because \text{Cancelling } 2\sin(2x) \cos(2x))\\ & = \dfrac{2 \sin^2(x)}{2 \sin(x) \cos(x)}\\ & = \dfrac{\sin(x)}{\cos(x)} ( \because \text{Cancelling } 2\sin(x))\\ & = \tan(x) \end{align}
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|
Proving a limit with the epsilon-delta definition I was looking to prove using the $\epsilon,\delta$ limit definition that $\lim_{x\to a}(\sqrt[3]{x})=\sqrt[3]{a}$, $(a>0)$. I'm not sure what sort of algebraic manipulation I should use on the expression $|\sqrt[3]{x}-\sqrt[3]{a}|$ (so I'll be able to continue with proving the limit).
Just a little hint would be great, thanks in advance.
|
You can use the identity
$$
x - a = \left( \sqrt[3]{x} - \sqrt[3]{a} \right) \left( \sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} \right),
$$
which is derived from the following identity:
$$
a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}).
$$
Then
$$
\forall x \in \mathbb{R} \setminus \{ a \}: \quad \sqrt[3]{x} - \sqrt[3]{a} = \frac{x - a}{\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}}}.
$$
Now, fix $ \epsilon > 0 $. Choose $ x \in \mathbb{R} $ so that
$$
|x - a| < \min \left( \frac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right).
$$
As $ a > 0 $, having $ |x - a| < \dfrac{a}{2} $ ensures that $ x > 0 $, and so
$$
\sqrt[3]{x^{2}} + \sqrt[3]{x} \cdot \sqrt[3]{a} + \sqrt[3]{a^{2}} > \sqrt[3]{a^{2}}.
$$
Next, having $ |x - a| < \sqrt[3]{a^{2}} \cdot \epsilon $ yields
$$
\left| \sqrt[3]{x} - \sqrt[3]{a} \right| < \epsilon.
$$
You can therefore set $ \delta := \min \left( \dfrac{a}{2},\sqrt[3]{a^{2}} \cdot \epsilon \right) $.
|
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|
Geometrical Inequalities I couldnt solve the following: we need to minimize $$\sqrt{\frac{(a+b-c)(b+c-a)(a+c-b)}{(a+b+c)}},$$ where a,b,c are sides of a triangle.
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This expression can be arbitrarily close to $0$. Let $a=2\varepsilon$, $b = c = 1/2 - \varepsilon$. There is a triangle with sides $a$, $b$, and $c$. As $\varepsilon$ goes to $0$, the expression approaches $0$. Specifically, we have
$$\sqrt{\frac{(a+b-c)(b+c-a)(a+c-b)}{(a+b+c)}}
= \sqrt{\frac{2\varepsilon \cdot (1 - 4\varepsilon)\cdot 2\varepsilon}{1}} \leq \sqrt{2\varepsilon \cdot 2\varepsilon} = 2\varepsilon.$$
This expression can be arbitrarily close to $0$.
The expression is maximized when $a=b=c$.
|
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|
If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$ If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$
I have tried to solve the equation first and find the values of the roots but it becomes very long process. Is there any easy process?
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Define $w_1=(1+\sqrt{3}i)/2$ and $w_2=(1-\sqrt{3}i)/2$ where $i=\sqrt{-1}$. Now observe that
\begin{align}
(a^3+1)=(a+1)(a-w_1)(a-w_2)
\end{align}
Similarily for other terms also. This helps you to write
\begin{align}
(a^3+1)(b^3+1)(c^3+1)(d^3+1)=P_1P_2P_3
\end{align}
where
\begin{align}
P_1&=(a+1)(b+1)(c+1)(d+1) \\
P_2&=(a-w1)(b-w_1)(c-w_1)(d-w_1) \\
P_3&=(a-w2)(b-w_2)(c-w_2)(d-w_2)
\end{align}
Now for any $x$, you can write your original quartic equation as
\begin{align}
f(x)&=x^4-x^3+2x^2+x+1 \\
&=(a-x)(b-x)(c-x)(d-x)
\end{align}
Put $x=-1$. You obtain
\begin{align}
f(-1)=(a+1)(b+1)(c+1)(d+1)=P_1
\end{align}
So you get $P_1=f(-1)=4$. Similarily $P_2=f(w1)=1+\sqrt{3}i$ $P_3=f(w2)=1-\sqrt{3}i$ and hence $P_1P_2P_3=4*(1+\sqrt{3}i)*(1-\sqrt{3}i)=16$ and your answer.
|
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|
$2n+1$ and $n^2+1$ are always coprime or their gcd is $5$ Using a spreadsheet, it can be inferred that when $n≡2[5]$, then $\gcd(n^2+1,2n+1)=5$, else $\gcd(n^2+1,2n+1)=1$.
Indeed, when $n≡2[5]$, $n^2+1$ and $2n+1$ can easily be shown to be multiples of $5$, so their gcd is at least $5$. But then, I can't see how to complete the proof.
|
Let $d=\gcd(2n+1,n^2+1)$. Then since $d\mid 2n+1$ and $d\mid n^2+1$, we must also have $$d\mid(n^2+1)-(2n+1)=n(n-2)\;.$$ Clearly $\gcd(n,2n+1)=1$, and $d\mid 2n+1$, so $\gcd(n,d)=1$, and therefore $d\mid n-2$. By definition $d\mid 2n+1$, so $d\mid(2n+1)-2(n-2)=5$, and therefore $d=1$ or $d=5$.
The problem doesn’t require you to do so, but you can easily check that $d=5$ if and only if $n\equiv2\pmod 5$, so that $5\mid n-2$.
|
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|
Limit of Functions Using $\lim_{n\to\infty} (1+\frac{x}{n}) ^n = e^x $ Can someone help me solve the following three exercices?
1) $\displaystyle \lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \Large\frac{8}{x} } $
2) $\displaystyle \lim_ {x \to \infty} \left( 1+ \frac{1}{9x^4 + 8x^2 +8 }\right) ^{8x^4 +9x^2 + 1} $
3) $\displaystyle \lim_{x\to 0 } \sqrt[\Large x]{1+9x}$
These aren't homework, but pre-exam (not formal) exercices
Thanks in advance
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I'll show you one, but they are all very similar:
$$\begin{align*}\lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \large \frac{8}{x} }&=\lim_{x\to \infty} \left(\left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{9x^2+x+\Large \frac1x} \right)^{\Large \frac{x^2 + \Large\frac{8}{x}}{9x^2+x+\frac1x} }\overset{(*)}{=}\\
&=\lim_{x\to \infty}e^{\Large \frac{x^2 + \frac{8}{x}}{9x^2+x+\frac1x} }=\lim_{x\to \infty}e^{\Large \frac{1 + \frac{8}{x^3}}{9+\frac1x+\frac1{x^3}} }=e^{1/9}\end{align*}$$
Where $(*)$ follows from the fact that $\lim_{x\to\infty}9x^2+x+\frac1x=\infty$.
Can you follow the solution? Can you apply it to the other two?
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/245068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $
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Note that
$$\frac{{1.3{\cdots}(2n - 1)}}{{2.4{\cdots}(2n)}} = \frac{{\frac{{(2n - 1)!}}{{2.4{\cdots}(2n - 2)}}}}{{{2^n}n!}} = \frac{{\frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}}}}{{{2^n}n!}} = \frac{{(2n - 1)!}}{{{2^{n - 1}}(n - 1)!}} \cdot \frac{1}{{{2^n}n!}} \to 0$$
using Stirling's approximation because $n$ is "large" here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/247360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How to show that $f\left( \frac{ x + y }{ 2 }\right ) \leq \frac{ f( x ) + f( y ) }{ 2 }$ when $f''(x) \geq 0$. I need to show that if $f: (a,b) \to \mathbb{ R }\text{ with}\;\; f''( x ) \geq 0$ for all $x \in (a,b)$, then $f\left( \frac{ x + y }{ 2 } \right) \leq \frac{ f( x ) + f( y ) }{ 2 }$.
I know that since $f''( x ) \geq 0$, then $f'(x)$ is monotone increasing. I'm not really sure where to go from here.
|
Since $f''(x) \geq 0$, $f'(x)$ is monotone increasing (as you already observed). Let $x, y \in (a, b)$ and WLOG $x < y$. Therefore, $x < \frac{x + y}{2} < y$. Now, applying MVT in $\left(x, \frac{x+y}{2}\right)$ and $\left(\frac{x + y}{2}, y\right)$, we get
$$
f'(c_1) = \frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}}\\
f'(c_2) = \frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}}
$$
for some $c_1 \in \left(x, \frac{x+y}{2}\right)$ and some $c_2 \in \left(\frac{x + y}{2}, y\right)$.
Now $c_1 \leq c_2$, therefore $f'(c_1) \leq f'(c_2)$. This implies,
$$
\frac{f\left(\frac{x + y}{2}\right) - f(x)}{\frac{y - x}{2}} \leq
\frac{f(y) - f\left(\frac{x + y}{2}\right)}{\frac{y - x}{2}} \\
f\left(\frac{x + y}{2}\right) \leq \frac{f(x) + f(y)}{2}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?
Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?
Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Suppose if the numbers were very big, I wouldn't have got the answer.
Do you know any ways to find the answer?
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If $x,y > 1$ are both odd, then $x+y > 2$ is even, a contradiction.
Then $y = 2$ and by inspection $x=5$ as $x = 3$ results in $x-y = 1$.
Or, we have that $x-2,x,x+2$ are all prime meaning we have three consecutive integers modulo $3$ which is only satisfied by the triple $(3,5,7) \implies y = 5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/250584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "100",
"answer_count": 5,
"answer_id": 4
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|
How do I show there are no elementary function solutions for the differential equation $f''(x)=f(\sqrt{x}), x>0$? How do I show there are no elementary function solutions for the differential equation $f''(x)=f(\sqrt{x}), x>0$ in the $C^2(0,\infty)$ space solutions?
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Expanding on Jack D'Aurizio's answer,
write the equation as
$f(x^2) = f''(x)$.
If $f(x) = \sum_{n=0}^{\infty} a_n x^n$,
$f(x^2) = \sum_{n=0}^{\infty} a_n x^{2n}$
and $f''(x) = \sum_{n=2}^{\infty} n (n-1)a_n x^{n-2}
= \sum_{n=0}^{\infty} (n+2) (n+1)a_{n+2} x^n
$
so $a_{2n} = (n+2) (n+1)a_{n+2}$
and $a_{2n+1} = 0$.
Setting $n = 0$,
$a_0 = 2a_2$.
Setting $n = 1$,
$a_2 = 6a+3 = 0$,
so $a+0 = 0$.
Setting $n = 2$,
$a_4 = 12 a_4$, so $a_4 = 0$.
If $n = 2k+1$, $a_{4k+2} = 0$.
In particular $a_6 = 0$.
For $n = 4$, $a_8 = 30a_6 = 0$.
Suppose there is a $n > 4$ for which $a_{2n} \ne 0$.
Let $m$ be the smallest such $n$.
Then, since $2m > m+2 > 6$,
$a_{2m} = (m+2)(m+1)a_{m+2} = 0$.
Therefore there is no such $m$,
and $a_n = 0$ for ann $n$.
Thus the only solution is $f(x) = 0$
if $f(x)$ has a power series expansion.
If $f(x) = a x^b$,
$f''(x) = a b (b-1) x^{b-2}$
and $f(\sqrt{x}) = a x^{b/2}$.
For this to be a solution,
$a = a b(b-1)$ and $b/2 = b-2$
or $b=4$ and $a=0$,
so there is no solution of this form.
There may be a non-zero solution not of these forms,
but I do not know of any.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/251882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$ I've been going through some series notes from my lecture and got stuck at this equality: $$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$$
Where $z,w\in\mathbb C$.
Rather than a proof, I'm looking more for a way of understanding this equality, so next time I see something similar I go "oh, right, I know that". Right now I only have vague ideas of why that would be true.
Thanks a bunch for any help!
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$$
(1+2+3+4+\cdots)\cdot\left(\begin{array} {} & \text{one} \\[6pt] + & \text{two} \\[6pt] + & \text{three} \\[6pt] + & \text{four} \\[6pt] + & \cdots \end{array}\right)
$$
$$
= \sum \left[ \begin{array}{cccc} 1\cdot\text{one}, & 2\cdot\text{one}, & 3\cdot\text{one}, & 4\cdot\text{one}, & \cdots\\[6pt]
1\cdot\text{two}, & 2\cdot\text{two}, & 3\cdot\text{two}, & 4\cdot\text{two}, & \cdots\\[6pt]
1\cdot\text{three}, & 2\cdot\text{three}, & 3\cdot\text{three}, & 4\cdot\text{three}, & \cdots\\[6pt]
1\cdot\text{four}, & 2\cdot\text{four}, & 3\cdot\text{four}, & 4\cdot\text{four}, & \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right]
$$
\begin{align}
& = \cdots\cdots\cdots +\sum\left[ \begin{array}{cccc} \cdot & \cdot & 3\cdot\text{one}, & \cdot & \cdots\\[6pt]
\cdot & 2\cdot\text{two}, & \cdot & \cdot & \cdots\\[6pt]
1\cdot\text{three}, & \cdot & \cdot & \cdot & \cdots\\[6pt]
\cdot & \cdot & \cdot & \cdot & \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right] \\[18pt]
& {}\qquad\qquad\qquad{}+ \sum\left[ \begin{array}{cccc} \cdot & \cdot & \cdot & 4\cdot\text{one}, & \cdots\\[6pt]
\cdot & \cdot & 3\cdot\text{two}, & \cdot & \cdots\\[6pt]
\cdot & 2\cdot\text{three}, & \cdot & \cdot & \cdots\\[6pt]
1\cdot\text{four}, & \cdot & \cdot & \cdot & \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array} \right] + \cdots\cdots\cdots
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find the radius of convergence of a power series expansion of the rational function $f(z)=\frac{(z^2)-1}{(z^3)-1 }$ I am having trouble figuring out the answer
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$$
\frac{z^2-1}{z^3-1} = \frac{(z-1)(z+1)}{(z-1)(z^2+z+1)} = \frac{z+1}{z^2+z+1}
$$
The denominator is $0$ when
$$z=\dfrac{-1\pm\sqrt{1^2-4\cdot1\cdot1}}{2} = \frac{-1\pm i\sqrt{3}}{2}. $$
If you want this in powers of $z$, i.e. powers of $(z-0)$, so the center is $0$, then the radius of convergence is the distance from the center, $0$, to the nearest point where the function fails to be well behaved, i.e from $0$ to either of $\dfrac{-1\pm i\sqrt{3}}{2}$. They're both at the same distance:
$$
\left| \dfrac{-1\pm i\sqrt{3}}{2} \right| = \frac14+\frac34=1.
$$
So that is the radius of convergence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/257337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
show that if $a,b,c \in \mathbb{R}^+$ show that if $a,b,c \in \mathbb{R}^+$ different from zero, then:
$$(a^2+b^2+c^2)\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq(a+b+c)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
I had no success in my attempts
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The direction of the inequality is flipped. Here is a proof of the correct direction.
Upon expansion we see it suffices to prove:
$$\frac{a^2}{b^2} + \frac{a^2}{c^2} + \frac{b^2}{a^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{c^2}{b^2} \ge \frac ab + \frac ac + \frac ba + \frac bc + \frac ca + \frac ca$$
Now note that for any positive real number $x$ we have $x^2 + \frac 1{x^2} \ge x + \frac 1x$ because the the function $f(x) = x + 1/x$ is monotonically increasing in $[1,\infty]$ and monotonically decreasing in $(0,1]$. Thus the result follows by applying this inequality on each of the terms of the above expression.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/261669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to prove $\sum\limits_{i =1}^{26} \frac{a_i}{\sum_{j =0}^{i} a_j^2} \leq \sqrt{26}$ Given arbitrary real numbers $a_i$,
Prove that
$$\sum_{i =1}^{26} \frac{a_i}{\sum_{j =0}^{i} a_j^2} \leq \sqrt{26}$$
where $a_0 = 1$
So it will look like:
$$\frac{a_1}{(1+a_{1}^2)} + \frac{a_2}{(1+ a_{1}^2 + a_{2}^2)} + \cdots + \frac{a_{26}}{(1+ a_1^2+ \cdots + a_{26}^2)}$$
|
Here is a proof for the more general inequality
$$
\sum_{i=1}^n \frac{a_i}{1 + \sum_{j=1}^i a_j^2} \leq \sqrt{n}.
$$
where $a_1, \ldots, a_n$ range over $\mathbb{R}$. The method is a little cumbersome so I would be interested in a more direct proof.
I assume this problem comes from a math competition and would be interested to know its source.
Let $S_n$ denote the sum. Clearly we can assume that each $a_i \geq 0$, as otherwise we could increase $S_n$ by swapping the sign of one of the variables. Now make the change of variables
$$
a_i = t_i \sqrt{1 + a_1^2 + \cdots + a_{i-1}^2}
$$
for each $i$ so that
$$
\frac{a_i}{1 + \sum_{j=1}^{i-1} a_j^2 + a_i^2} = \frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} \frac{t_i}{1 + t_i^2}
$$
and furthermore
$$
\frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} = \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}.
$$
Now we have
$$
S_n = \sum_{i=1}^n \frac{t_i}{1 + t_i^2} \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}.
$$
Grouping terms, we get
$$
S_n = \frac{t_1}{1 + t_1^2} + \frac{1}{\sqrt{1 + t_1^2}} \left( \frac{t_2}{1 + t_2^2} + \frac{1}{\sqrt{1 + t_2^2}} \left( \cdots \right)\right)
$$
We therefore define $f : [n] \to \mathbb{R}^+$ inductively by
$$
f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2}
$$
and
$$
f(k) = \max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + f(k+1) \frac{1}{\sqrt{1 + t_k^2}}.
$$
Then we have $S_n \leq f(1)$ (and in fact, $f(1) = \max_{t_1, \ldots, t_n} S_n$).
Note that if $M$ is an upper bound for $f(k+1)$, then
$$
\max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + M \frac{1}{\sqrt{1 + t_k^2}}
$$
is an upper bound for $f(k)$. It therefore suffices to show that
$$
f(k) \leq \sqrt{n+1-k}
$$
inductively in $k$, starting at $n$.
We have the base case
$$
f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2} \leq \frac{1}{2}.
$$
It therefore suffices for us to show
$$
\frac{t}{1 + t^2} + \sqrt{m} \frac{1}{\sqrt{1 + t^2}} \leq \sqrt{m+1}
$$
for every $m \geq 1$ uniformly in $t \geq 0$. With the trivial bound
$$
\frac{t}{1 + t^2} \leq \frac{t}{\sqrt{1 + t^2}}
$$
it suffices to show
$$
\frac{t + \sqrt{m}}{\sqrt{1 + t^2}} \leq \sqrt{m+1}.
$$
Squaring, it suffices to show
$$
\frac{t^2 + 2 t \sqrt{m} + m}{1 + t^2} \leq m + 1
$$
Or, equivalently,
$$
t^2 + 2 t \sqrt{m} + m \leq m + 1 + (m + 1) t^2
$$
or
$$
2 t \sqrt{m} \leq 1 + m t^2.
$$
But this is the AM-GM inequality for the pair $(1, mt^2)$, and the proof is complete.
Note that the method of proof showed that the inequality is not sharp for any $n$.
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"url": "https://math.stackexchange.com/questions/263180",
"timestamp": "2023-03-29T00:00:00",
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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.