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Prove this equality $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
This is how I did it:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4 \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=64$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=64$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$15 \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=60$$
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
Is this correct?
|
Close. As lemon pointed out, you started by assuming what was needed to be proven. This is easily fixed, however.
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=x \;\;\; |^{3}$$
$$2+11i+3\sqrt[3]{(2+11i)^2(2-11i)}+3\sqrt[3]{(2+11i)(2-11i)^2}+2-11i=x^3$$
$$4+3\sqrt[3]{(2+11i)(2-11i)} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3$$
$$3\sqrt[3]{4+121} \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3-4$$
$$15 \cdot (\sqrt[3]{2+11i}+\sqrt[3]{2-11i})=x^3-4$$
$$15 \cdot x=x^3-4$$
$$x^3-15x-4=0$$
$$(x-4)(x^2+4x+1)=0$$
Since both $\sqrt[3]{2+11i}$ and $\sqrt[3]{2-11i}$ have positive real parts, we know that their sum must have a positive part. Both roots of $x^2+4x+1$ are negative, as can easily be checked, so we have:
$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=4$$
I'm not sure if this is the easiest way to go, but it's the closest to what you already have.
|
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|
Primality Criterion for Specific Class of Proth Numbers Is this proof acceptable ?
Theorem :
Let $N = k\cdot 2^n+1$ with $n>1$ , $k<2^n$ , $3 \mid k $ , and
$\begin{cases}
k \equiv 3 \pmod {30} , & \text{with }n \equiv 1,2 \pmod 4 \\
k \equiv 9 \pmod {30} , & \text{with }n \equiv 2,3 \pmod 4 \\
k \equiv 21 \pmod {30} , & \text{with }n \equiv 0,1 \pmod 4 \\
k \equiv 27 \pmod {30} , & \text{with }n \equiv 0,3 \pmod 4
\end{cases}$
, thus
$N$ is prime iff $5^{\frac{N-1}{2}} \equiv -1 \pmod N$
Proof :
Necessity : $\textit{If}~ N ~\textit{is prime then} ~ 5^{\frac{N-1}{2}} \equiv -1 \pmod N $
Let $N$ be a prime , then by Euler criterion :
$5^{\frac{N-1}{2}} \equiv \left(\frac{5}{N}\right) \pmod N$
If $N$ is prime then $N \equiv 2,3 \pmod 5 $ and therefore : $\left(\frac{N}{5}\right)=-1$ .
Since $N \equiv 1 \pmod 4 $ according to the law of quadratic reciprocity it follows that : $\left(\frac{5}{N}\right)=-1$ .
Hence , $5^{\frac{N-1}{2}} \equiv -1 \pmod N $ .
Sufficiency : $\textit{If}~ 5^{\frac{N-1}{2}} \equiv -1 \pmod N~ \textit{then}~ N~ \textit{is prime}$
If $5^{\frac{N-1}{2}} \equiv -1 \pmod N$ then by Proth's theorem $N$ is prime .
|
You just took two (true) theorems and emerged them into one, and based your proof, just by assuming both theorems are true.
For me it is unacceptable.
Proving first part is just too easy, but it is always nice to also show why you used or took as granted the given formulas. Not many people heard about Legendre symbol.
Proof of second part is just in the (topic of) question itself.
It didn't take 8 lines for Euler or Proth to prove their theorems.
|
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|
Polynomial Factoring over a finite field Ok, so I'm trying to figure out how to factor polynomials over a finite field.
My polynomial is x^5 + x^2 + x + 1 and I have to factor over GF(2)
I know the answer is (x+1)^2 * (x^3 + x + 1), because both (x+1)^2 and (x^3 + x + 1) are irreducible. (x+1)^2 is an irreducible polynomial of order 2 and (x^3 + x + 1) is an irreducible polynomial of order 3. But what does that have to do with GF(2)?
I also know that if I were to factor the same polynomial over GF(3) the answer would be (x+1)(x^4 + 2x^3 + x^2 + 1). (x+1) is an irreducible polynomial of order 1 and (x^4 + 2x^3 + x^2 + 1) is an irreducible polynomial of order 4. But again, what does that have to do with GF(3)?
Does GF(2) mean my largest irreducible polynomial can only be an order of 3? And GF(3) my largest can only be an order of 4?
Can someone please verify this? Thanks!
|
If you multiply $(x+1)^2$ and $x^3+x+1$ you get $x^5+2 x^3+x^2+x+1$, and if you multiply $x+1$ and $x^4+2x^3+x^2+1$ you get $x^5+3x^4+3x^3+x^2+x+1$. You don't get $x^5+x^2+x+1$. So what's happening here?
The Galois field $GF(p)$ for a prime $p$ is the integers modulo $p$. When you factor a polynomial over $GF(p)$, you regard the coefficients as integers modulo $p$. This is why $x^5+2x^3+x^2+x+1 = x^5+x^2+x+1$ in the first case, and $x^5+3x^4+3x^3+x^2+x+1 = x^5+x^2+x+1$ in the second case. As you can see, the factorization depends on $p$.
There is also a definition of $GF(q)$ for prime powers $q$, but you can ignore it for now.
|
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|
tough factorisation problem How would you factorise this equation given that $x=7$ is a root of this equation
$$x^3 - 67x + 126 = 0.$$
Any help would be thoroughly appreciated.
|
$x^3-67x+126$ and the root is $x=7$. One writes $x=7$ as $x-7=0$ and convert the above equation in quadratic, by dividing $x^3-67x+126$ by $x-7$; so the quotient is $x^2+7x-18$. Now one factorises this $x^2+7x-18=x^2+9x-2x-18=x(x+9)-2(x+9)=(x-2)(x+9)$ then $\displaystyle(x-2)(x+9)=\frac{x^3-67x+126}{x-7}$, so $(x-2)(x+9)(x-7)$ is factorization of $x^3-67x+126$.
|
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|
Tricky Decomposition Decompose the following polynomial:
$A=x^4(y^2+z^2) + y^4(z^2+x^2) + z^4(x^2+y^2) +2x^2 y^2 z^2 $
(by the way I'm not quite sure about the tags and the title feel free to edit them if you wish)
|
Define the polynomial:
$$
f(a,b,c) = a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc
$$
Notice that $f$ is cyclic:
$$
f(a,b,c) = f(b,c,a) = f(c,a,b)
$$
Observe that since:
\begin{align*}
f(a,-a,c)
&= a^2(-a + c) + a^2(c + a) + 0 - 2a^2c \\
&= a^2(-a + c + c + a - 2c) \\
&= 0
\end{align*}
it follows by the Factor Theorem that $a+b$ is a factor of $f$. But since $f$ is cyclic, so are $b+c$ and $a+c$. Hence, for some constant $k$, we have:
$$
a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc = k(a + b)(b + c)(a + c)
$$
By comparing the coefficient of $a^2b$, observe that $k = 1$. Hence, we conclude that:
$$
A = f(x^2, y^2, z^2) = (x^2 + y^2)(y^2 + z^2)(x^2 + y^2)
$$
|
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|
Evaluate this infinite sum $\displaystyle \sum_{n=1}^{\infty} \left(x^{3n-2} - x^{3n+1}\right)$
We can't simplify anything any more. Except.
$\displaystyle \sum_{n=1}^{\infty} \left( \frac{x^{3n}}{x^2} - \frac{x^{3n}x^3}{x^2} \right)$
$\displaystyle \sum_{n=1}^{\infty} \frac{x^{3n}(1 - x^3)}{x^2}$
|
$\displaystyle S=\sum_{n=1}^{\infty} x^{3n-2} - x^{3n+1}=\dfrac{1-x^3}{x^2}\sum_{n=1}^\infty(x^3)^n$
If $|x^3|\le1,$ using Infinite Geometric Series Formula Derivation,
$\displaystyle S=\frac{1-x^3}{x^2}\cdot\dfrac{x^3}{1-x^3}=x$
|
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|
Zero as an Eigenvalue of stress tensor Given, $\sigma=\left[ \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right]$, I am interested in the eigenvalues as well as eigenvectors of the stress tensor. $\lambda_1=0, \lambda_2=\alpha\beta\sqrt{1+\beta^2}, \lambda_3=-\alpha\beta\sqrt{1+\beta^2}$. But i need to construct a system of mutually orthogonal vectors using the eigenvectors. I know that if I have $\vec{x_1}, \vec{x_1}, \vec{x_2}$, then $\vec{x_1}, \vec{x_2}, \vec{x_1}\times \vec{x_2}$ form a set of mutually orthogonal vectors. If $\vec{x_1}, \vec{x_1}, \vec{x_1}$, then $\vec{x_1}, \vec{x_2}, \vec{x_3}$ form a set of mutually orthogonal vectors, where $\vec{x_2}, \vec{x_3}$ are any orthogonal vectors lying in any plane perpendicular to $\vec{x_1}$. But how to find the eigenvectors and a set of mutually orthogonal vectors in the present case?
|
Assuming $\alpha\beta\ne 0$ we have:
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right)\left( \begin{array}{c}
\frac{1}{\beta} \\
1 \\
0 \end{array} \right) =0 \left( \begin{array}{c}
\frac{1}{\beta} \\
1 \\
0 \end{array} \right)$$
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 1 \end{array} \right)\left( \begin{array}{c}
\frac{-\beta}{\sqrt{1+\beta^2}} \\
\frac{1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right) =-\alpha \beta \sqrt{1+\beta^2} \left( \begin{array}{c}
\frac{-\beta}{\sqrt{1+\beta^2}} \\
\frac{1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right)$$
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right)\left( \begin{array}{c}
\frac{\beta}{\sqrt{1+\beta^2}} \\
\frac{-1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right) =\alpha \beta \sqrt{1+\beta^2} \left( \begin{array}{c}
\frac{\beta}{\sqrt{1+\beta^2}} \\
\frac{-1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right)$$
If $\alpha\beta=0$ the matrix is the zero matrix. In such a case any vector is an eigenvector with corresponding eigenvalue zero.
|
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|
What are the integer solutions of the system $a^2+b^2=c^2$, $a^3+b^3+c^3=d^3$? How to solve these equations to find the integer numbers (a, b, c, and d)?
$$a^2+b^2=c^2\tag{1}$$
$$a^3+b^3+c^3=d^3\tag{2}$$
I know one of solutions which is $a=3, b=4, c=5, d=6$, but I don't know if there are others solutions
|
$(3,4,5,6)$ is the only non-trivial solution involving strictly positive integers (by non-trivial I mean where all common factors are removed).
Proof:
$(a, b, c)$ is a Pythagorean triple (EDIT: and WLOG is a primitive triple since $k|a,b,c\implies k|d$), so it can be parametrised as $a = x^2 - y^2$, $b = 2xy$, $c = x^2 + y^2$ with $x > y \ge 1$ and $GCD(x, y) = 1$ (proof left as an exercise).
Then $d^3 = a^3 + b^3 + c^3 = \left(x^2-y^2\right)^3 + \left(2xy\right)^3 + \left(x^2+y^2\right)^3=2x^2\left(x^4+4xy^3+3y^4\right)$.
We first note that this is even, and so $d^3$ is even and hence divisible by 8. Then either $x$ is even (so that $2x^2$ provides the factor of 8) or $x^4+4xy^3+3y^4$ is divisible by 4. Note that these possibilities are not simultaneously possible, since that would imply that $y$ is even, contradicting the coprimality of $x$ and $y$.
If $x=2m$, then $d^3=8m^2\left(16m^4+8my^3+3y^4\right)$, which then implies that $m|16m^4+8my^3+3y^4$, and hence $m|y$ (i.e. $m$ divides evenly into $y$). If $m\neq 1$, then that would mean $x$ and $y$ share a common factor, which we've stated isn't the case, so $m=1$ which then gives $x=2$, and since $x>y\geq 1$ we must have $y=1$, which when you put it all into everything gives the $(3,4,5,6)$ solution.
On the other hand, if $x$ is odd (and greater than 1), then we must be able to pull a factor of $x$ out of $x^4+4xy^3+3y^4$ to make a perfect cube to equal $d^3$, which implies that $x|3y^4$. But since $x$ and $y$ are coprime, the only possibility is $x=3,y=1$. Putting these back into the parametrisation gives $a=8,b=6$ which is going to not give us a primitive/non-trivial solution.
We can relax our restriction down to $x\geq y\geq 1$, in which case $x=1,y=1$ is also a possibility, which gives $a=0,b=2$ which doesn't work for $a,b,c,d>0$ but does lead to the $\left(1,0,-1,0\right)$ solution with a bit of wrangling.
I suspect, but have no proof as yet, that these are the only two non-trivial solutions across all integers. The above proof may extrapolate neatly to negative values, or it may not.
|
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|
Evaluating $\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x$ Hint given: Write the integrand as an integral.
I'm supposed to do this as double integration.
My attempt:
$$\int_0^2 [\tan^{-1}y]^{\pi x}_{x}$$
$$= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}$$
$$= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}$$
$$= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
$$= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}$$
Carrying out this integration, I got, $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]$$
But I'm supposed to get $$2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5$$
Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you.
|
A more straight forward approach uses integration by parts.
Define:
\begin{align}
& I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2}
\end{align}
using partial fraction this reads:
\begin{align}
I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc}
\end{align}
taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done
|
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Why is $2x-1=7$ not $x=-4 \text{ or } x=4$ How would you explain why $3(2x-1)^2=147$, is $2x-1=7 \text{ or } 2x-1=-7$. But not $2x=8 \text{ so } x=4 \text{ or } x=-4$?
|
If $3(2x-1)^2=147=3\times7^2$, then you have an equation of the form $A^2=B^2$ with $A=2x-1$ and $B=7$, hence $A=\pm B$:
You can write that $A^2=B^2$ is equivalent to $A^2-B^2=0$, but $A^2-B^2=(A-B)\cdot(A+B)$, hence it's also equivalent to $(A-B)\cdot(A+B)=0$. But a product is zero if and only if at least one of the factor is zero, hence $A-B=0$ or $A+B=0$, that is, $A=B$ or $A=-B$.
Now, replace $A$ with $2x-1$, and $B$ with $7$, and you get:
$$2x-1=\pm 7$$
Which is an abuse of notation for $$(2x-1=+7)\ \ \ \mathrm{ or }\ \ \ (2x-1=-7)$$
So you have the two solutions $x=4$ and $x=-3$:
$$2x-1=+7\implies 2x=8\implies x=4$$
$$2x-1=-7\implies 2x=-6\implies x=-3$$
|
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Show sequance is monotonic Let $x>0$ (fixed) and $n$ be natural. Show that $$\displaystyle (x^n+x^{n-1}+...+1)^{\frac{1}{n}}$$ is monotonic.
I tried by induction but didn't work but intuition tells me it's decreasing.
|
For $x=1$, $\sqrt[n]{n}$ is incsreasing since $\frac{d}{{dy}}\left( {\sqrt[y]{y}} \right) = \sqrt[y]{y}\frac{{1 - \ln y}}{{{y^2}}} \leqslant 0$ for $e \geqslant y \geqslant 1$ and is decreasing for $y > e$. Indeed, the first four members of the sequence are 1, 1.41421, 1.44225 and 1.41421, so the sequence is not monotone, although it is monotone after a first couple of members.
If not, then ${a_n} = {\left( {\frac{{{x^{n + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{n}}}$, so
$\frac{d}{{dy}}{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}} = \underbrace {{{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)}^{\frac{1}{y}}}\frac{1}{{{y^2}}}}_{ \geqslant 0}\left( {\frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \left( {x - 1} \right) + \ln \frac{{{x^y}}}{{{x^{y + 1}} - 1}}} \right)$
Similarly as before we obtain, for $y$ sufficiently large,
$0 < x < 1 \Rightarrow \underbrace {y\left( {1 - \frac{1}{{1 - {x^{y + 1}}}}} \right)\ln x}_{ \leqslant 0} - \underbrace {\ln \left( {1 - {x^{y + 1}}} \right)}_{ \to 0} + \underbrace {\ln \left( {1 - x} \right)}_{ \leqslant 0}$
and
$x > 1 \Rightarrow \frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \frac{{{x^{y + 1}} - x}}{{{x^{y + 1}} - 1}} \geqslant \frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \frac{{{x^{y + 1}} - 1}}{{{x^{y + 1}} - 1}} = \frac{{y\ln x}}{{{x^{y + 1}} - 1}} \geqslant 0$.
So, $\frac{d}{{dy}}{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}}$ is either negative for sufficiently large $y$, or positive, so that $f\left( y \right) = {\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}}$ is either eventually descending or ascending. Since the sequence in question is the restriction ${\left. f \right|_\mathbb{N}}$, we conclude that the sequence is eventually decreasing for $0 < x < 1$ and always increasing for $x > 1$.
|
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Decide whether the $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational Working needs to be shown
$\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$
My guess is to multiply by $\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}$ then we have a rational number but is it enough to prove the rationality of a number?
|
If $\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}$ is rational, then so is $\dfrac{5}{\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}}$ and consequently $\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}\in \mathbb Q$.
The sum of two rational numbers is a rational number, thus $\left(\sqrt{\sqrt{5}+3}+\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{\sqrt{5}+3}-\sqrt{\sqrt{5}-2}\right)\in \mathbb Q$.
Proceed.
|
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How do I find the determinant of a 4x4 matrix when the diagonal is made up of variables? Evaluate: $\det(A)$, where $A=
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a\end{bmatrix}$
|
By adding appropriate multiples of row 4 to the other 3 rows, you can see that the determinant of your matrix is
$$
\left|
\begin{array}{cccc}
0 &1-a &1-a &1-a^2\\
0 & a-1 & 0 & 1-a\\
0 & 0 & a-1 & 1-a\\
1& 1 & 1 & a
\end{array}
\right|
$$
Then, using cofactor expansion along the first column, the determinant is
$$
-\left|
\begin{array}{ccc}
1-a &1-a &1-a^2\\
a-1 & 0 & 1-a\\
0 & a-1 & 1-a\\
\end{array}
\right|
$$
Every entry of the matrix here is a multiple of $(1-a)$, so this determinant is
$$
-(1-a)^3\left|
\begin{array}{ccc}
1 & 1 &1+a\\
-1 & 0 & 1\\
0 & -1 & 1\\
\end{array}
\right|
$$
Now you can use your favourite technique(s) for computing the determinant of a $3\times 3$ matrix.
|
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"url": "https://math.stackexchange.com/questions/1016943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does this system has a solution? Solve the system of equations
$$\begin{cases}2x_1+x_3+5x_4=0 \\x_1+2x_2-x_3=0\\x_1+x_2+2x_4=0\end{cases}$$
I created the matrix for this system and I found that it has many solution but I'm not sure if that true of not.
|
As you said, it has many solutions.
Working with the matrix we get
$ \left( \begin{array}{cccc}
2 & 0 & 1 & 5\\
1 & 2 & -1 & 0 \\
1 & 2 & 0 &2 \end{array} \right) \to$ $ \left( \begin{array}{cccc}
1 & 0 & 1/2 & 5/2\\
0 & 2 & -3/2 & -5/2 \\
0 & 2 & -1/2 & -1/2 \end{array} \right) \to$ $ \left( \begin{array}{cccc}
1 & 0 & 1/2 & 5/2\\
0 & 1 & -3/4 & -5/4 \\
0 & 0 & 1 &2 \end{array} \right)\\$ $\to\left( \begin{array}{cccc}
1 & 0 & 0 & 3/2\\
0 & 1 & 0 & 1/4 \\
0 & 0 & 1 &2 \end{array} \right)$
Follows that the solutions are of the form $a(6 ,1,8,-4)$, $a\in\mathbb{R}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3. Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3.
How do I start with this question? Rationalising the denominator? But how would I expand it?
|
Yes, you are right. You want to use the identity $$a^2-b^2=(a+b)(a-b)$$ In your case $a:=\sqrt{x^2-6x+13}$ and $b=2$. The expansion is $$\left(\sqrt{x^2-6x+13}-2\right)\cdot\left(\sqrt{x^2-6x+13}+2\right)=x^2-6x+13-4=x^2-6x+9$$ Now, this simplifies significantly with the term in the numerator (but do not forget to multiply the numerator with the conjugate as well).
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the integral $\int_{0}^{4}\sqrt{9t^2+t^4}dt$
Evaluate the integral $\displaystyle\int_{0}^{4}\sqrt{9t^2+t^4}dt$.
I know I have to substitute $u=9t^2+t^4$ and $\displaystyle\frac{du}{dt}=18t+4t^3$
But what should I do hereafter?
|
$\int{\sqrt{9t^2+t^4}}dt = \int{t\sqrt{9+t^2}}dt$
$u = 9+t^2 \implies du = 2t \cdot dt \implies t \cdot dt = \frac{1}{2}du$
$\frac{1}{2}\int{\sqrt{u}}du = \frac{1}{3}u^{\frac{3}{2}}+c = \frac{1}{3}(9+t^2)^{\frac{3}{2}}+c$
$\frac{1}{3}((9+16)^{\frac{3}{2}}-9^{\frac{3}{2}})=\frac{98}{3}$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
summation of $\sum^n_{k=0} (n-k)^2$ I'm trying to find the recurrence of
$$ T(n) = T (n-1) + n^2$$
After following the steps,
$$T (n) = T (n-1) + n^2 = T (n-2) + (n-1)^2 + n^2 $$
$$T (n) = T (n-2) + (n-1)^2 + n^2 = T(n-3) + (n-3)^2 + (n-1)^2 + n^2 $$
$$T (n) = T (n-3) + (n-3)^2 + (n-1)^2 + n^2 =T (n-4) + (n-4)^2 + (n-3)^2 + (n-1)^2 + n^2 $$
i generalize recurrence relation at the kth step of the recursion, which is
$$ T(n) = T (n-k) + \sum^n_{k=0} (n-k)^2$$
Just wondering,
What is the summation of
$$\sum^n_{k=0} (n-k)^2$$
is it the same as $$ n(n+1)(2n+1)/6$$
?
|
Hint: It is the same as $\sum_{k=0}^n k^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Chain rule with triple composition We are supposed to apply the chain rule on the following function $f$:
$$
f(x) = \sqrt{x+\sqrt{2x+\sqrt{3x}}}
$$
I assumed we could rewrite this as $$ f(x) = g(h(j(x))) $$
However, I was not sure how to define the functions $$g(x), h(x), j(x) $$
Any help would be appreciated.
|
1) $j(x) = 2x + \sqrt{3x}$
$j'(x) = 2 + \frac{\sqrt{3}}{2\sqrt{x}}$
2) $h(x) = x + \sqrt{2x + \sqrt{3x}} = x + \sqrt{j(x)}$
$h'(x) = 1 + \frac{1}{2*\sqrt{j(x)}} * j'(x)$ $=1 + \frac{1}{2*\sqrt{2x + \sqrt{3x}}} * (2 + \frac{\sqrt{3}}{2\sqrt{x}})$
3) $g(x) = \sqrt{x + \sqrt{2x + \sqrt{3x}}} = \sqrt{h(x)}$
$g'(x) = \frac{1}{2*\sqrt{h(x)}} * h'(x)$ $=\frac{1}{2*\sqrt{x + \sqrt{2x + \sqrt{3x}}}} * (1 + \frac{1}{2*\sqrt{2x + \sqrt{3x}}} * (2 + \frac{\sqrt{3}}{2\sqrt{x}}))$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1021755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find bases for eigenspaces of A $$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$
Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.
I don't really know where to start on this problem.
|
Consider the matrix $$\lambda I - A = \begin{pmatrix}\lambda-6 & -4 \\ 3 & \lambda +1\end{pmatrix}$$
Now, our eiegnavlues of $A$, are the solution to the equation $\det{(\lambda I - A)}=0$
\begin{align}\det{(\lambda I - A)}&=0 \\ \implies(\lambda -6)(\lambda+1)-(3)(-4) &=0 \\ \implies \lambda^2-5\lambda +6 &= 0 \\\implies(\lambda-2)(\lambda-3) &=0 \\ \implies \lambda_1=2 \text{ and } \lambda_2=3\end{align}
Thus $\lambda_1=2$ and $\lambda_2=3$ are the two eigenvalues corresponding to matrix $A$.
Since we have two distinct eigenvalues, we know that we will have two eigenvectors of $A$, $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, that corresponds to each eigenvector.
To find these eigenvectors, we must have that is satisfies the equation $(\lambda I - A)\bar{x}=\bar{0}$
For $\lambda_1 =2$: $$(2I-A)=\begin{pmatrix}-4 & -4 \\ 3 &3\end{pmatrix}$$
Which, reduces to $$\begin{pmatrix}1 &1 \\ 0 &0\end{pmatrix}$$
Thus we know $x_1 +x_2 = 0$. Let $x_2=t, \ t\in \mathbb{R} \implies x_1= -t$
Thus $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix} = t \begin{pmatrix}-1 \\ 1\end{pmatrix}, t \in \mathbb{R}$ is the eigenvector of $A$ corresponding to the eigenvalue $\lambda_1 =2$.
Thus the basis for the eigenspace of $A$ corresponding to $\lambda_1 = 2$, is given by $$E_{\lambda_1}=\bigg \{ \begin{pmatrix} -1 \\ 1\end{pmatrix} \bigg \}$$
For $\lambda_2 = 3$:
$$(3I - A) = \begin{pmatrix}-3 & -4 \\ 3 & 4\end{pmatrix}$$
which reduces to $$\begin{pmatrix}1 & \frac{4}{3} \\ 0 & 0\end{pmatrix}$$
Thus we know $x_1 + \frac{4}{3}x_2 =0$. Let $x_2 = s, s\in \mathbb{R} \implies x_1 = -\frac{4}{3}s$
Thus $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix} = s\begin{pmatrix}-\frac{4}{3} \\ 1\end{pmatrix}$ is the eigenvector of $A$ corresponding to the eigenvector $\lambda_2=3$.
Thus the basis for the eigenspace of $A$ corresponding to eigenvalue $\lambda_2 =3$ is:
$$E_{\lambda_2}=\bigg\{ \begin{pmatrix}-\frac{4}{3} \\ 1\end{pmatrix}\bigg \}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Does $a\ln(x^2 +y^ 2 )+b$ satisfy Laplace’s equation? I can't verify that $F(x,y) = a\ln(x^2 +y^ 2 )+b$ satisfies Laplace’s equation ($F_{xx}+F_{yy}=0$). Here is what I did:
\begin{align*}
F_x &= \frac{2ax}{x^2 + y^2}
&F_y &= \frac{2ay}{x^2+y^2}
\\
F_{xx} &= \frac{2a}{x^2+y^2} - \frac{4ax^2}{(x^2+y^2)^2}
&
F_{yy} &= \frac{2a}{x^2+y^2} - \frac{4ay^2}{(x^2+y^2)^2}
\end{align*}
and $F_{yy}\neq-F_{xx}$. So what should I do?
|
Let $F(x,y) = a \ln(x^2 + y^2) + b $. Then, Using quotient rule:
$$ F_x = \frac{ 2ax}{x^2 + y^2} \iff F_{xx} = \frac{2a(x^2+y^2)-2x(2ax)}{(x^2+y^2)^2} = \frac{2ay^2 - 2ax^2}{(x^2+y^2)^2}$$
Similarly,
$$ F_y = \frac{ 2ay}{x^2+y^2} \implies F_{yy} = \frac{2a(x^2+y^2) -2ay(2y)}{(x^2+y^2)^2} = \frac{2ax^2 - 2ay^2}{(x^2+y^2)^2}$$
Hence,
$$ F_{xx} + F_{yy} = 0 $$
|
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|
Modular arithmetic and using in well-ordering principle I need to prove the following, but I do not know how to go about it.
If $$ (*)\:\:\: x^{3} - y^{3}= 3^{n} $$
Then $$ x \equiv 0 (mod 3) \:\: and \:\:\: y \equiv 0 (mod 3)$$
In addition, to prove that the equation above (*), there is no solution in $ N^{+} $ (in other words $ \quad 0\neq x,y\in \mathbb N $ ).
By using the Well-ordering principle, i.e. Any non-empty subset of N, the set of natural numbers has a minimum element.
Edit
I can use these facts: $$ \forall x\in \mathbb N^{+} \:\:\: x^{3} \equiv x (mod 3)$$
And also:
$$ \forall x,y\in \mathbb N^{+} , \:\:\: if \:\:\: x^{3} - y^{3}\equiv 0 (mod 3) \:\:\: \\Then \:\:\: x \equiv y (mod 3) $$
Thanks
|
Let $x$ and $y$ be positive integers such that $x^3-y^3=3^n$, where $n\ge 3$. We first show that $3$ divides $x$ and $3$ divides $y$. Then later we use this to show that in fact there are no positive solutions.
In the OP, it is observed that $x\equiv y\pmod{3}$, so $x-y$ is divisible by $3$. Now we use the fact that $x^3-y^3=(x-y)(x^2+xy+y^2)$.
Note that if $x$ and $y$ are positive and not both equal to $1$, it follows that $x^2+xy+y^2\gt 3$, and therefore $x^2+xy+y^2=3^k$ for some $k\ge 2$. Now from the fact that $9$ divides $(x-y)^2$ and the identity
$$(x^2+xy+y^2)-(x-y)^2=3xy$$
we conclude that $9$ divides $3xy$. Thus $3$ divides $xy$. It follows that $3$ divides one of $x$ or $y$, and therefore since $x\equiv y\pmod{3}$, it divides both.
Now that we have the divisibility result, the rest follows by the least number principle, or equivalently by infinite descent.
It is easy to verify that there are no positive solutions with $n=0$, $n=1$, or $n=2$. Suppose that for some $n\ge 3$ there are positive solutions. Then there is a smallest $n\ge 3$ with this property. Let $x^3-y^3=3^n$. From our divisibility result, we have $x=3s$, $y=3t$ for some positive $s$ and $t$. But then $3^3s^3-3^3t^3=3^n$ and therefore $s^3-t^3=3^{n-3}$, contradicting the minimality of $n$.
Remark: It is not quite true that if $x$ and $y$ are integers such that $x^3-y^3$ is a power of $3$ then $3$ divides $x$ and $y$. There is the obvious example $x=1$, $y=0$. But there is also the more interesting example $x=2$, $y=-1$, for which we have $x^3-y^3=9$.
|
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|
How to integrate $\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx$ How do I integrate
\begin{equation}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx,
\end{equation}
which has arisen from a problem I'm working on? I've noticed I can do the following:
\begin{align}
\int_0^1\sqrt{-x^6+x^4-x^2+1}\:dx & =\int_0^1\sqrt{\left(-x^4\right)\left(x^2-1\right)-\left(x^2-1\right)}\:dx \\[3ex]
& = \int_0^1\sqrt{\left(-x^4-1\right)\left(x^2-1\right)}\:dx \\[3ex]
& = i\int_0^1 \sqrt{\left(x^4+1\right)\left(x+1\right)\left(x-1\right)}\:dx
\end{align}
But where do I go from here? Also, I'm a bit unsure about my last step above, i.e. not sure if it would be the right route to take.
Thanks in advance!
|
I think I can simplify it some, but I can't run it to ground.
Let $\theta = \cos^{-1} x$, so $x = \cos \theta$ and $dx = -\sin \theta \:d\theta$. Also $1 - x^2 = 1 - \cos^2 \theta = \sin^2 \theta$.
$$\begin{align}
\int_0^1 \sqrt{-x^6+x^4-x^2+1}\:dx & =
\int_{\frac{\pi}2}^0 -\sin \theta\sqrt{-\cos^6 \theta + \cos^4 \theta - \cos^2 \theta + 1} \:d\theta\\
& = \int_0^{\frac{\pi}2} \sin \theta\sqrt{(\cos^4 \theta + 1)(1 - \cos^2 \theta)} \:d\theta\\
& = \int_0^{\frac{\pi}2} \sin^2 \theta\sqrt{\cos^4 \theta + 1} \:d\theta\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sec^2(x) - \tan^2(x) = 1$ given that $\sin^2(x) + \cos^2(x) = 1$ I need to prove from $\sin^2(x) +\cos^2(x) = 1$ that $\sec^2(x) - \tan^2(x) = 1$.
|
$$ \sec^2(x)-\tan^2(x)=1 $$
$$ \frac{1}{\cos^2(x)}-\frac{\sin^2(x)}{\cos^2(x)}=1 $$
$$ \frac{1-\sin^2(x)}{\cos^2(x)}=1 $$
$$ 1-\sin^2(x)= \cos^2(x) $$
$$ 1= \sin^2(x)+\cos^2(x) $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the possible values of |A + B + C | $ |A |= |B | = |C | = 1 $ ,where A B and C are complex nos
$$ \frac{A^2}{BC}+ \ \frac{B^2}{ \ {CA}} \ +\ \frac{C^2}{ \ {AB}} + 1 = 0$$
Find the possible values of $ |A + B + C |$
Tried substituting cos(theta) + i sin(theta)
|
$$A=\cos a+i\sin a, B=\cos b+i\sin b, C=\cos c+i\sin c, a, b, c \in [0,2\pi).$$
From the given condition results:
$$\cos (2a-b-c)+\cos (2b-c-a)+\cos (2c-a-b)+1=0 $$
$$\sin (2a-b-c)+\sin (2b-c-a)+\sin (2c-a-b)=0 $$
or $$\cos x+\cos y+\cos z+1=0 $$
$$\sin x+\sin y+\sin z=0 $$ where
$$2a-b-c=x, 2b-c-a=y, 2c-a-b=z$$ with $$x+y+z=0$$
$$\cos \frac{x}{2}\cdot\cos \frac{y}{2}\cdot\cos \frac{z}{2}=0 $$
$$\sin \frac{x}{2}\cdot\sin \frac{y}{2}\cdot\sin \frac{z}{2}=0 $$
For $\cos \frac{x}{2}=0 $ and $\sin \frac{y}{2}=0$
result $x=(2k+1)\pi, y=2l\pi$
and
$2a-b-c=(2k+1)\pi, 2b-c-a=2l\pi.$
Find $a=c+\frac{(4k+2l+2)\pi}{3}, b=c+ \frac{(2k+4l+1)\pi}{3}.$
$$A+B+C=\cos (c+\frac{(4k+2l+2)\pi}{3})+\cos (c+\frac{(2k+4l+1)\pi}{3})+\cos c+$$
$$+i[\sin (c+\frac{(4k+2l+2)\pi}{3})+\sin (c+\frac{(2k+4l+1)\pi}{3})+\sin c]=$$
$$= \cos c-2\sin (c+(k+l)\pi)\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6}+$$
$$+i[\sin c+2\cos (c+(k+l)\pi)\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6})]$$
$$=\cos c-2(-1)^{k+l}\sin c\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6}+$$
$$+i[\sin c+2(-1)^{k+l}\cos c\cdot\cos((k-l)\frac{\pi}{3}+\frac{\pi}{6})]$$
$$|A+B+C|=1+4\cos^2 ((k-l)\frac{\pi}{3}+\frac{\pi}{6}) .$$
For $k-l =6p$ or $k-l =6p+4$ find $$|A+B+C|=1.$$
For $k-l =6p+1$ or $k-l =6p+2$ or $k-l =6p+3$ or $ k-l =6p+5$ find $$|A+B+C|=2.$$
|
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|
Proving the inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$ If $a,b,c$ are non-negative real numbers prove the following inequality $a^4+b^4+c^4+2abc(a+b+c)\ge \frac{(a+b+c)^4}9$.
|
Expanding, this is equivalent to (with $\sum$ denoting cyclic sums):
$$4\sum a^4 +3 abc\sum a \ge 2\sum ab(a^2+b^2)+3\sum a^2b^2 $$
which follows from combining Schur's inequality of fourth degree
$$3\sum a^4+3abc\sum a \ge 3\sum ab(a^2+b^2)$$
with the AM GMs:
$$\sum a^4 \ge \sum a^2b^2, \quad \sum ab(a^2+b^2) \ge 2\sum a^2b^2$$
|
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|
Exponential Function of Quaternion - Derivation The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a \left(\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2})\right)$$
I'm having a difficult time finding a derivation of this formula. I keep trying to derive it, but I end up getting different results. Would someone be able to point me to a proof of this formula or do the derivation here?
Note: I also don't understand why some people say $e^q = e^a e^{b i + c j + d k}$. Can you please explain this, too?
|
The definition of quaternionic exponential is given by the absolutely convergent series
$$
e^z=\sum_{k=0}^\infty\dfrac{z^k}{k!}
$$
It is well known that, from this definition, if $x, y$ commute we have $e^xe^y=e^ye^x=e^{x+y}$.
Since real quaternions commute with all other quaternions, for $a \in \mathbb{R}$ we have $e^{a+z}=e^ae^z \; \forall z\in \mathbb{H}$ so, if $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$, we have $e^z=e^ae^\mathbf{v}$, where $\mathbf{v}$ is an imaginary (or vector) quaternion.
Now we have:
claim
If $ \mathbf{v} \in \mathbb{H}_P$ is an imaginary quaternion, putting $\theta=|\mathbf{v}|$ we have:
$$
e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta}
$$
proof
We note that:
$$
\mathbf{v}^2= (b \mathbf{i}+c \mathbf{j} +d \mathbf{k})(b \mathbf{i}+c \mathbf{j} +d \mathbf{k})=
-b^2-c^2-d^2=-|\mathbf{v}|^2
$$
so:
$$
\mathbf{v}^2= -\theta^2
\quad,\quad \mathbf{v}^3= -\theta^2\mathbf{v}
\quad,\quad \mathbf{v}^4= \theta^4
\quad,\quad \mathbf{v}^5= \theta^4 \mathbf{v}
\quad,\quad \mathbf{v}^6= -\theta^6
\quad,\quad \cdots
$$
and the series become.
$$
\begin{split}
e^\mathbf{v}&=\sum_{k=0}^\infty\dfrac{\mathbf{v}^k}{k!}=\\
%
&=1+\dfrac{\mathbf{v}}{1!}-\dfrac{\theta^2}{2!}-\dfrac{\theta^2\mathbf{v}}{3!}+\dfrac{\theta^4}{4!}+\dfrac{\theta^4\mathbf{v}}{5!}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=1+\dfrac{\theta\mathbf{v}}{1!\,\theta}-\dfrac{\theta^2}{2!}-\dfrac{\theta^3\mathbf{v}}{3!\,\theta}+\dfrac{\theta^4}{4!}+\dfrac{\theta^5\mathbf{v}}{5!\,\theta}-\dfrac{\theta^6}{6!}+\cdots=\\
%
&=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{\mathbf{v}}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\
%
&=\cos\theta +\dfrac{\mathbf{v}}{\theta}\sin\theta
\end{split}
$$
So the exponential of a quaternion is:
$$
e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$
|
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|
Simplify - Help I am supposed to simplify this:
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
The answer is supposed to be this, but I can not seem to get to it:
$$x(x^2-1)(x^3+1)(5x^3-3x+2)$$
Thanks
|
$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$
Factor out the $(x^2-1)$ term to get
$$(x^2-1)\left((x^2-1) (x^3+1) (3x^2) + (x^3+1)^2(2x)\right)$$
then factor out the $(x^3+1)$ term to get
$$(x^2-1)(x^3+1)\left((x^2-1) (3x^2) + (x^3+1)(2x)\right)$$
then factor out the $x$ term to get
$$x(x^2-1)(x^3+1)\left((x^2-1) (3x) + (x^3+1)(2)\right)=x(x^2-1)(x^3+1)(5x^3-3x+2)$$
|
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|
Where did I go wrong in this integration $\int\frac{\ln(1-e^x)}{e^{2x}}\,dx$ Here is the closest I've come to the answer
Link to Wolfram equality not giving true as output
NB! I marked where I'm unsure in $\color{red}{red}$ color. And please don't get startled because of the wall of equation, it's simply a more tedious integration in my opinion.
$$\int\frac{\ln(1-e^x)}{e^{2x}}\,dx$$
let: $e^x= u \iff dx=\frac{1}{u}du$
$$\int\frac{\ln(1-u)}{u^3}\,du = -\frac{\ln(1-u)}{2u^2} - \frac{1}{2}\int\frac{1}{u^2(1-u)}\,du$$
This can be written as
$$-\frac{\ln(1-u)}{2u^2}-\frac{1}{2}\int\left(\color{red}{\frac{A}{u}+\frac{B}{u^2}-\frac{C}{1-u}}\right)\,du\qquad\qquad(1)$$
where $A,B,C$ is determined through setting
$$\frac{1}{u^2(1-u)} = \color{red}{\frac{A}{u}+\frac{B}{u^2}-\frac{C}{1-u}}$$
By multiplying with LHS denominator and factoring RHS I get
$$1 = (-A-C)t^2+(A-B)t+B \implies B = A = 1 \land C = -1$$
This put into $(1)$ gives us
$$-\frac{\ln(1-u)}{2u^2}-\frac{1}{2}\int\left(\frac{1}{u}+\frac{1}{u^2}-\frac{-1}{1-u}\right)\,du$$
and
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}+\int\frac{1}{1-u}\,du\right)$$
another substitution $w = 1-u \iff du = -dw$ gives us
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}-\int\frac{1}{w}\,dw\right)$$
Finally this is equivalent to
$$-\frac{1}{2}\left(\frac{\ln(1-u)}{u^2} + \ln(u) - \frac{1}{u}-\ln(w)\right) + C$$
Substitution back to x and factoring gives us
$$-\frac{1}{2e^{2x}}\left(\ln(1-e^x) + \color{red}{e^{2x}\cdot x} - e^x - e^{2x}\ln(1-e^x)\right) + C$$
|
$$I=\int\frac{\ln(1-u)}{u^3}\,du$$
$$I=\frac{\ln(1-u)}{-2u^2}-\int\frac{1}{2u^2(1-u)}$$
Now, $$\frac{1}{u^2(1-u)}=\frac{1}{u^2}+\frac1{u}+\frac1{(1-u)}$$
So:
$$I=\frac{1}{2}\left(-\frac{\ln(1-u)}{u^2}-\left(-\frac1u+\ln u-\ln|1-u|\right)\right)+c\\
I=\frac{1}{2e^{2x}}\left(-\ln(1-e^x)+e^{x}-e^{2x}x+e^{2x}\ln|1-e^x|\right) + c$$
|
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|
Finding Fourier cosine series of sine function I am trying to find Fourier cosine series of following function, but think that I am messing up somewhere.
$$
f(x)=\sin \bigg ( \frac{\pi x}{l} \bigg )
$$
Fourier cosine series can be written as
$$
f(x)=\frac{a_0}{2} + \sum\limits_{n=1}^\infty a_n \cos \bigg ( \frac{n \pi x}{l} \bigg )\\
a_n=\frac{2}{l} \Bigg [\int\limits_{0}^l \sin \bigg ( \frac{\pi x}{l} \bigg )\cos \bigg ( \frac{n \pi x}{l} \bigg)\,dx \Bigg]\\
=\frac{-1}{l} \Bigg [\frac{\cos(n(1+\pi))}{\frac{n(1+\pi)}{l}} + \frac{\cos(n(1-\pi))}{\frac{n(1-\pi)}{l}} - \frac{1}{\frac{n(1+\pi)}{l}} + \frac{1}{\frac{n(1-\pi)}{l}} \Bigg]\\
=\Bigg [\frac{-\cos(n(1+\pi))}{n(1+\pi)} + \frac{\cos(n(1-\pi))}{n(1-\pi)} + \frac{1}{n(1+\pi)} - \frac{1}{n(1-\pi)} \Bigg]
$$
The problem is I am getting cosine terms in the expansion, which is against my thinking that cosine series expansion of sine function should be zero. Am I messing up in the algebra part or thinking? Can someone comment? Thanks!
|
With
$$f(x)=\sin\left(\dfrac{\pi x}{l}\right),$$
you want to find $a_n$. The formula says:
$$a_n=\dfrac{2}{l}\int_0^l \sin\left(\dfrac{\pi x}{l}\right) \cos\left(\dfrac{n\pi x}{l}\right)dx.$$
With $u=\pi x/l$, you get:
$$a_n=\dfrac{2}{\pi}\int_0^\pi \sin(u) \cos(nu)\, du.$$
One of these identities can help you:
$$\int\sin (a_1x) \cos(a_2x)\, dx= -\dfrac{\cos((a_1-a_2)x)}{2(a_1-a_2)}-\dfrac{\cos((a_1+a_2)x)}{2(a_1+a_2)}+C.$$
With $a_1=1$ and $a_2=n$:
$$\int_0^\pi\sin (u) \cos(nu)\, du = -\dfrac{\cos((1-n)\pi)}{2(1-n)}-\dfrac{\cos((1+n)\pi)}{2(1+n)} + \dfrac{1}{2(1-n)}+\dfrac{1}{2(1+n)}$$
If $n$ is even:
$$\int_0^\pi\sin (u) \cos(nu)\, du = -\dfrac{-1}{2(1-n)}-\dfrac{-1}{2(1+n)} + \dfrac{1}{2(1-n)}+\dfrac{1}{2(1+n)}=\dfrac{2}{(1-n)(1+n)}$$
If $n$ is odd:
$$\int_0^\pi\sin (u) \cos(nu)\, du = -\dfrac{1}{2(1-n)}-\dfrac{1}{2(1+n)} + \dfrac{1}{2(1-n)}+\dfrac{1}{2(1+n)}=0$$
Special cases:
$$\int_0^\pi\sin (u)\, du =2$$
$$\int_0^\pi\sin (u) \cos(u)\, du =0$$
Now, you can find the expression for $a_n$.
|
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|
Find the area of a triangle if its two sides measure $6 in.$ and $9 in.$, and the bisector of the angle between the sides is $4\sqrt{3}$ in. Find the area of a triangle if its two sides measure $6$ in. and $9$ in., and the bisector of the angle between the sides is $4\sqrt{3}$ in. I'm thinking of using the formula $A$=$\frac{1}{2}bh$ I can't find the base or height, I used the Angle bisector formula which is $l=$$\frac{\sqrt{ab[(a+b)^2-c^2]}}{a+b}$ So i found out C which i think is the base should i multiply it by $2$? because I think it's the half. From here I'm lost
|
Let $|BC|=a$, $|AC|=b$, $|AB|=c$,
angle bisector $|AD|=\beta_a$ in $\triangle ABC$.
Given $b=6,\ c=9,\ \beta_a=4\sqrt3$, find the area $S_{ABC}$.
Using known expression for the length of the angle bisector,
\begin{align}
\beta_a&=bc\left(
1-\frac{a^2}{(b+c)^2}
\right)
\tag{1}\label{1}
,
\end{align}
the missing side length is found as
\begin{align}
a&=(b+c)\sqrt{1-\frac{\beta_a^2}{bc}}
=5
\tag{2}\label{2}
,
\end{align}
and the area is found as usual by Heron’s formula
\begin{align}
S_{ABC}&=
\tfrac14\,\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=10\sqrt2
\tag{3}\label{3}
.
\end{align}
Or, in terms of the given lengths,
\begin{align}
S_{ABC}&=
\tfrac14\,\beta_a\,(b+c)\,\sqrt{4-\beta_a^2\,(\tfrac1b+\tfrac1c)^2}
=10\sqrt2
\tag{4}\label{4}
\end{align}
|
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|
Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
|
We don't seem to have 15 yet:
${2\cdot 2 + 2 \choose 2} = 15$
If we're going to use $\Gamma$ and the like, we can get $30$ too:
$2 \cdot {\Gamma(2+2) \choose 2} = 30$
No $19$ or $27$ or $29$ yet either (though this keeps getting more absurd):
$\lfloor{2+\sqrt{2}}\rfloor^{\lfloor{2+\sqrt{2}}\rfloor} = 27$
$\lfloor \mathrm{Exp}(2)\rfloor \cdot \lfloor{2 + \sqrt{2}}\rfloor - 2 = 19$
$\lfloor \mathrm{Exp}(2)\rfloor \cdot (2 + 2) + \lfloor{\sqrt{2}}\rfloor = 29$
I think every number has appeared once now (some rather more questionably than others).
|
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|
Indefinite Integral of Reciprocal of Trigonometric Functions How to evaluate following integral
$$\int \frac{\mathrm dx}{\sin^4x+\cos^4x\:+\sin^2(x) \cos^2(x)}$$
Can you please also give me the steps of solving it?
|
$$\int \frac{1}{\sin^4x+\cos^4x+\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{\sin^4x+\cos^4x\:+2\sin^2\left(x\right)\cos^2\left(x\right)-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{(\sin^2x+\cos^2x)^2-\sin^2\left(x\right)\cos^2\left(x\right)}dx$$
$$=\int \frac{1}{1-\frac{1}{4}\sin^2\left(2x\right)}dx\quad\because \sin2x=2\sin x\cos x$$
|
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|
Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
|
Here is another way to look at it: Use the FOIL method on the right hand side.
$(x-5-2)(x-5+2)$ becomes $x^2-5x+2x-5x+25-10-2x+10-4$
Combine and cancel your like terms and you're left with $x^2-10x+25-4$, which of course can be rewritten as $(x-5)^2-4$.
|
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|
Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$ I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?
$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$
Thanks in advance.
|
$$\mathrm{cosec} A+\mathrm{cotan} A=\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\\=\frac{1+\cos A}{\sin A}\frac{\cos A+\sin A-1}{\cos A+\sin A-1}\\=\frac{\cos A+\sin A-1+\cos^2 A+\sin A\cos A-\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(1)}\frac{\sin A-1+\cos^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(2)}\frac{\sin A-\sin^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(3)}\frac{1-\sin A+\cos A}{\cos A+\sin A-1},$$ where:
in $(1)$ we cancel $\cos A$ and $-\cos A$ in the numerator;
in $(2)$ we use $\sin^2A+\cos^2 A=1;$
in $(3)$ we divide numerator and denominator by $\sin A.$
|
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|
Using mathematical induction to prove $\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$ This induction problem is giving me a pretty hard time:
$$\frac{1}1+\frac{1}4+\frac{1}9+\cdots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
I am struggling because my math teacher explained us that in this case ($n^2$) when we prove that $n+1$ satisfies the property we have to write it like this: $$ \frac{1}1+\frac{1}4+\frac{1}9+\cdots+ \frac{1}{n^2} +\frac{1}{(n+1)^2} < \frac{4(n+1)}{(2(k+1)+1)}$$
and I probably got lost in the process.
|
$$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}<\frac{4n}{2n+1}$$
fist proof for one: $1< \frac{4}{3}$,
for k-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{k^2}<\frac{4k}{2k+1}$$
for k+1-> $$\frac{1}1+\frac{1}4+\frac{1}9+\ldots+\frac{1}{n^2}+\frac{1}{(k+1)^2}<\frac{4k+4}{2k+3}$$
Now,
$$\frac{4k}{2k+1}+\frac{1}{(k+1)^2}<\frac{4k+4}{2k+3}$$
if we demostrate that:
$$-\frac{4k}{2k+1}-\frac{1}{(k+1)^2}+\frac{4k+4}{2k+3}>0$$
then it is demonstrate
|
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|
Proving $x^2 - y^2 + z^2 \gt (x - y + z)^2$ Prove that
$$x^2 - y^2 + z^2 > (x - y + z)^2$$
where: $x < y <z$ for all natural numbers.
Thank for help.
|
Hints:
$(x-y+z)^2-z^2=(x-y)(x-y+2z) > (x-y)(x-y+2y)=(x-y)(x+y)\\=x^2-y^2$
|
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|
Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ How can we prove that:
$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$
|
\begin{align}J&=\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx\\
U&=\int_0^1\frac{\log(1+x)\log^3 x}{1-x}dx,V=\int_0^1\frac{\log(1-x)\log^3 x}{1+x}dx\\
J&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln^3 x\ln(1-x)\ln(1+x)\Big]_0^1-\frac{1}{3}\int_0^1 \left(\frac{1}{1+x}-\frac{1}{1-x}\right)\ln^3 x\,dx\\
&=\frac{1}{3}\Big(U-V\Big)\\
\end{align}
From Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques
One obtains:
\begin{align*}U&=-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\\
V&=\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)\\
J&=\boxed{\frac{3}{4}\zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)}
\end{align*}
|
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|
Integrate the function Find the following integral,
$$I_n=\displaystyle\int (\arctan \theta)^n\ d\theta$$
Thinking that maybe there is some reduction formula, I tried using integration by parts but unable to derive it. Any help will be appreciated.
Actually the problem that I have been struggling over for quite some time is of finding $I_2$. If it can be solved, that will also be of great help.
|
The basic method for obtaining $\mathcal{I}_{1}$ is integration by parts:
$$\begin{align}
\mathcal{I}_{1}
&=\int\arctan{(x)}\,\mathrm{d}x\\
&=x\arctan{(x)}-\int\frac{x}{x^2+1}\,\mathrm{d}x\\
&=x\arctan{(x)}-\frac12\ln{\left(x^2+1\right)}+\color{grey}{constant}.\\
\end{align}$$
The strategy for obtaining $\mathcal{I}_{2}$ is similar, except this time we have to integrate by parts twice. Then, an explicit formula for the anti-derivative may be obtained in terms of a non-elementary special function known as the Clausen function:
$$\begin{align}
\mathcal{I}_{2}
&=\int\arctan^2{(x)}\,\mathrm{d}x\\
&=x\arctan^2{(x)}-\frac12\ln{\left(x^2+1\right)}\arctan{(x)}-\int\frac{x\arctan{(x)}-\frac12\ln{\left(x^2+1\right)}}{x^2+1}\,\mathrm{d}x\\
&=x\arctan^2{(x)}-\frac12\ln{\left(x^2+1\right)}\arctan{(x)}-\int\frac{x\arctan{(x)}}{x^2+1}\,\mathrm{d}x\\
&~~~~~ +\frac12\int\frac{\ln{\left(x^2+1\right)}}{x^2+1}\,\mathrm{d}x\\
&=x\arctan^2{(x)}-\frac12\ln{\left(x^2+1\right)}\arctan{(x)}-\frac12\ln{\left(x^2+1\right)}\arctan{(x)}\\
&~~~~~ +\frac12\int\frac{\ln{\left(x^2+1\right)}}{x^2+1}\,\mathrm{d}x+\frac12\int\frac{\ln{\left(x^2+1\right)}}{x^2+1}\,\mathrm{d}x\\
&=x\arctan^2{(x)}-\ln{\left(x^2+1\right)}\arctan{(x)}+\int\frac{\ln{\left(x^2+1\right)}}{x^2+1}\,\mathrm{d}x\\
&=x\arctan^2{(x)}-\ln{\left(x^2+1\right)}\arctan{(x)}+\int\ln{\left(\sec^2{\theta}\right)}\,\mathrm{d}\theta;~~x=\tan{\theta}\\
&=x\arctan^2{(x)}-\ln{\left(x^2+1\right)}\arctan{(x)}-2\int\ln{\left(\cos{\theta}\right)}\,\mathrm{d}\theta\\
&=x\arctan^2{(x)}-\ln{\left(x^2+1\right)}\arctan{(x)}+2\ln{(2)}\,\theta-\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}+\color{grey}{constant}\\
&=x\arctan^2{(x)}-\ln{\left(x^2+1\right)}\arctan{(x)}+2\ln{(2)}\arctan{(x)}-\operatorname{Cl}_{2}{\left(\pi-2\arctan{(x)}\right)}+\color{grey}{constant}.\\
\end{align}$$
|
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|
Show $ 5 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ≥ (1+a)(1 +b)(1+c)$ when $abc =1$ Just an inequality that I couldn't solve:
For $a,b,c$ positive reals such that $abc = 1$, show
$$5 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge (1 + a)(1 + b)(1 + c).$$
WhatI have done so far: I have shown this is equivalent, after some algebra, to $3+a^2c+b^2a + c^2b \ge ab+bc+ac + a + b + c$, and I have absolutely no idea where to go from there.
|
Substitute $a = u/v, b = v/w, c = w/ u$ to get rid of the constraint $abc=1$, and you have the equivalent inequality
$$\sum_{cyc} u^3v^3 + 3u^2v^2w^2 \ge uvw \sum_{cyc} uv(u+v)$$
Now let $x = uv, y = vw, z = wu$ and we get in these new variables the inequality:
$$\sum_{cyc} x^3 + 3xyz \ge \sum_{cyc} xy(x+y)$$
which is the familiar Schur's inequality.
|
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|
Give $x,y,z: \begin{cases}x^2+y^2+z^2=5\\x-y+z=3\end{cases}$. Find Min, Max of $P=\frac{x+y-2}{z+2}$ Give $x,y,z: \begin{cases}x^2+y^2+z^2=5\\x-y+z=3\end{cases}$
Find Min, Max of $P=\frac{x+y-2}{z+2}$
Please help me ?
|
edit 1:
It is easy to go directly:
$x^2+(x+z)^2+z^2=5 \implies x= \pm \dfrac{\sqrt{1+6z-3z^2}-z+3}{2},y=\pm \dfrac{\sqrt{1+6x-3x^2}+z-3}{2},1-\dfrac{2}{\sqrt{3}}\le z\le 1+\dfrac{2}{\sqrt{3}}$
$P= \dfrac{\pm2\sqrt{1+6z-3z^2}-2}{z+2}$
edit2:
$P=-\dfrac{2(1+\sqrt{1+6z-3z^2})}{z+2},P'=-\dfrac{2(\sqrt{-3z^2+6z+1}+9z-5) }{\sqrt{-3z^2+6z+1}(z^2+4z+4)}=0,z_1=\dfrac{4-\sqrt{2}}{7},P_{min}=P(z_1)$,
$P=\dfrac{2(\sqrt{1+6z-3z^2}-1)}{z+2},P'=\dfrac{2(\sqrt{-3z^2+6z+1}-9z+5) }{\sqrt{-3z^2+6z+1}(z^2+4z+4)}=0,z_2=\dfrac{4+\sqrt{2}}{7},P_{max}=P(z_2)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I prove this seemingly simple trigonometric identity $$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$
Show that, $8bc=a[4b^2 + (b^2-c^2)^2]$
I tried to solve this for hours and have gotten no-where. Here's what I've got so far :
$$
\\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})
\\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta-\phi)}
\\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$
|
$\color{red}{c^2=\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi\tag{1}}$
$\color{blue}{b^2=\tan^2\theta+\tan^2\phi+2\tan\theta\tan\phi\tag{2}}$
$(1)-(2)$ gives,
$\begin{align}\left(c^2-b^2\right) & =\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi-\tan^2\theta-\tan^2\phi-2\tan\theta\tan\phi\\ &=2(1+\sec\theta\sec\phi-\tan\theta\tan\phi)\\&=2\left(1+\dfrac{1}{\cos\theta\cos\phi}-\dfrac{\sin\theta\sin\phi}{\cos\theta\cos\phi}\right)\\&=\dfrac{2}{\cos\theta\cos\phi}(1+\cos(\theta+\phi))\\&=\dfrac{4\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}\end{align}$
$\color{darkgreen}{\therefore\left(c^2-b^2\right)^2=\dfrac{16\cos^4\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2\phi}\tag{3}}$
$\color{brown}{4b^2=\dfrac{4\sin^2(\theta+\phi)}{\cos^2\theta\cos^2 \phi}=\dfrac{16\sin^2\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}\tag{4}}$
$$\boxed{4b^2+\left(b^2-c^2\right)^2=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$
$c=\dfrac{\cos\theta+\cos\phi}{\cos\theta\cos\phi}=\dfrac{2\cos\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{\cos\theta\cos\phi}$
$b=\dfrac{\sin(\theta+\phi)}{\cos\theta\cos\phi}=\dfrac{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}$
$a=2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)$
$$\boxed{\dfrac{8bc}{a}=\dfrac{32\sin\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)\cos^2\theta\cos^2 \phi}=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.
Therefore the nth term of given series is :
$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)
By using partial fraction :
and solving for A,B and C we get A = 1/2, B = -1, C =1/2
Putting the values of A,B and C in (1) we get :
$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$
But by putting $r =1,2,3, \cdots$ I am not getting the answer. Please guide how to solve this problem . Thanks.
|
You can get the result to pop out if you play around with the indices of the infinite series and expand a couple terms. $$\sum_{r=1}^\infty\frac{2r-1}{r(r+1)(r+2)} = \sum_{r=1}^\infty\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)} \\ = \frac{1}{2}\sum_{r=1}^\infty \frac{1}{r}-\sum_{r=1}^\infty\frac{1}{r+1}+\frac{1}{2}\sum_{r=1}^\infty\frac{1}{(r+2)} \\ = \frac{1}{2}\left(1+\frac{1}{2}+\sum_{r=3}^\infty \frac{1}{r}\right)-\left(\frac{1}{2}+\sum_{r=3}^\infty\frac{1}{r}\right)+\frac{1}{2}\sum_{r=3}^\infty\frac{1}{r} \\ = \frac{1}{2}\left(1+\frac{1}{2}\right)-\frac{1}{2}+\sum_{r=3}^\infty \frac{1}{r}\left(\frac{1}{2}-1+\frac{1}{2} \right)$$ The answer should be clear from here.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. I then set $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$. However, I afterwards realized that I had squared an inequality to get this. The following is what my proof looked like before I realized this mistake:
Let $\varepsilon > 0$. Choose $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$ and let $n > N$. Then $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. Thus $2n\varepsilon > 1 - \varepsilon^2$, $\varepsilon^2 + 2n\varepsilon > 1$, $\varepsilon^2 + 2n\varepsilon + n^2 = (\varepsilon + n)^2 > 1 + n^2$, $\varepsilon + n > \sqrt{1 + n^2}$, and $\varepsilon > \sqrt{n^2 + 1} - n$. Hence $| \sqrt{n^2 + 1} - n - 0 | = \sqrt{n^2 + 1} - n <
\varepsilon.$ Therefore, $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$.
How should I find $N$? Also, are there any other errors in this proof?
|
For any positive $c$,
$(n+\frac{c}{2n})^2
=n^2+c+\frac{c^2}{4n^2}
> n^2+c
$,
so
$\sqrt{n^2+c}
<n+\frac{c}{2n}
$.
Therefore,
if $c > 0$,
$n
<\sqrt{n^2+c}
<n+\frac{c}{2n}
$
so
$\lim_{n \to \infty} \sqrt{n^2+c}-n
= 0
$.
You can prove this when $c < 0$
similarly.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.
Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
$$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$
Along my proof I am stuck at the above section where it would be shown that:
$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$ is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.
|
The solution must be a quadratic polynomial (because its first order difference is a linear polynomial) so it is obtained with the Lagrangian formula on three points. Taking $(0,0),(1,1),(2,5)$,
$$0\frac{(n-1)(n-2)}{(0-1)(0-2)}+1\frac{(n-0)(n-2)}{(1-0)(1-2)}+5\frac{(n-0)(n-1)}{(2-0)(2-1)}=\frac{n(3n-1)}2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate the sum Please help me to evaluate the following
$$\sum\limits_{n=1}^\infty \frac{\left(\frac{3-\sqrt{5}}{2}\right)^n}{n^3}$$
Have no idea how to evaluate. Any trick please?
Thanks
|
Differentiating: $\displaystyle f(z) = \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right)$
We have, $\displaystyle f'(z) = \frac{\operatorname{Li}_2(z)}{z} - \frac{\operatorname{Li}_2(1-z)}{1-z} +\frac{\operatorname{Li}_2\left(1 - \frac{1}{z}\right)}{1 - \frac{1}{z}}.\left(\frac{1}{z^2}\right)$
Using the reflection formula and Abel Identity as indicated here we have,
$\displaystyle \operatorname{Li}_2(z)-\operatorname{Li}_2\left(1 - \frac{1}{z}\right)=\zeta(2)-\ln z\ln(1-z)+\frac12\ln^2 z$
$\displaystyle \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1 - \frac{1}{z}\right)=-\frac12\ln^2z$
Since, $$\displaystyle \begin{align}f'(z) &= \frac{\operatorname{Li}_2(z)}{z} - \frac{\operatorname{Li}_2(1-z)}{1-z} +\frac{\operatorname{Li}_2\left(1 - \frac{1}{z}\right)}{1 - \frac{1}{z}}.\left(\frac{1}{z^2}\right) \\ &= \frac{\operatorname{Li}_2(z)-\operatorname{Li}_2\left(1-\frac{1}{z}\right)}{z}-\frac{\operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)}{1-z} \\ &=\frac{\zeta(2)-\ln z\ln(1-z)}{z}+\frac{\ln^2 z}{2z} + \frac{\ln^2 z}{2(1-z)} \tag{1}\end{align}$$
We can differentiate $\displaystyle g(z) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}$ and see that it is equal to $f'(z)$, and it agrees with $f(z)$ at $z=1$, i.e., $g(1) = f(1) = \zeta(3)$.
Alternatively, direct integration of $(1)$ yields $g(z)$.
Hence, $\displaystyle \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}$
Now, to compute $\operatorname{Li}_3(\xi)$, where, $\displaystyle \xi = \frac{3-\sqrt{5}}{2}$ we observe, $\displaystyle \xi^2 - 3\xi + 1= 0 \implies \frac{-\xi}{1-\xi} = -(1-\xi)$
and,
$\displaystyle \begin{align} &f(1-\xi) \\&= \operatorname{Li}_3(\xi) + \operatorname{Li}_3(1-\xi)+\operatorname{Li}_3\left(\frac{-\xi}{1-\xi}\right) \\&= \operatorname{Li}_3(\xi) + (\operatorname{Li}_3(1-\xi) + \operatorname{Li}_3(-(1-\xi))) \tag{2}\\&= \operatorname{Li}_3(\xi) + \frac{1}{4}\operatorname{Li}_3((1-\xi)^2) \\&= \frac{5}{4}\operatorname{Li}_3(\xi) \end{align}$
In step $(2)$, we used: $\displaystyle \operatorname{Li}_3(z)+\operatorname{Li}_3(-z) = \frac{1}{4}\operatorname{Li}_3(z^2)$
Equating and simplifying with $\xi = \phi^{-2}$ yields $\displaystyle \operatorname{Li}_3(\phi^{-2})= \frac{4}{5}\zeta(3)+\frac{2}{3}\log^3\phi-\frac{2\pi^2}{15}\log\phi.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
|
Just apply the following reduction formula inductively.
$$\frac{x^n-1}{x-1} = x^{n-1}+\frac{x^{n-1}-1}{x-1}$$
Derivation
$$\begin{array}{lll}
\frac{x^n-1}{x-1}&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^n - x^{n-1}+x^{n-1}-1}{x-1}\\
&=&\frac{x^{n-1}(x-1)+x^{n-1}-1}{x-1}\\
&=&x^{n-1}+\frac{x^{n-1}-1}{x-1}\\
\end{array}$$
|
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|
Prove using induction $2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$
Show that
$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$
I'm not really sure how to get started on this problem, but here is what I have done so far:
Base case $n(1)$:
$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$
But not sure where to go from here. Tips?
|
For simplicity, please note that $2^{3n}=8^n$.
First, show that this is true for $n=1$:
*
*$\frac{8^1-1}{7}=1\in\mathbb{N}$
Second, assume that this is true for $n$:
*
*$\frac{8^n-1}{7}=k\in\mathbb{N}$
Third, prove that this is true for $n+1$:
*
*$\frac{8^{n+1}-1}{7}=\frac{8\cdot8^n-1}{7}$
*$\frac{8\cdot8^n-1}{7}=\frac{8\cdot8^n-8+7}{7}$
*$\frac{8\cdot8^n-8+7}{7}=\frac{8(8^n-1)+7}{7}$
*$\frac{8(8^n-1)+7}{7}=\frac{8\cdot7k+7}{7}$ assumption used here
*$\frac{8\cdot7k+7}{7}=\frac{7(8k+1)}{7}$
*$\frac{7(8k+1)}{7}=8k+1\in\mathbb{N}$
|
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|
Why is $\sum \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$ convergent? We have the following series:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$$
According to WolframAlpha it is convergent but I don't see why. Intuitively, the expression under the sum must be assymptotically similair to something like $\frac{1}{n^\alpha}$, and clearly $\sqrt{n^2}$ dominates the denominator. In the numerator the expression looks similair either to $\sqrt{n^3}$ but $\frac{\sqrt{n^3}}{\sqrt{n^2}}={\sqrt{n}}$ which gives as assymptotic similarity to sequence, the sum of which is divergent. On the other hand, if $\sqrt{n}$ dominates the numerator we get $\frac{\sqrt{n}}{\sqrt{n^2}}=\frac{1}{\sqrt{n}}$ so again assymptotic similarity to sequence, which gives divergent sum.
So why is this sum convergent anyway?
|
You may write, as $n$ is great:
$$
\begin{align}
\frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}&=\frac{(\sqrt{n^3+n}-\sqrt{n^3})(\sqrt{n(n+1)}+1)}{n(n+1)-1}\\\\
&=\frac{n^{5/2} }{n^2}\frac{\left(\sqrt{1+\frac{1}{n^2}}-1\right)\left(\sqrt{1+\frac{1}{n^2}}+\frac1n\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\
&=\frac{n^{5/2} }{n^2}\frac{\left(\frac{1}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\left(1+\mathcal{O}\left(\frac{1}{n}\right)\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\
& \sim \frac{n^{5/2}}{n^4}=\frac{1}{n^{3/2} }
\end{align}
$$ and your series is convergent by the comparison test.
|
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|
Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3^2=5\cdot 2^1$$
$$2^2+4^2=5\cdot 2^2$$
$$2^2+6^2=5\cdot 2^3$$
|
Consider the following two cases:
$$n=2k\implies 5\cdot 2^{2k}=(1+4)\cdot 2^{2k}=2^{2k}+2^{2k+2}=\left(2^{k}\right)^2+\left(2^{k+1}\right)^2$$
$$n=2k+1\implies 5\cdot 2^{2k+1}=10\cdot 2^{2k}=(1+9)\cdot 2^{2k}=2^{2k}+9\cdot 2^{2k}=\left(2^{k}\right)^2+\left(3\cdot 2^{k}\right)^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help with $\int \frac{1}{(\sin x + \cos x)}$ Kindly solve this question
$$\int \frac{1}{(\sin x + \cos x)} dx$$
I reached up to
$$\frac{(1+\tan^2x)}{1-\tan^2x + 2\tan x}$$
|
Hint :
\begin{align}
\int \frac{\mathrm dx}{\sin x + \cos x} &=\int \frac{\mathrm dx}{\sin x + \cos x}\cdot \frac{\sin x - \cos x}{\sin x - \cos x}\\
&=\int \frac{\sin x - \cos x}{\sin^2 x - \cos^2 x}\mathrm dx\\
&=\int \frac{\sin x}{1 - 2\cos^2 x}\mathrm dx-\int \frac{\cos x}{2\sin^2 x -1}\mathrm dx\\
&=\int \frac{1}{2\cos^2 x-1}\mathrm d(\cos x)-\int \frac{1}{2\sin^2 x -1}\mathrm d(\sin x)\\
\end{align}
|
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|
Evaluating $ \int \frac{1}{\sin x} dx $ Verify the identity
$$\sin x = \frac {2 \tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$$
Use this identity and the substitution $t = \tan\frac{x}{2}$ to evaluate the integral of $$ \int \frac{1}{\sin x} dx $$
My answer:
$$ A = \frac{x}{2} \Rightarrow \sin 2A= \frac{2\tan A}{1+\tan^2 A} = \frac{2\tan A}{\sec^2 A} = 2 \tan A\cos^2 A = 2 \sin A \cos A = \sin 2A $$
Since $x=A/2$, $\sin 2A = \sin x$
Let $t=\tan \frac{x}{2}$
$$ \int \frac{1}{\sin x} dx = \int \frac{2t}{1+t^2} dt $$
Let $u= 1+t^2$, $ du = 2t\,dt$
$$\int \frac{2t}{1+t^2} dt = \int \frac{2t}{u}\cdot\frac{1}{2t} du = \int \frac{1}{u} \,du = \ln u + C \\ = \ln(1+t^2) + C = \ln\left(1+\tan^2 \frac{x}{2} \right) + C $$
Then what do I do?
How do I show this is equal to $-\cos x + C$ ?
|
$$I=\int\frac{dx}{\sin x}=\int\frac{1+\cos x}{\sin x}\frac{dx}{1+\cos x}=\int\frac{1+\cos x}{\sin x}d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\ln\left|\frac{\sin x}{1+\cos x}\right|+C=\ln\left|\tan {\frac{x}{2}}\right|+C$$
|
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|
Just a proof of algebra If $a+b+c=0$,
Show that
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this problem but can't find a solution. Please help me.
|
Assuming $a,b,c$ are non-zero & distinct
$$F=\dfrac{c}{a-b}\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]$$
$$=1+\frac c{a-b}\cdot\frac{b^2-bc+ca-a^2}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a)-c(b-a)}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a-c)}{ab}$$
$$=1-\frac{c(b+a-c)}{ab}$$
As $b+a=-c,$
$$F=1-\frac{c(-c-c)}{ab}=1+2\cdot\frac{c^2}{ab}$$
Now using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
$a+b+c=0\implies a^3+b^3+c^3=3abc$
Now, divide both sides by $abc$
|
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|
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$.
We show that it is also divisible by $11$ when $n = k + 2$
$2^{3k+5} + 5\cdot 3^{k+2}$
$32\cdot 2^3k + 5\cdot 9 \cdot3^k$
$32\cdot 2^3k + 45\cdot 3^k$
$64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$)
$(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$
The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?
|
$\begin{align}2^{3}\equiv -3\pmod {11} & \implies 3^n\equiv 2^{3n}\pmod {11} \ \color{blue}{(\text{since}\ n\ \text{is even})}\\& \implies 2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}+5\cdot 2^{3n}\pmod {11}\\&\implies2^{3n-1}+5\cdot 3^n\equiv 2^{3n-1}(1+5\cdot 2)\pmod {11}\\&\implies 2^{3n-1}+5\cdot 3^n\equiv 0\pmod {11}\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.$
So $\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.$
Now $\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi$
Now How can I calculate $\sin 2\phi.$
Help me, Thanks
|
Let $$A = (1+\sin(\theta))(1+\cos(\theta))$$ and $$B = (1-\sin(\theta))(1-\cos(\theta))$$ Observe that $$A+B = 2+2\sin(\theta)\cos(\theta)$$ $$AB = (\sin(\theta)\cos(\theta))^2$$
Substituting $A = 5/4$, we now have a quadratic in $B$ that is easily solved.
|
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|
If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
|
$$a^a \ b^b \;?\; a^b \ b^a \\
\frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\
\left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\
\left(\frac{a}{b}\right)^{a-b} \;?\; 1 $$
if $a \ge b$, then $c = \frac{a}{b} \ge 1$, and $d = a-b \ge 0$. Thus $c^d \ge 1$, so $?$ is $\ge$.
if $a < b$, then:
$$\left(\frac{a}{b}\right)^{a-b} \\
= \left(\frac{b}{a}\right)^{b-a}$$
and we re-use the first result by symmetry.
|
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|
Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$
Evaluate
$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$
I found the following
*
*$\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$
*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$
Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation
$$8x^3+4x^2-4x-1=0$$
And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.
For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution
$$x\mapsto x^3$$
and getting another equation
$$8x^9+4x^6-4x^3-1=0$$
However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.
Also, it's given that the sum evaluates to a radical of the form
$$\sqrt[3]{\frac{1}{d}(a-b\sqrt[b]{c})}$$
where $a, b, c \: \text{&} \: d \in \mathbb Z$
Can somebody please help me with this question?
Thanks!
|
let $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$
then we have
$$\begin{cases}
x^3+y^3+z^3=-\dfrac{1}{2}\\
(xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\
(xyz)^3=\dfrac{1}{8}
\end{cases}$$
use this identity
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$
so
$$\begin{cases}
(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz=-\dfrac{1}{2}\\
(xy+yz+xz)^3-3(xy+yz+xz)[xyz(x+y+z)]+3x^2y^2z^2=-\dfrac{1}{2}\\
xyz=\dfrac{1}{2}
\end{cases}$$
let $$u=x+y+z, v=xy+yz+xz$$
then we have
$$\begin{cases}
u^3-3uv+2=0\\
4v^3-6uv+5=0
\end{cases}$$
so we have
$$\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}$$
so
$$4\left(\dfrac{u^3+2}{3u}\right)^3-2u^3+1=0\Longrightarrow 4u^9-30u^6+75u^3+32=0$$
let $t=u^3$,so we have
$$4t^3-30t^2+75t+32=0$$
let $t=\dfrac{5}{2}-a$,then
$$4\left(\dfrac{5}{2}-a\right)^3-30\left(\dfrac{5}{2}-a\right)^2+75\left(\dfrac{5}{2}-a\right)+32=0$$
$$\Longrightarrow 4a^3=\dfrac{189}{2}\Longrightarrow a=\dfrac{3\sqrt[3]{7}}{2}$$
so
$$u=x+y+z=\sqrt[3]{\cos{\dfrac{2\pi}{7}}}+\sqrt[3]{\cos{\dfrac{4\pi}{7}}}+\sqrt[3]{\cos{\dfrac{6\pi}{7}}}=\sqrt[3]{\dfrac{1}{2}\left(5-3\sqrt[3]{7}\right)}$$
|
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|
A reduction formula for $\int_0^1 x^n/\sqrt{9 - x^2}\,\mathrm dx$
Let $$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2\sqrt2$$
I've found that $\displaystyle I_0 = \sin^{-1}\left(\frac{1}{3}\right)$, but that's it.
I'm struggling to go any further. Anyone have any hints?
|
$$I_n = \int_0^1 \frac{x^n}{\sqrt{9 - x^2}}\,\mathrm dx= \int_0^1 x^{n-1}\cdot\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx$$
Using Integration By Parts
$$\frac{x}{\sqrt{9 - x^2}}\,\mathrm dx=\,\mathrm dv\iff-\sqrt{9-x^2}=v$$
$$x^{n-1}=u\iff (n-1)x^{n-2}\,\mathrm dx=\,\mathrm du$$
$$\begin{align}
I_n
&= -x^{n-1}\cdot\sqrt{9-x^2}\Bigg|_0^1+(n-1)\int_0^1x^{n-2}\sqrt{9-x^2}\,\mathrm dx\tag{1}\\
&= -2\sqrt2+(n-1)\int_0^1\frac{x^{n-2}(9-x^2)}{\sqrt{9-x^2}}\,\mathrm dx\tag{2}\\
&= -2\sqrt2+9(n-1)\int_0^1\frac{x^{n-2}}{\sqrt{9-x^2}}\,\mathrm dx-(n-1)\int_0^1\frac{x^{n}}{\sqrt{9-x^2}}\,\mathrm dx\tag{3}\\
&= -2\sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}\tag{4}\\
\end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2\sqrt2$$
|
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|
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
|
\begin{align}
&\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx
=2\int_{0}^{1}\frac{(x^2-1)\ln{x}}{1+x^4}dx \\
=&\ \sqrt2 \int_{0}^{1}\ln x \ d\left( -\tanh^{-1}\frac{\sqrt2x}{1+x^2}\right)
\overset{ibp}={\sqrt2} \int_0^1 \frac1x \tanh^{-1}\frac{\sqrt2x}{1+x^2}dx\\
=& \ {\sqrt2} \int_0^1 \int_0^{\pi/4}
\frac{2(x^2+1)\cos t}{(x^2+1)^2-4x^2\sin^2t}dt\ dx\\
= & \ {\sqrt2} \int_0^{\pi/4} \tan^{-1}\frac{2x\cos t}{1-x^2}\bigg|_{x=0}^{x=1}\ dt= {\sqrt2} \int_0^{\pi/4} \frac\pi2 dt
=\frac{\pi^2}{4\sqrt{2}}
\end{align}
|
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|
Find $x$ such that $x \equiv7\pmod {37}$ and $x^2 \equiv 12\pmod {37^2}$ Find $x$ such that $x \equiv7 \pmod {37}$ and $x^2 \equiv 12\pmod {37^2})$
My attempt: Given $x \equiv7\pmod {37}$
so $37|(x-7)$ so $37^2|(x-7)^2$
so $x^2-14x+49 \equiv 0\pmod {37^2}$
as $12-14x+49 \equiv 0\pmod{37^2}$ [as $x^2 \equiv 12\pmod {37^2}$]
so $14x\equiv 61\pmod {37^2}$ now I can find a set of solution using Euclidean algorithm
I would like some one to verify if this is correct as it seems longer and also is there a better and shorter method
|
I would do $$(37n+7)^2=12\pmod{37^2}\\14n*37+49=12\pmod{37^2}\\14n+1=0\pmod{37}$$
which has a smaller modulo
|
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|
What's $\int \frac{1}{\sqrt{25-x^2}}$ What is $$\int \frac{1}{\sqrt{25-x^2}}$$ WolframAlpha says $\sin^{-1}(\frac{x}{5})$ while I got $\frac{1}{5}\sin^{-1}(\frac{x}{5})$. What is correct?
Thanks in advance.
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Let $x = 5u \to dx = 5du \to \displaystyle \int \dfrac{dx}{\sqrt{25-x^2}} = \displaystyle \int \dfrac{5du}{\sqrt{25-25u^2}} = \displaystyle \int \dfrac{du}{\sqrt{1-u^2}} =\arcsin u + C = \arcsin \left(\frac{x}{5}\right) + C$
|
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|
Inequality $\frac{x^3+y^3}{x-y}>4$ Let $x>y>0$ and $xy\geq 1$. Prove that $$\frac{x^3+y^3}{x-y}>4.$$
Of course we can factor $(x^3+y^3)=(x+y)(x^2-xy+y^2)$, but it is not very useful. For fixed $x-y$, we can try to find the minimum value of $x,y$ such that $xy\geq 1$, and show that $\frac{x^3+y^3}{x-y}>4$ for those values. But that amounts to solving the quadratic $x(x-k)= 1$, and the answer would be quite ugly to substitute into the inequality.
|
$$x^3+y^3 - 4(x-y) \ge 2(x^2 - xy +y^2) - 4(x-y) $$
$$= 2(x^2 - 2xy +y^2 - 2(x-y) + xy) \ge 2(x-y-1)^2$$
Where, we used $x+y \ge 2\sqrt{xy} \ge 2$ and $xy \ge 1$
|
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|
Trigonometric equation, find $\sin \theta $ Find $\sin \theta $ if $a$ and $c$ are constants
$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$
|
after simpßlification we find this here
$$2\,\sin \left( \theta \right) \left( \cos \left( \theta \right)
\right) ^{5}ac- \left( \cos \left( \theta \right) \right) ^{8}+{a}^{
2} \left( \cos \left( \theta \right) \right) ^{6}- \left( \cos
\left( \theta \right) \right) ^{6}{c}^{2}+2\, \left( \cos \left(
\theta \right) \right) ^{6}- \left( \cos \left( \theta \right)
\right) ^{4}{a}^{2}-2\,\sin \left( \theta \right) {a}^{2}c\cos
\left( \theta \right) -{a}^{3} \left( \cos \left( \theta \right)
\right) ^{2}+a{c}^{2} \left( \cos \left( \theta \right) \right) ^{2}
-a \left( \cos \left( \theta \right) \right) ^{2}+{a}^{3}
=0$$
i think it is implossible to find an explicit formula for $\theta$ try a numerical method
|
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|
How can I write this power series as a power series representation? How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
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Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots) $$
$$ =(1+x)(1+2y+3y^2+4y^3 +\ldots)$$
$$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$
$$ = (1+x)\left( \frac{y}{1-y}\right)'$$
Edit:
$$ = (1+x) \frac{1}{(1-y)^2} $$
$$ = \frac{1+x}{(1-x^2)^2} $$
$$ = \frac{1}{(1-x)(1+x^2)}.$$
|
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|
Diophantine equation : two products of linear factors differ by a constant Recently, I was asked the following question by a friend : find
all $a,b,c,a',b',c',k \in {\mathbb Z}$ with $k\neq 0$ such that the identity
$$
(X-a)(X-b)(X-c)+k=(X-a')(X-b')(X-c')
$$
holds in ${\mathbb Z}[X]$. Does anyone have an idea ?
Note that there is a certain translation invariance for the solutions :
$(a,b,c,a',b',c',k)$ is a solution iff $(a+m,b+m,c+m,a'+m,b'+m,c'+m,k)$ is,
for any $m\in{\mathbb Z}$.
|
These are not all solutions, but one infinite family.
For example, for any integers $x,y,z$, take $a = xyz$, $b = -x(x+y)z$, $c = -y(x+y)z$, $k = x^2 y^2 (x+y)^2 z^3$. Note that $ab + ac + bc = 0$. If $p(X) = (X-a)(X-b)(X-c)$, we have
$p(0) = -x^2 y^2 (x+y)^2 z^3$ and $p'(0) = 0$, and then
$$p(X) + k = X^2 (X + (x^2 + xy + y^2) z )$$
EDIT: The same value of $(x^2 + x y + y^2) z$ can occur for different $x,y,z$, so
this produces different solutions with $a,b,c$ distinct. For example,
try $x,y,z = -3,8,1$ and $x', y', z' = -3,1,7$: we get
$$ (X+24)(X-15)(X+40) = (X+21)(X+42)(X-14)-2052$$
EDIT: Let $q(a,b,c) = (X-a)(X-b)(X-c)$, so you want to solve $q(a,b,c) +k = q(a',b',c')$. Note that this is true for some $k$ iff
$\dfrac{d}{dX} q(a,b,c) = \dfrac{d}{dX} q(a',b',c')$, and that is equivalent to
$$ \eqalign{a + b + c &= a' + b' + c'\cr
ab + ac + bc &= a'b' + a'c' + b'c'\cr}$$
Now given two integers $u,v$, we may try to solve
$$ \eqalign{a + b + c &= u\cr
ab + ac + bc &= v\cr}$$
Eliminating $c$ gives us
$$ 3 (2a + b - u)^2 + (3 b - u)^2 = {4} u^2 - 12 v $$
Given $u$ and $v$, the equation $3 y^2 + z^2 = 4 u^2 - 12 v$ has finitely many integer solutions $(y,z)$. We want solutions where $z \equiv -u \mod 3$
(so $b = (z+u)/3$ is an integer), and where $y \equiv u - b \mod 2$ (so
$a = (y+u-b)/2$ is an integer).
For example, with $u=79$ and $v = 1920$ there are $24$ integer solutions of $3 y^2 + z^2 = 1924$. Of these, $12$ satisfy $z \equiv - u \mod 3$, and all these
satisfy $y \equiv u - b \mod 2$. The corresponding $(a,b,c)$ values are the six permutations of $(15,24,40)$ and the six permutations of $(12,31,36)$.
Thus these correspond to
$$ (X - 15)(X - 24)(X - 40) + k = (X - 12)(X - 31)(X - 36)$$
where in this case $k = 15\cdot 24 \cdot 40 - 12 \cdot 31 \cdot 36 = 1008$.
The next step would be to look at this in terms of factorization in $\mathbb Z[\sqrt{-3}]$.
|
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|
Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$ The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.
I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?
|
1 comes from long division of $x^2+2$ by $x^2-1.$ when you divide $x^2 + 2$ by $x^2 - 1$ you get the quotient $1$ and the remainder $3.$ that is $$x^2 + 2 = 1(x^2-1)+3 \mbox{ or }{x^2 + 2 \over x^2 -1} = 1 + {3 \over x^2 - 1}$$
this is very similar to how you do division in integers. for example when you divide $123$ by $7,$ the quotient to be 17(how many times 7 goes into 123) and remainder(what is left over) to be 4. we can write this in two ways:
$$123 = 7*17 + 4 \mbox{ or } {123 \over 7} = 17 + {4 \over 7}$$
|
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|
Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
0\end{align}
$$
Define $x = \frac{4^k - 1}{3}$. So far I have:
$$k_1 \to 1 \Longrightarrow x_1 \to 1
\\
k_2 \to 2 \Longrightarrow x_2 \to 5
\\
k_3 \to 3 \Longrightarrow x_3 \to 21
\\
k_4 \to 4 \Longrightarrow x_4 \to 85$$
But then it's evident that
$$4^{k_n} = x_{n+1} - x_n$$
I don't know if this helps, these are ideas floating in my head.
|
Note that $x^k - y^k$ is always divisible by $x-y$. (In fact, $x^k - y^k = (x - y)(x^{k-1} + x^{k-2}y + x^{k-3}y^2 + \cdots + y^{k-1})$ Specialize for $x=4, y=1$.
|
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|
Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ After seeing the neat little identity $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ somewhere on MSE, I tried generalising this to higher consecutive powers in the form
$\sum_{k=0}^a\epsilon_k(n+k)^p=C$, where $C$ is a constant and $\epsilon_k=\pm1$. I discovered a relatively simple algorithm to generate these patterns: simply take $(n+2^p-1)^p$ and subtract $(n+2^p-2)^p$ (using $n\to n-1$) to get a polynomial of degree $p-1$. Take this difference and subtract $(n+2^p-3)^p-(n+2^p-4)^p$ (using $n\to n-2$) to get a polynomial of degree $p-2$. Repeat this process until $n^p$ is reach. The first few examples of this are
$$
\begin{align}
n^0&=1\\
(n+1)^1-n^1&=1\\
(n+3)^2-(n+2)^2-(n+1)^2+n^2&=4\\
(n+7)^3-(n+6)^3-(n+5)^3+(n+4)^3-(n+3)^3+(n+2)^3+(n+1)^3-n^3&=48
\end{align}
$$
Upon doing this for the next several powers and checking OEIS, it would appear the constant corresponding to the power $p$ is $$\large C_p=2^{\frac{p(p-1)}2} p!$$
However, this is an observation only, and I have no idea how to go about proving this. The only thing I notice is that $\frac{p(p-1)}{2}=\sum\limits_{k=1}^{p-1}k$, but I don't know how to use this fact. Does any one know how to prove this observation?
|
Note that since $R$ and $1$ commute,
$$
R^{2^k}-1=\left[\sum\limits_{j=0}^{2^k-1}R^j\right](R-1)\tag{1}
$$
Therefore,
$$
\begin{align}
\prod_{k=0}^{n-1}\left(R^{2^k}-1\right)x^n
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right](R-1)^nx^n\tag{2a}\\
&=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right]n!\tag{2b}\\
&=\left[\prod_{k=0}^{n-1}2^k\right]n!\tag{2c}\\[6pt]
&=2^{n(n-1)/2}\,n!\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $(1)$
$\text{(2b)}$: the $n^\text{th}$ forward difference of $x^n$ is $n!$
$\text{(2c)}$: on a constant sequence, $R^j=1$
$\text{(2d)}$: $\sum\limits_{k=0}^{n-1}k=n(n-1)/2$
|
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|
How to integrate $f(x) = \frac{1}{a + b \cos x + c \sin x }$ over $x \in (0,\pi/2)$
Conjecture 1
$$ \begin{align*} I_{T}=\int_0^{2\pi} \frac{\mathrm{d}x}{a + b \cos x + c
\sin x} & = \frac{2\pi}{\rho} \tag{1} \\ I_{T/4} = \int_0^{\pi/2} \frac{\mathrm{d}x}{a +
b \cos x + c \sin x} & = \frac{2}{\rho} \arctan \left(\frac{\rho}{a+b+c}\right)\tag{2}
\end{align*} $$
Where $\rho^2 = a^2-b^2 -c^2$.
*
*Is this result correct, and does it cover all cases?
*Does the result above cover all cases - what if the denominator becomes a perfect square?
*When does the integral converge?
I am mainly posting this so that i can close all the duplicates in
the future. Integrals related to these seems to pop up regularly.
|
Assuming $b^2+c^2<a^2$ and expressing $\sin(x)$ and $\cos(x)$ in terms of $\tan(x/2)$ we have:
$$I_{T/4}=2\int_{0}^{1}\frac{dt}{a(1+t^2)+b(1-t^2)+c(2t)}=2\int_{0}^{1}\frac{dt}{(a-b)t^2+2ct+(a+b)}.$$
Since the discriminant of $p(t)=(a-b)t^2+2ct+(a+b)$ is negative by the initial assumptions, $p(t)$ does not vanish on $[0,1]$. Assuming $a>b$, we have:
$$\begin{eqnarray*}I_{T/4}&=&2(a-b)\int_{0}^{1}\frac{dt}{(a-b)^2 t^2 + 2c(a-b)x + (a^2-b^2)}\\&=&2(a-b)\int_{0}^{1}\frac{dt}{\left((a-b)t+c\right)^2+\rho^2}\\&=&2\int_{0}^{a-b}\frac{dt}{(t+c)^2+\rho^2}=2\int_{c}^{a-b+c}\frac{dt}{t^2+\rho^2}\\&=&\left.\frac{2}{\rho}\arctan\frac{t}{\rho}\right|_c^{a+b-c},\end{eqnarray*}$$
so:
$$ I_{T/4}=\frac{2}{\rho}\left(\arctan\frac{a+b-c}{\rho}-\arctan\frac{c}{\rho}\right)$$
and since $\arctan x-\arctan y=\arctan\frac{x-y}{1+xy}$ the last formula can be put into the form:
$$ I_{T/4}=\frac{2}{\rho}\arctan\frac{\frac{a+b-2c}{\rho}}{1+\frac{c(a+b-c)}{\rho^2}}.$$
|
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|
Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.
|
Set $\dfrac1x=h^2$ where $h>0$ to get
$$\lim_{ x \to \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}$$
$$=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}}}}=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\frac1h}}=\sqrt{\frac1{h^2}+\sqrt{\frac{h+1}{h^2}}}$$
$$=\sqrt{\frac1{h^2}+\frac{\sqrt{h+1}}h}=\sqrt{\frac{1+h\sqrt{h+1}}{h^2}}=\frac{\sqrt{1+h\sqrt{h+1}}}h$$
$$\implies\lim_{ x \to \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\lim_{h\to0}\frac{\sqrt{1+h\sqrt{h+1}}-1}h$$
$$=\lim_{h\to0}\frac{1+h\sqrt{h+1}-1}{h(\sqrt{1+h\sqrt{h+1}}+1)}=\frac{\sqrt{0+1}}{\sqrt{1+0\sqrt{0+1}}+1}$$
|
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|
Why is $\cos(x/2)+2\sin(x/2)=\sqrt5 \sin(x/2+\tan^{-1}(1/2))$ true? According to Wolfram Alpha the following equality holds:
$$\cos\left(\frac{x}{2}\right)+2\sin\left(\frac{x}{2}\right)=\sqrt5 \sin\left(\frac{x}{2}+\tan^{-1}\left(\frac{1}{2}\right)\right)$$
I also checked it numerically. Why is it true?
|
Using $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,
\begin{align}
\sin\left(x + \tan^{-1} \left(\frac 1 2\right) \right)
&= \sin x \cos \left(\tan^{-1}\left(\frac 1 2\right)\right) +
\cos x \sin \left(\tan^{-1}\left(\frac 1 2\right)\right) \\
&= \sin x \frac 2 {\sqrt 5} + \cos x \frac 1 {\sqrt 5}.
\end{align}
|
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|
Value of $\lim\limits_{z \to 0}\bigl(\frac{\sin z}{z}\bigr)^{1/z^2}$ Find the value of $$\lim\limits_{z \to 0}\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$
So I took a log: $$\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ If I could expand it something like $\log(1+x)$ .. any hints ?
|
$\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-......$
Therefore, we have
$\lim_{z \to 0}({\frac{\sin{z}}{z}})^{1/z^2}=\lim_{z \to 0}(\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-......}{Z})^{1/z^2}$
=$\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......)^{1/z^2}$
=$(\lim_{z \to 0}(1-\frac{z^2}{3!}+\frac{z^4}{5!}-......))^{1/z^2}$ (the limit of expression inside the bracket is taken)
This gives:
$(1-\frac{0^2}{3!}+\frac{0^4}{5!}-......))^{1/z^2}=1^{1/z^2}=\underline{1}$
|
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|
General Term of a Sequence What would be the best way in finding a general term $a_n, n>3$ for the recursive sequence?
$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}$$
where $a_1 = 1 ; a_2 = 2 ; a_3 = 24$ ;
|
Given the recursive sequence:
$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}\tag{E}$$
$(\bf E)$ can be rearranged to:
$$\frac{a_n}{a_{n-1}}=6\frac{a_{n-1}}{a_{n-2}}-8\frac{a_{n-2}}{a_{n-3}}\tag{1}$$
Seeing a pattern in $(1)$ let a new term $t_n$ be: $$t_n=\frac{a_n}{a_{n-1}}\tag{2}$$
Now $t_n$ follows the recursive sequence:
$$t_n=6t_{n-1}-8t_{n-2}\tag{3}$$
The characteristic equation of $t_n$ is:
$$r^2=6r-8\iff r^2-6r+8=0\implies r=2,4\tag{4}$$
So $t_n$ is given by the relation:
$$t_n=a2^n+b4^n\tag{5}$$
Also we know that $t_2=2,t_3=12$.
So after substituting:
$$\begin{align}4a+16b=2\quad\mid&\quad8a+64b=12\tag{6}\\ a=-0.5\quad\mid&\quad b=0.25\tag{7}\end{align}$$
Putting back values of $a$ and $b$:
$$t_n=4^{n-1}-2^{n-1}\tag{8}$$
Now from $(2)$ and $(8)$:
$$a_n=(4^{n-1}-2^{n-1})a_{n-1}\tag{9}$$
Following this pattern:
$$\begin{align}a_n&=(4^{n-1}-2^{n-1})(4^{n-2}-2^{n-2})...(4^{2-1}-2^{2-1})\underbrace{a_1}_{a_1=1}\tag{10}\\a_n&=\left(\prod_{k=1}^{n-1}(4^{k}-2^{k})\right)\tag{11;$n\ge2$}\\\end{align}$$
|
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|
Arithmetic Error in Calculation of the Limit of a Given Function I consider a function $f(x)$ which is equal to $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6}$
While trying to evaluate the $\lim_{x \to 6} f(x)$
It is true that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6}$
It is also true that $\dfrac{a}{\frac{b}{c}} = \dfrac{a\cdot c}{b}$
I apply this property to $f(x)$ and come to the conclusion that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6} = \dfrac{3x-18}{x}-\dfrac{x-6}{2}$, which is of course incorrect.
Additionally, I believe I'm taking an improper approach to solving this limit.
|
Note: $$\frac a{\frac bc} = \frac{ac}b$$
You arrive at only $ac$.
Also, $$\dfrac{\frac ab}{c} = \frac{a}{bc}$$
So $$ \dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6} = \frac{3}{x(x-6)} - \frac 1{2(x-6)} = \frac{2\cdot 3}{2x(x-6)}-\frac{x}{2x(x-6)} = \dfrac{6-x}{2x(x-6)}$$
$$= \frac{-(x-6)}{2x(x-6)} = \frac{-1}{2x}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A hint to help me integrate $ \int {x^5+1\over x^4+x^3+x^2} \, dx $ Numerator is of a higher degree than the denominator, so after division I get the following:
$$ \int {x^5+1\over x^4+x^3+x^2} \, dx = \frac 12x^2-x + \int {1\over x^2+x+1} \, dx $$
|
\begin{align}
& \underbrace{x^2+x+1=\left(x^2+x+\frac 1 4 \right)+\frac 3 4}_{\text{completing the square}} = \left( x+\frac 1 2 \right)^2 + \frac 3 4 \\[10pt]
= {} & \frac 3 4 \left( \frac 4 3 \left( x+\frac 1 2 \right)^2 + 1 \right) = \frac 3 4 \left( \left( \frac{2x+1}{\sqrt{3}} \right)^2 +1 \right) = \frac 3 4 (u^2 + 1)
\end{align}
and $du=\dfrac{2}{\sqrt{3}}\,dx$.
And $\displaystyle\int\frac{du}{u^2+1}=\arctan u + C$.
|
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|
Pythagorean type diophantine equation.
How to find all solutions to
$$ a^2+b^2+c^2+d^2=e^2+2$$
where all variables $a$ to $e$ are positive integers and $e^2 \equiv 1 \mod 8$
I tried using parameterization similar to pythagoras equation, but no success so far. Any help will be appreciated.
Thanks!
|
For such equations, you can use the standard approach.
One approach is to use equations Pell. For the beginning will talk about a more simple way.
I. Eq.1
$$a^2+b^2+c^2+d^2=(2q+1)^2+2$$
Solutions have the form:
$$a=x + k$$
$$b=x + p$$
$$c=x + s$$
$$d=x - z+1$$
$$q=x$$
where,
$$x = -1 -z+ (z^2 - k p - k s - p s),\quad z = k+p+s$$
for any $k,p,s$.
II. Eq.2
$$a^2+b^2+c^2+d^2=(4q+1)^2+2$$
Solutions have the form:
$$a=2(y + k)$$
$$b=2(y + p)+1$$
$$c=2(y + s)+1$$
$$d=2(y - z)-1$$
$$q=y$$
where,
$$y = -k + 2z + 2(z^2 - k p - k s - p s),\quad z = k+p+s$$
for any $k,p,s$.
III. Eq.3
$$a^2+b^2+c^2+d^2=(8q+1)^2+2$$
Solutions have the form:
$$a=16k^2+4p^2+4s^2+8pk+8ks+4ps+2s+2p-1$$
$$b=16k^2+4p^2+4s^2+8pk+8ks+4ps+4s+4p+8k$$
$$c=16k^2+4p^2+4s^2+8pk+8ks+4ps+4s+6p+4k+1$$
$$d=16k^2+4p^2+4s^2+8pk+8ks+4ps+6s+4p+4k+1$$
$$q=4k^2+p^2+s^2+2pk+2sk+ps+s+p+k$$
$k,p,s$ - integers asked us.
|
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|
Area of circle and parabola I shall determine the surface area of the intersection of a circle with Radius $R= 1 $ and the set of points above the parabola $f(x) = 2x^2$. Only the positive $x,y$-plane is of interest.
My approach is to parameterize it with polar coordinates as $x=R\cos(\phi)$ and $y=R\sin(\phi)$, thus $\vec \gamma(\phi) = \begin{pmatrix}
R \cos(\phi) \\
R \sin(\phi) \\
0 \\
\end{pmatrix}$
Now I struggle with setting bounds of the integral, something like $0 \le R \le 1 $ and $? \le \phi \le ?.$
I have determined the point of intersection as $x=0.62481$, thus $\phi = 51.332°$ or $\phi = 0.896$ in rad. How do I continue?
|
I will assume the circle is centered at the origin since you did not state.
We will have the equation of the circle: $x^2+y^2=1$
For the parabola, we have $y=2x^2$
To get the points of intersection, we just have to set y=$2x^2$ in the equation of the parabola.
$x^2+4x^4=1$
Solving the equation, we get $x=0.68,-0.68$ as you said.
You should note that $y=2x^2$ is above the x-axis and thus we can take only the upper semicircle, and thus $y=\sqrt{(1-x^2)}$
Now, to continue, all we have to do is compute the integral $\int_{0}^{0.68} (\sqrt{(1-x^2)}-2x^2\,dx$
We set $x=\sin(\theta)$ so $dx=\cos(\theta)\,d\theta$ When $x=0,\theta=0$ and $x=0.68,\theta=0.74$ so the limits of integration are set.
We now calculate the indefinite integral:
\begin{align}
& \int \sqrt{1-x^2}-2x^2\,dx=\int (\sqrt{1-\sin^2 \theta} -2\sin^2(\theta))\cos(\theta) \, d\theta \\[8pt]
= {} & \int \cos^2(\theta) \,d\theta-2\int \sin^2 \theta \cos(\theta)\,d\theta=\frac{1}{2} \int 1+\cos(2\theta)\,d\theta \\[6pt]
& {} -2\cdot\frac{1}{4}\int \cos(\theta)-\cos(3\theta)\,d\theta=\frac{1}{2}\theta + \frac{1}{4} \sin(2\theta)-\frac{1}{2}\sin(\theta)+\frac{1}{6}\sin(3\theta)
\end{align}
Now, putting the values at the limit of integration, we get:
Area=0.41
|
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|
How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$? How to compute $\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}}$?
My partial solution:
$$
\int_{0}^{+\infty} \frac{dx}{e^{x+1} + e^{3-x}} = \int_{0}^{+\infty} \frac{dx}{e^{3-x}(1 + e^{2x-2})} \\
= \int_{0}^{+\infty} \frac{e^{x-3}dx}{1 + e^{2x-2}}.
$$
Thank you very much.
|
It is natural to let $y=e^x$. Then $dy=e^x\,dx$, so $dx=\frac{1}{y}\,dy$. Thus we want
$$\int_1^\infty \frac{1}{y}\cdot \frac{1}{e\cdot y+e^3 \cdot \frac{1}{y}}\,dy,$$
which simplifies to
$$\int_1^\infty \frac{1}{e}\cdot\frac{1}{y^2+e^2}\,dy.$$
Let $y=et$, and we are nearly finished.
|
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|
How does polynomial long division work?
I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$
|
Some trick on how to do this quickly: $3x^3 - 2x^2 + 4x - 3 = 3x(x^2+3x+3) - 11x^2 - 5x - 3 = 3x(x^2+3x+3) - 11(x^2+3x+3) + 28x + 30 = (3x-11)(x^2+3x+3) + 28x+30$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does the limit behavior of this function take over at 35? I've been working with this function on an semi-related question:
$$f(N)=\left\lfloor \frac{10N}{\lceil \frac{3}{4} N \rceil} \right\rfloor$$
It's clear that $10\leq f(N) \leq 13$, and that $\displaystyle\lim_{n\rightarrow \infty} f(N)=13$.
It's also true that $f(N)=13\,, \forall N\geq 36$.
Why is $35$ the last integer at which $f$ deviates from $13$?
Should be something simple, but I got stuck at
$$\frac{10}{13}N < \left\lceil \frac{3}{4}N \right\rceil$$.
|
Let
$N = 4n+k$
where $0 \le k \le 3$.
I will show that
$f(N)
= 13+\left\lfloor \frac{1-\frac{3k}{n}}{3+\frac{k}{n}} \right\rfloor
$.
Therefore,
if $n \ge 9$,
$f(N) = 13$.
Since
$\lceil \frac{a}{b} \rceil
= \lfloor \frac{a+b-1}{b} \rfloor
$
if $a$ and $b$
are integers with
$a \ge 0$
and $b \ge 1$,
$\begin{array}\\
f(N)
&=\left\lfloor \frac{10N}{\lceil \frac{3}{4} N \rceil} \right\rfloor\\
&=\left\lfloor \frac{10(4n+k)}{\lceil \frac{3(4n+k)}{4} \rceil} \right\rfloor\\
&=\left\lfloor \frac{40n+10k}{\lceil \frac{12n+3k}{4} \rceil} \right\rfloor\\
&=\left\lfloor \frac{40n+10k}{\lceil 3n+\frac{3k}{4} \rceil} \right\rfloor\\
&=\left\lfloor \frac{40n+10k}{3n+\lceil \frac{3k}{4} \rceil} \right\rfloor\\
&=\left\lfloor \frac{40n+10k}{3n+\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
&=\left\lfloor \frac{40+\frac{10k}{n}}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
&=\left\lfloor 13+\frac{40+\frac{10k}{n}-13(3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
&=13+\left\lfloor \frac{1+\frac{10k}{n}-13(\frac1{n}\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
&=13+\left\lfloor \frac{1+\frac1{n}(10k-13(\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
\end{array}
$
For $k=0, 1, 2, 3$,
$\lfloor \frac{3k+3}{4} \rfloor)
=0, 1, 2, 3
=k
$
so
$10k-13(\lfloor \frac{3k+3}{4} \rfloor)
=10k-13k
=-3k
$.
Therefore
$\begin{array}\\
f(N)
&=13+\left\lfloor \frac{1+\frac1{n}(10k-13(\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\
&=13+\left\lfloor \frac{1+\frac1{n}(-3k)}{3+\frac{k}{n}} \right\rfloor\\
&=13+\left\lfloor \frac{1-\frac{3k}{n}}{3+\frac{k}{n}} \right\rfloor\\
\end{array}
$
|
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|
Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me to use the geometric series:
$$\sum_{n=0}^{\infty} (\frac{1-(\frac{1}{3})^{n+1}}{\frac{2}{3}}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
which can even be simplified further to:
$$\sum_{n=0}^{\infty} ((1-(\frac{1}{3})^n\cdot\frac{1}{3})(\frac{3}{2})) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{3}{2}-(\frac{1}{3})^n\cdot\frac{1}{2}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{1}{2}(3-(\frac{1}{3})^n) \cdot (\sum_{k=0}^{n}2^{n-k})$$
This is as far as I know what to do. The solution by the way is:
$$\sum_{k=0}^{\infty} (\frac{1}{3})^k \sum_{k=0}^{\infty} (\frac{1}{2})^k$$ which could be simplified again with the geometric series.
However I have now idea what to do with $\sum_{k=0}^n2^{n-k}$ though we could write it as $\sum_{k=0}^n 2^n \sum_{k=0}^n 2^{-k}$ this would not make much sense as the former sum would converge against infinity.
|
$$\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{1}{2^{n-k}}=\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{2^k}{2^{n}}=\sum^{\infty}_{n=0}\frac{1}{2^{n}}\sum^{n}_{k=0}\Big(\frac{2}{3}\Big)^k$$
Using the geometric series result for a convergent geometric series yields
$$\sum^{\infty}_{n=0}\frac{1}{2^{n}}\sum^{n}_{k=0}\Big(\frac{2}{3}\Big)^k=\sum^{\infty}_{n=0}\frac{1}{2^{n}}\cdot\frac{1-\Big(\frac{2}{3}\Big)^{n+1}}{1-\frac{2}{3}}=3\cdot\sum^{\infty}_{n=0}\frac{1-\Big(\frac{2}{3}\Big)^{n+1}}{2^{n}}=3\cdot\sum^{\infty}_{n=0}\Big(\frac{1}{2^n}-\frac{2}{3}\cdot\frac{1}{3^n}\Big)=3\cdot\sum^{\infty}_{n=0}\frac{1}{2^n}-2\cdot\sum^{\infty}_{n=0}\frac{1}{3^n}=3\cdot\frac{1}{1-1/2}-2\cdot\frac{1}{1-1/3}=6-3=3$$
|
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|
How to show $I_p(a,b) = \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$
Show that $$I_p(a,b) = \frac{1}{B(a,b)}\int_0^p u^{a-1}(1-u)^{b-1}~du\\= \sum_{j=a}^{a+b-1}{a+b-1 \choose j} p^j(1-p)^{a+b-1-j}$$ when $a,b$ are positive integers.
I have no idea how to proceed. Please help.
|
Hint: Use $\frac{1}{B(a,b)}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}=\frac{(a+b-1)!}{(a-1)!(b-1)!}$ and integration by parts to evaluate the integral:
$$I=\int_{0}^{1}u^{a-1}(1-u)^{b-1} = \left.\frac{1}{a}u^a(1-u)^{b-1}\right|_{0}^{p}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I= \frac{1}{a}p^a(1-p)^{b-1}+\frac{b-1}{a}\int_{0}^{p}u^{a}(1-u)^{b-2}\,du, $$
$$ I = \frac{1}{a}p^a(1-p)^{b-1} + \frac{b-1}{a(a+1)}p^{a+1}(1-p)^{b-2} +\frac{(b-1)(b-2)}{a(a+1)}\int_{0}^{p}u^{a+2}(1-u)^{b-3}\,du = \ldots $$
|
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|
How to solve $\int_0^\pi \frac{x\sin x}{1+ \sin^2 x}dx$? I have solved
$$\int_0^\pi \frac{x\sin x}{1+ \cos^2 x}dx=\pi^2/4$$. But I cannot solve it with the same way that I used for the upper problem. Can you help me to solve this problem?
$$\int_0^\pi \frac{x\sin x}{1+ \sin^2 x}dx$$
|
We are going to evaluate the integral
$\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x \tag*{} $
by substitution, odd and even functions.
Let $\displaystyle y=x-\frac{\pi}{2},$ then
$\displaystyle \quad \int_{0}^{\pi} \frac{x \sin x}{1+\sin ^{2} x} d x \\$
$\displaystyle=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{y \cos y}{1+\cos ^{2} y} d y+\frac{\pi}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos y}{1+\cos ^{2} y} d y$
$\displaystyle =0+\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d(\sin y)}{2-\sin ^{2} y}$
$\displaystyle =-\pi \int_{0}^{1} \frac{d z}{z^{2}-2} \quad \text {, where } z=\sin y$
$\displaystyle =-\frac{\pi}{2\sqrt{2}} \ln \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|_{0}^{1}$
$\displaystyle = \frac{\pi}{\sqrt{2}} \ln (\sqrt{2}+1)$
|
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|
Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
|
No L'Hospital, too messy. You are almost there, rewrite the expression as
$$
\frac{(x-2)x}{(x+2)(x-2)(1 +\sqrt{\frac{(x-2)^2(x+1)}{(x-2)^2}}}
$$
what do you get?
|
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|
$(3/4) \cdot (8/9) \cdot (15/16) \cdot \ldots$ How to prove that limit is $1/2$ How can one prove $$\frac34 \cdot \frac89 \cdot \cdots \cdot \frac{n^2-1}{n^2}$$ has a limit $1/2$?
I tried some manipulations with the terms with no success and have just no idea,
Thanks
|
Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations.
$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$
After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$
$$\frac{n+1}{2}\cdot\frac{1}{n}$$
Limit of this function tending to infinity $= 1/2$.
This question has been ansewred million times in this site.
Goodluck
|
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|
How can we say two algebraic expressions are "equal" if one is undefined at certain points and the other isn't? I'm trying to understand why it is that we can say $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{(x-1)} = x+1$ but then have it also be the case that the two functions $f(x) = \frac{x^2-1}{x-1}$ and $g(x)=x+1$ are not equal, since $f$ is undefined when $x=1$.
That is, if it is true that $\frac{x^2-1}{x-1} = x+1$ then why can we not just simplify and say that $f(x)=x+1$ so that $f(1) = (1) + 1 = 2$ ?
|
Let $f(x) = \frac{x^2-1}{x-1}$. The domain of $f(x)$ is $x \in \mathbb{R}-\{1\}$.
Actually, $$\frac{x^2 -1}{x-1} = x+1,$$ for all $x \neq 1$.
Hence, $f(x) \neq x+1$, because $1$ is in the domain of $x+1$.
|
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|
Invert a $2\times 2$ Matrix containing trig functions Invert the $2\times 2$ matrix:
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
My thought was to append the $2\times 2$ identity matrix to the right of the trig matrix and use row operations to get the answer.
I have to show all steps, so I cannot just flip it and call it a day.
|
Several people have posted ways to do this, but none did it the way the original posted proposed, i.e. row operations. Nor any with trigonometric identities. So here are those two methods:
\begin{align}
& \begin{bmatrix} \cos\theta & -\sin\theta & 1 & 0 \\ \sin\theta & \cos\theta & 0 & 1
\end{bmatrix} \\
\text{Now put }\\
\text{$(\sin\theta)\cdot$1st row${}-(\cos\theta)\cdot$2nd row } \\
\text{in the 2nd row: } & \begin{bmatrix} \cos\theta & -\sin\theta & 1 & 0 \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\
\text{Then put }\\
(-\sin\theta)\cdot\text{2nd row}+\text{1st row } \\
\text{in the 1st row: } &
\begin{bmatrix} \cos\theta & 0 & 1-\sin^2\theta & \sin\theta\cos\theta \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\[8pt]
= {} & \begin{bmatrix} \cos\theta & 0 & \cos^2\theta & \sin\theta\cos\theta \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\[6pt]
\text{Then multiply the 1st row by $1/\cos\theta$: } \\
\text{and the 2nd by $-1$: } &
\begin{bmatrix} 1 & 0 & \cos\theta & \sin\theta \\ 0 & 1 & -\sin\theta & \cos\theta \end{bmatrix}
\end{align}
That does it via row operations.
Now let's try trigonometric identities. First prove that
$$
\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}
\begin{bmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{bmatrix}
=\begin{bmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix}.
$$
The look at what happens when $\beta=-\alpha$ and you'll see the answer.
|
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|
Rewrite a circle's equation to easily see centre and radius $$x^{2}+y^{2}-5x-15y+30=0$$
I'm supposed to rewrite this equation so that you can easily see the centre and radius of the circle. I don't even know where to start. According to Mathematica the centre is $(5/2, 15/2)$ and the radius is $\sqrt{65/2}$.
|
Answer:
$$x^2 +y^2 -5x-15y+30 = 0$$
$$x^2 -5x +(\frac{5}{2})^2 +y^2 -15x +(\frac{15}{2})^2 -(\frac{5}{2})^2-(\frac{15}{2})^2+30 = 0$$
$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 = \frac{225+25}{4}-30$$
$$(x-\frac{5}{2})^2+(y-\frac{15}{2})^2 =(\sqrt{\frac{65}{2}})^2$$
Center$$ (\frac{5}{2},\frac{15}{2})$$and the radius $$= \sqrt{\frac{65}{2}}$$
|
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|
convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$ Test convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$
Attempt: Since, $\lim_{n \rightarrow \infty} \dfrac {1}{n} \in (0,1) \implies \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx \geq \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx$
If $I = \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx = \dfrac {1}{2} \ln (1+ \dfrac {1}{n^2})$
Now, for $\ln (1+\dfrac {1}{n^2}) <(1+\dfrac {1}{n^2})$
Since, $\sum_{n=1}^\infty (1+\dfrac {1}{n^2})$ is convergent $\implies \sum_{n=1}^\infty \ln (1+\dfrac {1}{n^2})$ is convergent as well.
But, I don't think this result is of any use as $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx \geq \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx$
Please guide me on moving from here.
Thank you very much for your help.
|
First note that the denominator is there just to confuse you. For sufficiently small $x$,
$$1\ge\frac1{1+x^2}\ge\frac12,$$
so we can disregard that altogether. Now
$$\sum_{n=1}^\infty\int_0^{1/n}\sqrt x\,dx=\frac23\sum_{n=1}^\infty \frac 1{n^{3/2}},$$
which converges.
|
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|
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\
\arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\
\text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$
I can't find the right words to explain why this is true?
Edit #1 (25 Jan, 20:10 UTC):
The following is a truth clause:
$$
\begin{array}{ll}
\frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\
\\
\frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0
\end{array}
$$
By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m \text{ (m is a constant)}\\
\\
\arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\
\\
\text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Using Calculus, why is that true for every $x \in [-1,1]$?
Edit #2:
A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.
Thus the equations from "Edit #1" should be as follows:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\
\\
\arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\
\\
A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
|
More simple....
From $\cos \alpha=\sin \left(\dfrac{\pi}{2}-\alpha\right)$ we have:
$$
\cos y=x \Rightarrow \sin\left(\dfrac{\pi}{2}-y\right)=x \Rightarrow
$$
$$
\Rightarrow
\begin{cases}
\arccos x=y \\
\arcsin x= \dfrac{\pi}{2}-y
\end{cases}
\Rightarrow
$$
$$
\Rightarrow
\arccos x+\arcsin x=\dfrac{\pi}{2}
$$
|
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|
Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+\cdots+1 )$
$\implies n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n^2$
$\implies \dfrac {1}{ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) } \geq \dfrac {1}{n^2}$
I guess this result is of not much use. Can somebody please tell me a direction to move ahead in this problem?
Thank you very much for your help .
|
The other way to look at it is, denoting $\displaystyle H_n = \sum\limits_{k=1}^{n} \frac{1}{k}$
The tail of the series: $$\displaystyle \frac{\frac{1}{n+1}}{H_{n+1}}+\frac{\frac{1}{n+2}}{H_{n+2}}+\cdots +\frac{\frac{1}{n+r}}{H_{n+r}} \ge \frac{\sum\limits_{k=1}^{r}\frac{1}{n+k}}{H_{n+r}} = \frac{H_{n+r}-H_{n}}{H_{n+r}} = 1-\frac{H_n}{H_{n+r}}$$
Since, $\displaystyle \lim\limits_{r \to \infty} 1-\frac{H_n}{H_{n+r}} = 1$
The series diverges by Cauchy Criteria.
|
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|
If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
|
If $ab+bc+ca < \dfrac{-1}{2} \Rightarrow 2(ab+bc+ca) + 1 < 0 \Rightarrow 2(ab+bc+ca) + a^2+b^2+c^2 < 0 \Rightarrow (a+b+c)^2 < 0$. Contradiction.
|
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|
Find the value of $ ( ab + bc + ca )^2 $ If $a,b,c$ are real numbers which satisfy
$a^2+b^2+ab = 9$
$b^2+c^2+bc = 16$
$c^2+a^2+ca = 25$
find the value of $ ( ab + bc + ca )^2 $
|
let
since
$$\begin{cases}a^2-2a\cdot b\cos{\dfrac{2\pi}{3}}+b^2=9\\
b^2-2bc\cos{\dfrac{2\pi}{3}}+c^2=16\\
c^2-2ca\cos{\dfrac{2\pi}{3}}+a^2=25
\end{cases}
$$
then we let
$$PA=a,PB=b,PC=c,\angle APB=\angle BPC=\angle APC=\dfrac{2\pi}{3}$$
By cosine
$$|AB|^2=25,|BC|^2=16,|AC|^2=9$$
so $\angle C=\dfrac{\pi}{2}$,
then we have
$$S_{ABC}=S_{APC}+S_{BPC}+S_{APB}=\dfrac{1}{2}\sin{\dfrac{2\pi}{3}}(ab+bc+ac)$$
$$\Longrightarrow \dfrac{\sqrt{3}}{4}(ab+bc+ac)=\dfrac{1}{2}|AC||BC|=6$$
so
$$ab+bc+ac=(8\sqrt{3})^2=192$$
|
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|
Antiderivative of $\frac{\sqrt{4-x}}{x\sqrt{x}}$ I need help to find the antiderivative of the function $\displaystyle x \, \mapsto \, \frac{\sqrt{4-x}}{x\sqrt{x}}$ on $]0,4[$. I have tried the change of variables $u = \sqrt{4-x}$ but it didn't help. Which change of variables (or trick) am I not thinking of ?
|
Let us try this:
$$\frac{4-x}x=u^2\implies x=\frac4{u^2+1}\implies dx=-\frac{8u}{(u^2+1)^2}\,du$$
and then
$$\int\frac1x\sqrt\frac{4-x}x\;dx=\int\frac{u^2+1}4\cdot u\cdot \left(-\frac{8u}{(u^2+1)^2}du\right)=-2\int\frac{u^2}{(u^2+1)^2}du=$$
$$=-\arctan u+\frac u{u^2+1}+C\longrightarrow-\arctan\sqrt\frac{4-x}x+\frac{\sqrt\frac{4-x}x}{1+\frac{4-x}x}=$$
$$=\frac{\sqrt{x(4-x)}}4-\arctan\sqrt\frac{4-x}x+C$$
|
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|
If $a^2+ab+b^2=c^2+cd+d^2$ then $a+b+c+d$ is not prime. Let $a,b,c,d$ be positive integers such that $a^2+ab+b^2=c^2+cd+d^2$. Show that $a+b+c+d$ is not prime.
My proof looks like this:
$(a+b)^2 - ab=(c+d)^2-cd$
$(a+b)^2 - (c+d)^2=ab-cd$
$(a+b+c+d)(a+b-c-d)=ab-cd$
$a+b+c+d=\frac{ab-cd}{a+b-c-d}$
I'd like to have $a+b+c+d$ as product (of integers) not quotient
|
It is necessary to use the formula. Diophantine equation $a^2+b^2=c^2+d^2$
For example this.
You can write a similar equation and solutions: $$a^2+ac+c^2=x^2+xy+y^2$$
Solutions have the form: $$a=q^2+k^2-p^2+kq$$
$$c=q^2+k^2+2p^2+kq-3pk-3pq$$ $$x=q^2-2k^2-p^2+3pk-2qk$$ $$y=k^2-2q^2-p^2+3pq-2qk$$
This means that:
$$a+c+x+y=q^2+k^2-p^2-2qk=(q-k-p)(q-k+p)$$
Easy number can always choose.
$$(q-k-p)(q-k+p)=1*19$$
Then for example: it turns out one of the numbers is negative. If we change the signs. $x=-x$ ; $y=-y$ . You get a multiple of 3 and the square. So simple number can be when $q=p-k+1$
This means that at least one number must be again negative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1122457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Solve the PDE by the method of characteristics. I am trying to figure out where my solution went wrong. I am off by a factor of two.
$$ u_x + u_y + u = e^{x+2y}$$
I first found that the characteristic curves are determined by $$\frac{dy}{dx} = 1 \implies y-x = C.$$
I then solved the ODE $$\frac{du}{dx} + u = e^{x+2y}$$
I found $u = \frac12 e^{x+2y} + e^{-x}K(C)$ giving $$u = \frac{e^{x+2y}}{2} + e^{-x}F(y-x)$$ as the general solution where $F$ is an arbitrary function. Where does my work go wrong?
The end solution should be half of what it currently is.
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$$\begin{align}
u_x + u_y + u &= e^{x + 2y} \\
\implies u_x + u_y &= e^{x + 2y} - u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)\\
\end{align} $$
Setting $u = u(x(s),y(s))$ we find
$$\begin{align}
\frac{d}{ds} u &= \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial y} \cdot \frac{dy}{ds} \\
&= \frac{\partial u}{\partial x} \cdot 1 + \frac{\partial u}{\partial y} \cdot 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\
&= e^{x + 2y} - u
\end{align}$$
Where $(1)$ comes from our original PDE at $(*)$. Equating, we find
$$\begin{align}
\frac{dy}{ds} &= 1 \\
\frac{dx}{ds} &= 1 \implies \frac{dx}{dy} = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\
\frac{du}{ds} &= e^{x + 2y} - u \implies \frac{du}{dy} + u = e^{x + 2y} \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\
\end{align}$$
Solving $(2)$ and $(3)$
$$x(y) = x_0 + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$
$$\begin{align}
\frac{du}{dy} + u &= e^{x + 2y} \\
&= e^{x_0 + 3y} \\
\implies (e^{y}u)' &= e^{x_0 + 4y} \\
\implies e^{y}u &= \frac{e^{x_0 + 4y}}{4} + f(x_0) \\
\implies u &= e^{-y} \bigg(\frac{e^{x_0 + 4y}}{4} + f(x_0) \bigg) \\
&= \frac{e^{x_0 + 3y}}{4} + e^{-y}f(x_0) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) \\
\end{align}$$
and using $(4) \implies x_0 = x - y$ we find
$$u = \frac{e^{x + 2y}}{4} + e^{-y}f(x - y)$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1124412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$.
I am not sure how to approach this problem. Should I divide this problem into multiple cases based on whether $a$,$b$,$c$ and odd/even or is there a more general solution?
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$0\le\dfrac12\left((a-b)^2+(b-c)^2+(c-a)^2\right)=\dfrac12\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)=a^2+b^2+c^2-ab-bc-ca\implies a^2+b^2+c^2\ge ab+bc+ca$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1125709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
Solving integral $\int \arcsin x \cos x dx$ Can anyone give me a hint how to solve $$\int \arcsin(x)\cos(x)dx ~?$$
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$\int\sin^{-1}x\cos x~dx$
$=\int\sin^{-1}x~d(\sin x)$
$=\sin^{-1}x\sin x-\int\sin x~d(\sin^{-1}x)$
$=\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
Let $x=\sin u$ ,
Then $dx=\cos u~du$
$\therefore\sin^{-1}x\sin x-\int\dfrac{\sin x}{\sqrt{1-x^2}}dx$
$=\sin^{-1}x\sin x-\int\sin\sin u~du$
$=\sin^{-1}x\sin x-\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}u}{(2n+1)!}du$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}u}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^n}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\cos^{2k}u}{(2n+1)!}d(\cos u)$
$=\sin^{-1}x\sin x+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\cos^{2k}u}{(2n+1)!k!(n-k)!}d(\cos u)$
$=\sin^{-1}x\sin x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\cos^{2k+1}u}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sin^{-1}x\sin x+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!(1-x^2)^{k+\frac{1}{2}}}{(2n+1)!k!(n-k)!(2k+1)}+C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1128772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find $\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$ Find $$\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}$$
with $n \in \mathbb{N}$.
My tried:
I think that, I need to find the value of
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
because:
$$\begin{align} I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}} &= \frac{x}{(1+x^n)\sqrt[n]{1+x^n}} \Bigg|_0^1-\int_0^1 x \ d(1+x^n)^{-1-\frac{1}{n}} \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1 x\left(-1-\frac{1}{n}\right)(1+x^n)^{-2-\frac{1}{n}}(nx^{n-1}) \ dx \\ &=\frac{1}{2\sqrt[n]{2}}-\int_0^1(-n-1)\frac{x^n}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx\\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)\int_0^1\frac{x^n+1-1}{(1+x^n)^2\sqrt[n]{1+x^n}} \ dx \\ &=\frac{1}{2\sqrt[n]{2}}+(n+1)I_1-(n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}.\end{align} $$
So that,
$$ (n+1)\int_0^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{1}{2\sqrt[n]{2}}+nI_1$$
But how to find the following integral ?
$$I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}$$
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Observe that
$$
\left(\frac{1}{(1+x^n)^{1/n}}\right)'=-\frac{x^{n-1}}{(1+x^n)^{1+1/n}} \tag1
$$
giving
$$
\begin{align}
\left(\frac{x}{(1+x^n)^{1/n}}\right)'=\left(x \times\frac{1}{(1+x^n)^{1/n}}\right)'=\frac{1}{(1+x^n)^{1/n}}-\frac{x \times x^{n-1}}{(1+x^n)^{1+1/n}}=\frac{1}{(1+x^n)^{1+1/n}}
\end{align}
$$
then
$$
\begin{align}
I_1=\int_{0}^1 \frac{dx}{(1+x^n)\sqrt[n]{1+x^n}}=\int_{0}^1 \frac{dx}{(1+x^n)^{1+1/n}}=\left[\frac{x}{(1+x^n)^{1/n}}\right]_{0}^{1}=\frac{1}{2^{1/n}}
\end{align}
$$ and you deduce the value of your initial integral:
$$
\int_{0}^1 \frac{dx}{(1+x^n)^2\sqrt[n]{1+x^n}}=\frac{2n+1}{2(n+1)}\frac{1}{2^{1/n}} .\tag2
$$
|
{
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"url": "https://math.stackexchange.com/questions/1128875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$
So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate
both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$
Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$
We can write $f'(x)$ as $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$
So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$
So $$\displaystyle f'(x) = = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$
So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.
In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it
will Cross $\bf{X-}$ axis at least one time.
My question is can we solve it any other way, i. e without using Derivative test.
Help me , Thanks
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Let
$$f(x)=1+\frac x1+\cdots+\frac{x^7}{7}$$
Assume that there are two different roots $a_1$ and $a_2, (a_1<a_2)$ of the equation $f(x)=0$ so by the mean value theorem we get for $a_3\in(a_1,a_2)$
$$0=f(a_1)-f(a_2)=(a_1-a_2)f'(a_3)\implies f'(a_3)=1+a_3+a_3^2+\cdots+a_3^6=0$$
and using the geometric sum
$$1+x+\cdots+x^6=\frac{1-x^7}{1-x},\quad x\ne1$$
we see that $a_3$ doesn't exist so it exists at most one root of $f(x)=0$ and since the polynomial is odd then it exists at least one root. Hence the unicity of the root.
|
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"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{3}}}\right)$. I can put the squareroot sign over the whole fraction, but that still doesn't really help me get an actual number for theta.
ALSO: It is a rule that I add $\pi$ to theta if $a < 0$. Why? Looking at the diagram of the triangle form by the complex vector $z$ and the real axis, the 'triangle' is in the second quadrant. Hence, the theta we find with $\theta = \tan^{-1}(\frac{b}{a})$ is the angle closes to the real axis on the left-side. However, we measure angle going counter-clockwise. Shouldn't we do the following computation to get the angle going clock-wise: $\pi - \tan^{-1}(\frac{b}{a})$?
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Note
$$\frac{1}{2}\sqrt{2 - \sqrt{3}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1 - \cos \frac{\pi}{6}}{2}} = \sin \frac{\pi}{12} $$
and similarly $$\frac{1}{2}\sqrt{2 + \sqrt{3}} = \cos \frac{\pi}{12}.$$ Therefore $$z = 2\left(-\sin \frac{\pi}{12} + i\cos \frac{\pi}{12}\right) = 2e^{i(\frac{\pi}{2} + \frac{\pi}{12})} = 2e^{i\frac{7\pi}{12}}.$$
Since $-\pi < \frac{7\pi}{12} \le \pi$, the principal argument of $z$ is $\frac{7\pi}{12}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.