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Trigonometric equation, find $\sin \theta $ Find $\sin \theta $ if $a$ and $c$ are constants
$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$
|
I think this relust is ugly.let
$$t=\sin{x}$$
$$1-\left(c-a\cdot\dfrac{t}{\pm \sqrt{1-t^2}}\right)^2=\dfrac{t^2(1-t^2)^2}{a^2-(1-t^2)^2}$$
so
$$(1-t^2)-(\pm c\sqrt{1-t^2}-at)^2=\dfrac{t^2(1-t^2)^3}{a^2-(1-t^2)^2}$$
then
$$[a^2-(1-t^2)^2][(1-t^2)-(c^2(1-t^2)-\pm 2act\sqrt{1-t^2}]=t^2(1-t^2)^3$$
|
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|
Probability of choosing a defective appliance Amongst $15$ appliances there are $10$ functional and $5$ defective. We choose $3$ appliances at random. What is the probability of
A) all of them being functional
B) exactly one being defective
The total number of possible choices is $15\cdot 14\cdot 13$, number of choices containing only functional appliances is $10\cdot 9\cdot 8$ so
$$P(A)=\frac{10\cdot 9\cdot 8}{15\cdot 14\cdot 13}$$
As for B), I tried a similar approach. We have $5$ possibilities of choosing a defective appliance and choose the other two from the functional ones so
$$P(B)=\frac{5\cdot 10\cdot 9}{15\cdot 14\cdot 13}$$
This is wrong however. I know that I am missing something but I am not sure why exactly this approach does not work. Thanks.
|
The number of ways to choose $1$ out of $5$ defective appliances is $\binom{5}{1}=5$
The number of ways to choose $2$ out of $10$ functional appliances is $\binom{10}{2}=45$
The number of ways to choose $3$ out of $15$ any type of appliances is $\binom{15}{3}=455$
Hence the probability of choosing $1$ defective and $2$ functional is $\frac{5\cdot45}{455}=\frac{45}{91}\approx0.5$
|
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|
Find the radius of convergence of the Taylor series of $f(x) = \frac{x-3}{x+2}\ln(5+x)$ at $x=0$ I need find radius of convergence for Taylor series in $x = 0$ (over $\mathbb{R}$) and find $x$'s at which series converges to $f$
$$f(x) = \frac{x-3}{x+2}\ln(5+x)$$
My solution
$\ln(5+x) = -\ln(5)\ln(1+\frac{x}{5}) = -\ln(5)\sum_{n \geq 1} \frac{(-1)^{n+1}}{5^nn}x^n, - \frac{1}{5} < x \leq \frac{1}{5}$
$\frac{x-3}{x+2} = \frac{x-3}{2} \sum \frac{ (-1)^n}{2^n}x^n, -\frac{1}{2} < x < \frac{1}{2}$
$\implies$ answer for both questions is intersection: $\frac{1}{5} < x \leq \frac{1}{5}$?
|
There are several mistakes in your computation. First of all
$$
\log(5+x)=\log5+\log\Bigl(1+\frac x5\Bigr).
$$
The Taylos series of $\log(1+x/5)$ converges for $|x/5|<1$, that is, $-5<x<5$. Similarly the Taylor series of $1/(x+2)$ converges for $|x|<2$.
|
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|
What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$ What is the value of $$\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?$$
EDIT
I bumped into this series when inserting $\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)$ and $\overrightarrow{r}=\left(\begin{array} {c}1\\1\\0\end{array}\right)$ into $$\frac{1}{|\overrightarrow{r}-\overrightarrow{r_1}|}=\sum_{l=0}^\infty\frac{r_{<}^l}{r_{>}^{l+1}}P_{l}(cos\theta)\;.$$
See (Series representation of $1/|x-x'|$ using legendre polynomials).
So I knew the result, $\sqrt{\frac{2}{3}}$, but wished to find out what other approaches there would be to evaluate the series.
It transpires that I overlooked the relatively standard evaluation of $\sum_{n=0}^{\infty}\binom{2n}{n}x^n$ as being equal to $\sqrt{\frac{1}{1-4x}}$, following a calculation similar to that given by achille-hui below, which requires some complex function theory, in particular when proving that $\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi}\left(e^{i\theta}+e^{-i\theta}\right)^{2n}d\theta$, or that of alex.jordan below, requiring no more than Taylor expansion.
|
$$
\begin{align}
\sum\binom{2n}{n}x^n
&=\sum\frac{1}{n!}\frac{(2n)!}{n!}x^n\\
&=\sum\frac{1}{n!}2^n(2n-1)(2n-3)\cdots(3)(1)x^n\\
&=\sum\frac{1}{n!}\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\cdots\left(\frac32\right)\left(\frac12\right)(4x)^n\\
&=\sum\frac{1}{n!}\left(-{\frac{2n-1}2}\right)\left(-{\frac{2n-3}2}\right)\cdots\left(-{\frac32}\right)\left(-{\frac12}\right)(-4x)^n\\
&=\sum\frac{1}{n!}f^{(n)}(0)(-4x)^n\\
\end{align}
$$
where $f(z)=(z+1)^{-1/2}$. Now interpret as a Taylor series and evaluate at $x=-{\frac18}$ (using the corresponding $z=\frac12$).
|
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|
Inequality $\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \frac{32}{7}$ If $a,b,c,d$ are positive real numbers such that $a^2+b^2+c^2+d^2 = 1$,
Prove that: $$\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \dfrac{32}{7}$$
I saw this problem is very similar to the problem I have got but with different condition on the variables. The problem in the link suggests a power series expansion of the LHS followed by establishing an inequality of the type: $$\sum_{n=0}^{\infty}(bcd)^n+(cda)^n+(dab)^n+(abc)^n$$
and establishing inequality
$(bcd)^n+(cda)^n+(dab)^n+(abc)^n\ge (K(a^2+b^2+c^2+d^2))^n$
for a positive constant $K$. Also I couldn't imitate the solution provided in the link for my problem. Is there a general method for solving these type of problems ?
|
We make use of the inequality: $\displaystyle \sum\limits_{cyc} abc \le \frac{1}{16}\left(\sum\limits_{cyc} a\right)^3$ several nice proofs are given here.
(The cyclic sum is taken over $a,b,c,d$)
We have: $\displaystyle \sum\limits_{cyc} (abc)^2 \le \frac{1}{16}\left(\sum\limits_{cyc} a^2\right)^3 = \frac{1}{16}$
and $\displaystyle \max\{abc,bcd,cda,dab\} < \left(\frac{a^2+b^2+c^2+d^2}{3}\right)^{3/2} = \frac{1}{3\sqrt{3}}$.
We need a positive constant $c > 0$, such that: $\displaystyle \frac{1}{1-x} \le \frac{8}{7} + c\left(x^2 - \frac{1}{64}\right)$
for $x \in \left(0,\frac{1}{3\sqrt{3}}\right)$ atleast.
Since, $\displaystyle \frac{8}{7} + c\left(x^2 - \frac{1}{64}\right) - \frac{1}{1-x} = \left(x - \frac{1}{8}\right)\left(c\left(x+\frac{1}{8}\right) - \frac{8}{7(1-x)}\right)$
Then, $\displaystyle x \le \frac{1}{8} \implies c \le \frac{64}{7(1-x)(1+8x)} = g(x)$ and the minima of $g(x)$ is attained at the point $x = \frac{7}{16}$ in te interval $(0,1)$ and is monotone decreasing the interval $\left(0,\frac{7}{16}\right)$. Thus, we may take $c = g(1/8) = \dfrac{256}{49}$.
Then, we see that $\displaystyle \frac{8}{7} + \dfrac{256}{49}\left(x^2 - \frac{1}{64}\right) \ge \frac{1}{1-x}$ for $x \in \left(0,\frac{3}{4}\right)$.
Thus, $\displaystyle \sum\limits_{cyc} \frac{1}{1-abc} \le \frac{32}{7} + \dfrac{256}{49}\sum\limits_{cyc}\left((abc)^2 - \frac{1}{64}\right) \le \frac{32}{7}$
|
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|
Is there any formula for $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$? I need to turn this $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$ and I completely lost my logic on this one.
When trying to get $D$ with $D=b^2-4ac$ formula, I get $D < 0$.
Can someone explain me how to get these $-5$ and $+3$. Should I do this in my mind or is there another formula for this?
Thank you.
|
Hint: Completing the square you have
$$16\Big[x^2 - \frac{1}{2}x\Big] +1 = 16\Big[\left(x - \frac{1}{4}\right)^2 - \frac{1}{16}\Big] +1 = 16\left(x - \frac{1}{4}\right)^2 - 1 +1 = (4x - 1)^2$$
Now think of $(4x -1)^2 - 4^2$ and use $a^2-b^2 = (a - b)(a+b)$.
|
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|
Minimizing Sum of Reciprocals Find the minimum value, in terms of $k$ of $\frac{1}{x_1}+…+\frac{1}{x_n}$ if $x_1^2+x_2^2+…+x_n^2=n$ and $x_1+x_2+…+x_n=k$, where $\sqrt{n} < k \leq n$.
I tried the am-hm, but how to relate with the sum of squares?
|
Given that $\sum\limits_{j=1}^nx_j^2=n$ and $\sum\limits_{j=1}^nx_j=k$
we want to minimize $\sum\limits_{j=1}^n\dfrac1{x_j}$.
That is, we need to find $x_j$ so that for all $\delta x_j$ where $\sum\limits_{j=1}^nx_j\,\delta x_j=0$ and $\sum\limits_{j=1}^n\delta x_j=0$, we also have $\sum\limits_{j=1}^n\dfrac{\delta x_j}{x_j^2}=0$.
Orthogonality implies that there exist $a$ and $b$ so that for all $j$, $ax_j+b=x_j^{-2}$. Thus,
$$
ax_j^3+bx_j^2-1=0\tag{1}
$$
This implies there are only $3$ distinct values for $x_j$, say $v_1$, $v_2$, and $v_3$.
The equations $\sum\limits_{j=1}^nx_j^2=n$ and $\sum\limits_{j=1}^nx_j=k$ imply that
$$
\begin{bmatrix}
v_1^2&v_2^2&v_3^2\\
v_1&v_2&v_3\\
1&1&1
\end{bmatrix}
\begin{bmatrix}
m_1\\m_2\\m_3
\end{bmatrix}
=
\begin{bmatrix}
n\\k\\n
\end{bmatrix}\tag{2}
$$
where $m_1$, $m_2$, and $m_3$ are the counts of each value.
Since the coefficient of $x_j$ in $(1)$ is $0$ (and the constant term is not $0$), Vieta's Formulas say that
$$
\frac1{v_1}+\frac1{v_2}+\frac1{v_3}=0\tag{3}
$$
This means that, for an interior critical point, one of the $v_j$ must be negative. Thus, the critical point must be on an edge. Let's consider the edge where $m_3=0$. In that case, we have
$$
\begin{bmatrix}
v_1^2&v_2^2\\
v_1&v_2\\
1&1
\end{bmatrix}
\begin{bmatrix}
m_1\\m_2
\end{bmatrix}
=
\begin{bmatrix}
n\\k\\n
\end{bmatrix}\tag{4}
$$
Let $m_1=m$, then the bottom equation implies $m_2=n-m$.
Let $v_1=v$, then the middle equation implies $v_2=\frac{k-mv}{n-m}$.
The top equation implies
$$
\begin{align}
&mv^2+(n-m)\left(\frac{k-mv}{n-m}\right)^2=n\\
&\implies mnv^2-2kmv+(mn+k^2-n^2)=0\tag{5}
\end{align}
$$
We can solve $(5)$ for $v$:
$$
v=\frac{km\pm\sqrt{m(n-m)(n^2-k^2)}}{mn}\tag{6}
$$
Using $(6)$, we get
$$
\begin{align}
&\frac mv+\frac{(n-m)^2}{k-mv}\\
&=\frac{nm^2}{km\pm\sqrt{m(n-m)(n^2-k^2)}}+\frac{n(n-m)^2}{k(n-m)\mp\sqrt{m(n-m)(n^2-k^2)}}\tag{7}
\end{align}
$$
Note that $(7)$ gives the same result if we use the upper choice of $\pm$ and $\mp$ as when we use the lower choice and substitute $m\mapsto n-m$. Thus, we can choose the $+$ in each $\pm$ and the $-$ in each $\mp$.
Taking the derivative of $(7)$ yields
$$
\frac{n^4(n^2-k^2)\sqrt{m(n-m)(n^2-k^2)}}{2\left(km+\sqrt{m(n-m)(n^2-k^2)}\right)^2\left(k(n-m)-\sqrt{m(n-m)(n^2-k^2)}\right)^2}\tag{8}
$$
Since $(8)$ is always positive, $(7)$ must be increasing. Thus, there is no internal critical point, so the critical point must be an edge. Because of a division by $0$, the "solution" for $m=0$, which is $x_j=\frac kn$, fails to satisfy $(4)$. Thus, we choose the closest we can, which is $m=1$. This gives the minimum to be
$$
\frac{n(n-1)}{k-\sqrt{\frac{n^2-k^2}{n-1}}}+\frac{n}{k+\sqrt{(n-1)(n^2-k^2)}}\tag{9}
$$
attained with $x_1=\dfrac kn+\sqrt{(n-1)\left(1-\frac{k^2}{n^2}\right)}$ and $x_j=\dfrac kn-\sqrt{\frac1{n-1}\left(1-\frac{k^2}{n^2}\right)}$ for $2\le j\le n$.
|
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|
Integrate using residue theorem This was a question on my complex analysis take home final. Since the semester is over and grades have been posted I believe I can post it now.
Let $a > 0$ and $b > 0$. Verify that
$$\int_{-\infty}^{\infty} \frac{dx}{e^{a x}+e^{-b x}} = \frac{\pi}{(a+b) \sin{\left (\frac{a \pi}{a+b} \right )}} $$
(Using Residue Theorem)
I believe we have to choose a rectangular contour which contains exactly one of the
singularities.
|
Rewrite the integrand as
$$\frac{e^{\left (\frac{b-a}{2} x \right )}}{2 \cosh{\left (\frac{a+b}{2} x \right )}} $$
Thus consider the contour integral
$$\oint_C dz \frac{e^{\left (\frac{b-a}{2} z \right )}}{2 \cosh{\left (\frac{a+b}{2} z \right )}} $$
where $C$ is the rectangle with vertices $\pm R \pm i 2 \pi/(a+b)$. The contour integral is then
$$\int_{-R}^R dx \frac{e^{\left (\frac{b-a}{2} x \right )}}{2 \cosh{\left (\frac{a+b}{2} x \right )}} + i \int_0^{2 \pi/(a+b)}dy \frac{e^{\left (\frac{b-a}{2} (R+i y) \right )}}{2 \cosh{\left (\frac{a+b}{2} (R+i y) \right )}} \\ + \int_{R}^{-R} dx \frac{e^{\left (\frac{b-a}{2} (x+i 2 \pi/(a+b)) \right )}}{2 \cosh{\left (\frac{a+b}{2} (x+i 2 \pi/(a+b) \right )}} + i \int_{2 \pi/(a+b)}^0 dy \frac{e^{\left (\frac{b-a}{2} (-R+i y) \right )}}{2 \cosh{\left (\frac{a+b}{2} (-R+i y) \right )}}$$
It should be clear that the second and fourth integrals vanish as $R \to \infty$, as both $a$ and $b$ are positive.
The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i \pi/(a+b)$. Thus, we have
$$\left [1+e^{i \pi (b-a)/(a+b)} \right ]\int_{-\infty}^{\infty} dx \frac{e^{\left (\frac{b-a}{2} x \right )}}{2 \cosh{\left (\frac{a+b}{2} x \right )}} = i 2 \pi \frac{e^{i \pi/2 \left (\frac{b-a}{a+b} \right )}}{(a+b) i}$$
The result follows after a little algebra.
|
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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 ?
|
You were in the good track. Starting with $$A=\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$ $$\log(A)=\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ Now, as Ian did, use the Taylor expansion $$\sin(z)=z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+O\left(z^8\right)$$ $$\frac{\sin z}z=1-\frac{z^2}{6}+\frac{z^4}{120}-\frac{z^6}{5040}+O\left(z^7\right)$$ Now, consider $$\log(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O\left(y^4\right)$$ and use $$y=\frac{z^2}{6}-\frac{z^4}{120}+\frac{z^6}{5040}$$ and arrive to $$\log\left(\frac{\sin z}z\right)=-\frac{z^2}{6}-\frac{z^4}{180}-\frac{z^6}{2835}+O\left(z^8\right)$$ $$\log(A)=-\frac{1}{6}-\frac{z^2}{180}-\frac{z^4}{2835}+O\left(z^6\right)$$ that is to say $$A=e^{-\frac{1}{6}-\frac{z^2}{180}-\frac{z^4}{2835}+\cdots}=e^{-\frac{1}{6}}e^{-\frac{z^2}{30}-\frac{2z^4}{945}}$$ Use that $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ and replace $$y=-\frac{z^2}{30}-\frac{2z^4}{945}$$ and you should finally arrive to $$A=e^{-\frac{1}{6}}\Big(1-\frac{z^2}{180}-\frac{17 z^4}{50400}+\cdots\Big)$$ which also shows how the limit is approached.
If you plot the original function and the last approximation on the same graph, you will probably be amazed to see how close are the curves for $-\frac {\pi}2 \leq x \leq\frac {\pi}2$.
|
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|
How many lines bisect both perimeter and area of a 3-4-5 triangle?
How many lines exist that divide both the perimeter and the area of a triangle with sides $3$, $4$ and $5$ into half?
|
If the line passes through the sides of length $3$ and $4$,
and its intersection with side $3$ is $x$ units from the acute angle on that side,
then the line cuts off a right triangle of base $3-x$ and height $3+x$.
The area of this triangle is $\frac 12 (9-x^2) \leq \frac 92.$
Setting this equal to $3$, we would have $x = \pm\sqrt3,$
but the construction of this line requires $0\leq x\leq 1,$
so there is no such line that cuts the triangle's area in half.
If the line passes through the sides of length $4$ and $5$,
and its intersection with side $4$ is $x$ units from the acute angle on that side,
then it cuts off a triangle with base $x$ and height $\frac 35 (6-x)$.
The area of this triangle is $\frac 12 \cdot x \cdot \frac 35 (6-x) = \frac{3}{10} x(6-x)$,
which takes a maximum value $\frac{27}{10}<3$ at $x=3,$
so no such line can cut the triangle's area in half.
The remaining case is a line through sides $3$ and $5$.
Let the line intersect side $3$ at a point $x$ units from the right angle.
Then it cuts off a triangle of base $3-x$ and height $\frac 45 (3+x),$
which has area $\frac 25 (9-x^2).$
Setting this equal to $3$, we find that $x = \pm\sqrt{\frac32},$
but $0\leq x\leq 2$ by the construction of the line,
so we have one solution, $x = \sqrt{\frac32}.$
|
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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.
|
it is equivalent to $\frac{\frac{6-x}{2x}}{x-6}=-\frac{x-6}{2x(x-6}$
since $\frac{3}{x}-\frac{1}{2}=\frac{6}{2x}-\frac{x}{2x}$
|
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|
Three digit number $ABC$ with $ABC = A + B^2 + C^3$ Is there a trick for solving this problem about number of digits?
$ABC$ is a three-digit natural number, such that $ABC = A + B^2 + C^3$.
According to above equation what is $ABC$ ?
|
The following is not really a trick, but at least it's a computation which can be done by hand.
First note that $10B\ge B^2$ (since $0\le B\le9$). Hence
$$100A+10B+C=A+B^2+C^3\implies 99A+C<C^3$$
and since $A\ge1$, this gives us immediately that $C\ge5$.
Now, assume that $B=0$. Then we have $100A+C=A+C^3$, but since none of $5^3=125$, $6^3=216$, $7^3=343$, $8^3=512$, and $9^3=729$ are in the range $[100n-10,100n+10]$, there are no solutions with $B=0$, so from here on we assume $B\ne 0$
Now we have $10B>B^2+A$, so $100A<C^3$, and thus $A$ is at most the first digit of $C^3$, but since we also know $A$ must be at least the first digit of $C^3$ (easy to see from the original formulation of the problem), we have that $A$ is exactly the first digit of $C^3$.
Now all we have to do is solve $5$ quadratic equations for $B$ (one for each possible value of $C$) and see if there are integer solutions:
$C=5$:$$100+10B+5=1+B^2+125\implies B=3,7$$
$C=6$:$$200+10B+6=2+B^2+216\implies B=5\pm\sqrt{13}$$
$C=7$:$$300+10B+7=3+B^2+343\implies B=5\pm i\sqrt{14}$$
$C=8$:$$500+10B+8=5+B^2+512\implies B=1,9$$
$C=9$:$$700+10B+9=7+B^2+729\implies B=5\pm i\sqrt{2}$$
At long last this gives us the $4$ solutions $ABC\in\{135,175,518,598\}$.
|
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|
Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given statement with 1.2 in it.)
$\sin^3 a + 3\sin a * \cos a (\sin a + \cos a) + \cos^3 a = 1.728$
$\sin^3 a + 3\sin a * \cos a (1.2) + \cos^3 a = 1.728$
$\sin^3 a + \cos^3 a = 1.728 - 3\sin a * \cos (a) *(1.2)$
Now I am stuck. What do I do next? Any hints?
|
You could use this
$$\eqalign{
& {(\sin a + cosa)^3} = \sin {a^3} + cos{a^3} + 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr
& \Rightarrow {\sin ^3}a + co{s^3}a = {(\sin a + cosa)^3} - 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr
& and \cr
& {(\sin a + cosa)^2} = 1 + 2\sin a \cdot cosa \cr
& \Rightarrow \sin a \cdot cosa = {{{{(\sin a + cosa)}^2} - 1} \over 2} \cr
& then : \sin a + cosa = k \cr
& \Rightarrow {\sin ^3}a + co{s^3}a = {k^3} - {3 \over 2}k({k^2} - 1) = - {k \over 2}({k^2} - 3) \cr} $$
|
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|
How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't get any more, and I can't use derivatives or multiple limits since we didn't study these yet.
Thanks
|
Since $x\ge0$, $\sqrt{x^2+x}\ge x$, and since $x\ge1$, $\sqrt{x^2+x}\ge \sqrt2$. Thus,
$$
\frac{x+y}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1
$$
and
$$
\frac1{\sqrt{x^2+x}+\sqrt{y^2+y}}\le\frac1{2\sqrt2}
$$
we have
$$
\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1+\frac1{2\sqrt2}\lt2
$$
|
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|
Solving for $r$ in ${12\choose{r}}=924$ I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of cases.
|
Use the definition $$_nC_r = \frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$\frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2\cdot 3 \cdot 7 \cdot 11$ which means $$\frac{12!}{924} =\require{cancel} \frac{1\cdot 2\cdot \cancel{3} \cdot \cancel{4} \cdot 5 \cdot 6 \cdot \cancel{7} \cdot 8 \cdot 9 \cdot 10 \cdot \cancel{11} \cdot 12}{\cancel{2^2}\cdot \cancel{3} \cdot \cancel{7} \cdot \cancel{11}} \\ = 2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 10 \cdot 12$$ So now you have $$2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 10 \cdot 12=r!(12-r)!$$ Now you can quickly eliminate $r=0,1,2,3,4,5,7,8,9,10,11,12$ because if $r$ is one of those numbers, then either $r!$ or $(12-r)!$ will have a factor of $7$, and will not cancel with the LHS, hence you cannot have equality. This leaves you with $r=6$ as the only option.
|
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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.
|
You have terms like $(\frac{1}{2^n} + \frac{1}{3}\cdot \frac{1}{2^{n-1}} + \cdots + \frac{1}{3^n})$ for each $n$. We can collect terms in $n$ instead of $k$ - we get a series $\frac{1}{3^k} \sum_{i = 0}^\infty 2^{-i}$ for each $k$.
E.g., pick a value of $k$. For $\frac{1}{3^k}$, there exists exactly one term also containing $2^{n-k}$ for any one given positive $n-k$. Collect all these terms to create the sum. We get
$$
\sum\sum(\cdots) = \sum_{k = 0}^\infty \frac{1}{3^k}\sum_{n=0}^\infty \frac{1}{2^n} \\
= 2\sum_{k = 1}\frac{1}{3^k} \\
= 2 \frac{1}{1 - \frac{1}{3}} \\
= 3
$$
|
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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.
|
HINT: Take $x=t+2$ and take limit as $t\to 0$ also notice that $\sqrt{t^2}=|t|$ and than take $\lim_{t\to0^+}$ and $\lim_{t\to 0^-}$
|
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|
Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
|
$$\cos^2B+\cos^2C+2\cos A\cos B\cos C =1+\cos^2B-\sin^2C+2\cos A\cos B\cos C$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
Again, $\cos(B+C)=\cos(\pi-A)=-\cos A$
Using Werner Formula,
$$2\cos A\cos B\cos C=\cos A[2\cos B\cos C]$$
$$=\cos A[\cos(B-C)+\cos(B+C)]=\cos A[\cos(B-C)-\cos A]$$
I hope the rest should not be too tough to deal with
|
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|
Easiest way to calculate determinant 5x5 witx x I would like to calculate this determinant:
\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}
|
The matrix is tridiagonal, and the determinant can be computed recursively as follows.
Let $f_n$ be the determinant of the $n\times n$ submatrix in the upper left of your matrix, and let $f_0=1$. Write the matrix as
\begin{vmatrix}a_1&b_1&0&0&0\\c_1&a_2&b_2&0&0\\0&c_2&a_3&b_3&0\\0&0&c_3&a_4&b_4\\0&0&0&c_4&a_5\end{vmatrix}
and this recurrence holds: $f_n=a_nf_{n-1}-c_{n-1}b_{n-1}f_{n-2}$. (The procedure is described on this Wikipedia page.) For this particular matrix,
$f_2=x^2-4$,
$f_3=x(x^2-4)-6x=x^3-10x$,
$f_4=x(x^3-10x)-6(x^2-4)=x^4-16x^2+24$, and
$f_5=x(x^4-16x^2+24)-4(x^3-10x)=x^5-20x^3+64x$.
|
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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.
|
\begin{align}
\left[\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right]^{-1} & = \frac{1}{{\left|\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right|}}\cdot\left(\begin{matrix}\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)\end{matrix}\right)\tag{1} \\[2em]
& = 1\cdot \left(\begin{matrix}\cos\left(\theta\right) & \sin\left(\theta\right) \\
-\sin\left(\theta\right) & \cos\left(\theta\right)\end{matrix}\right),\tag{2}
\end{align}
because
\begin{align}\left|\begin{matrix}\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right|&=\cos^2\left(\theta\right)-\left\{-\sin^2\left(\theta\right)\right\}=\sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1.\tag{3}
\end{align}
|
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|
Does the limit $\lim\limits_{x \to\infty}x^2 \sin\frac{1}{x}$ not exist? I used $u$ substitution for the limit $$\lim_{x \to\infty}x^2 \sin\frac{1}{x}$$ and got the limit does not exist by saying $u=\frac{1}{x}$. Is this correct and if so would that mean $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2}$ and $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x}$ also don't exist?
|
Since $\sin\frac{1}{x}\to0$ and $x^2\to\infty$, we use l'Hôpital's theorem:
$$\lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x^2}}=\lim_{x\to\infty}\frac{-\frac{1}{x^2}\cos\frac{1}{x}}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{x}{2}\cos\frac{1}{x}=\infty$$
since $\cos\frac{1}{x}\to1$ and $\frac{x}{2}\to\infty$, as $x \rightarrow \infty.$
|
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|
Find the general solution of the following first order differential equation $\frac{dy}{dx}=\frac{8x^3+3x^4}{y^4}$ I have multiplied both sides by $y^4$
which gives me $\frac{dy}{dx}y^4=x^3(3x+8)$
Then do I integrate both sides with respect to x?
$\int\frac{dy}{dx}yx^4dx=\int{x^3(3x+8)dx}$
Am I still on the right track?
So I get $\frac{yx^5}{5}=\frac{3x^5}{5}+2x^4+c$
Do I just solve for y after this?
|
By separation of variables, you get $$y^4 \,dy = (3x^4 + 8x^3)\,dx$$
Now, integrate each side of the equation:
$$\int y^4 \,dy = \int (3x^4 + 8x^3)\,dx$$
$$\iff \frac 15 y^5 = \frac 35x^5 + 2x^4 + c$$
|
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|
Modular exponentiation How do you solve:
$$5^{{9}{^{13}}^{17}} \equiv x\pmod {11}$$
I've been trying with this but no luck.
I get to ${{9}{^{13}}^{17}} \equiv x\pmod {11}$ from $5^3 * 5^3 * 5^3 = 64 \equiv 9\pmod {11}$. But volfram alpha disagrees with all that. So just a nudge in the right direction please.
thanks!
|
As an alternate way $$5^5=3125$$ $3124\equiv 0\pmod {11}$ (use the criterion for divisibility with $11$), so $$5^5\equiv 1 \pmod {11}$$ On the other hand, $9\equiv 4\pmod 5$ and $13\equiv 1\pmod 2\Rightarrow 13^{17}\equiv 1\pmod 2$, so $9^{13^{17}}\equiv 4\pmod 5$. It follows that $$5^{9^{13^{17}}}\equiv 5^4\pmod {11}$$ $5^4=625,\ 625\equiv 9\pmod {11}$ (apply again the divisibility criterion by $11$), so the conclusion is $$5^{9^{13^{17}}}\equiv 9\pmod {11}.$$
|
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|
Area between two functions My question is from Apostol's Vol. 1: One-variable calculus with introduction to linear algebra textbook.
Page 94. Exercise 16. Let $f(x)=x-x^2$, $g(x)=ax$. Determine $a$ so that the region above the graph of $g$ and below the graph of $f$ has area $\frac{9}{2}.$
My attempt at a solution. I tried following: $$a(S)=\int_{x_1}^{x_2}[f(x)-g(x)]\mathrm dx=\int_{x_1}^{x_2}x\mathrm dx-\int_{x_1}^{x_2}x^2\mathrm dx-a\int_{x_1}^{x_2}x\mathrm dx=\frac{x_2^2-x_1^2-a(x_1^2-x_2^2)}{2}-\frac{x_2^3-x_1^3}{3}=9/2$$ but this approach gets me nowhere, I graphed both functions, and I realize that $a$ must be negative but without boundaries, on integral, how should I find out what $a$ is, because $a$ changes as $x_1,x_2$ change. Please explain or give some hints.
|
By graphing the two functions, we can see that the region represented by the volume is:
$$f(x) - g(x) = x-x^2 - ax$$
In order to find the points of intersection, we must set the two graphs equal to one another:
$$x-x^2 = ax \Leftarrow\Rightarrow \ x^2 + x(a - 1) = 0 \Leftarrow\Rightarrow\ x(x + a - 1) =0$$
Therefore, $x = 0 , 1 - a$
Constructing our integral:
$$\int_{0}^{1-a} (x-x^2 - ax)dx = \left[\frac{x^2}{2} - \frac{x^3}{3} -\frac{ax^2}{2}\right]_{0}^{1-a} = \frac{(1-a)^2}{2} - \frac{(1-a)^3}{3} - \frac{a(1-a)^2}{2} = \frac{9}{2}$$
Simplifying we get:
$$\frac{(1-a)^3}{6} = \frac{9}{2}\Leftarrow\Rightarrow 2(1-a)^3 = 54 \Leftarrow\Rightarrow (1-a)^3 = 27$$
Therefore, $a = -2$
|
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|
How to show without calculator that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$ By wolfram alpha, I get
$\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor=2996$.
How to prove that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$ without calculator or wolfram alpha?
Thank in advances.
|
Prove that:
$$\log_{10}999^{999}+\log_{10}2<\log_{10}1000^{999}=2997$$
In other word:
$$\log_{10}2<\log_{10}1000^{999}-\log_{10}999^{999}=\log_{10}\left(\frac{1000}{999}\right)^{999}$$
so:
$$2<\left(\frac{1000}{999}\right)^{999}=\left(1+\frac{1}{999}\right)^{999}$$
It's true by Bernoulli's inequality.
Next we should prove $3 \cdot 999-1=2996<\log_{10}999^{999}$. It's equal:
$$3-\frac{1}{999}<\log_{10}999$$
or:
$$10^{3-\frac{1}{999}}=1000 \cdot 10^{-\frac{1}{999}}<999$$
$$10^{-\frac{1}{999}}<\frac{999}{1000}=\left(1-\frac{1}{1000}\right)$$
$$10^{-1}<\left(1-\frac{1}{1000}\right)^{999}$$
It's also true by Bernoulli's inequality.
|
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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 .
|
Hint. Recall that, as $n \to +\infty$, we have
$$1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\sim \ln n$$
then
$$n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)\sim n\ln n$$
and
$$
\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)} \sim \sum \frac{1}{n \ln n},
$$ your initial series is then divergent as is the latter series.
|
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|
How to show $\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\pi/2$ How to show that
$$\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi}{2}$$
The problem is that I don't know what is
$$\lim\limits_{x\to\infty}{\mathrm{arcsec}\ x}$$
|
Assuming that you intended to write
$$
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi}{2}
$$
First note that
$$
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx= \int_1^\infty \frac{x}{x^2\sqrt{x^2-1}}dx
$$
Let $u=\sqrt{x^2-1}$, then $du=\frac{x}{\sqrt{x^2-1}}dx$. So now
$$
\int_0^\infty \frac{1}{u^2+1}du
$$
$$=\lim\limits_{\beta\to \infty} \arctan \beta- \lim\limits_{\alpha\to 0^+} \arctan \alpha
$$
$$=\frac{\pi}{2}- 0= \frac{\pi}{2}
$$
As already mentioned, your original integral does not converge.
|
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|
How do I simplify this expression about factorization? I am trying to simplify this
$$\frac{9x^2 - x^4} {x^2 - 6x +9}$$
The solution is
$$\frac{-x^2(x +3)}{x-3} = \frac{-x^3 - 3x^2}{x-3} $$
I have done $$\frac{x^2(9-x^2)}{(x-3)(x-3)} = \frac{x^2(3-x)(3+x)}{(x-3)(x-3)} $$
but I do not find a way to simplify
How can I simplify to get the answer?
I have to use
$${(a +b)(a -b)} = {a^2 - b^2} $$
$${(a +b)^2} = {a^2 + 2ab+b^2} $$
$${(a -b)^2} = {a^2 - 2ab+b^2} $$
|
Notice that $(3-x)=-(x-3)$. Simplify accordingly.
|
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|
Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$ Prove that if $ a,b,c $ are non-negative real numbers such that $a+b+c = 3$, then $$ abc(a^2 + b^2 + c^2) \le 3 $$
My attempt :
I tried AM-GM inequality, tried to convert it to $a+b+c$, but I think I cannot get the manipulation of $abc$.
|
Let:
$$ M_p = \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p} $$
for $p>0$ and $M_0 = \sqrt[3]{abc}$. Since $M_p$ is a log-concave and increasing function,
$$ M_0^3 M_2^2 \leq M_{\frac{4}{5}}^5 \leq M_1^{5} = 1 $$
and the claim follows.
|
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|
What bound does the Hamming bound give you for the largest possible size of a $t$-error-correcting code of length $2t + 1$? Let $\mathbb{A}$ = $\{0, 1\}$ and suppose $t$ is a positive integer. What bound does the Hamming bound give you for the largest possible size of a $t$-error correcting code of length $2t+1$?
I have equation
$$
|C|\le \frac{q^n}{{n\choose0}+{(q-1)}{n\choose1}+\cdots+{(q-1)^t}{n\choose{t}}}
$$
with addition to the $2^n=\sum_{r=0}^n{n\choose{r}}$
I get answer $|C| \le 1$.
Is this correct?
|
In the equation you gave for the Hamming bound, $n$ is the length of the code and $q$ is the size of the alphabet. In your case, $n=2t+1$ and $q=|\mathbb{A}|=2$. Substituting $q=2$ in the bound:
$$
|C|
\le \frac{2^n}{\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{t}}
= \frac{2^{n}}{\binom{2t+1}{0} + \binom{2t+1}{1} + \ldots + \binom{2t+1}{t}}.
$$
To evaluate the denominator, you can use the other equation you listed:
$$\begin{align}
2^n &= \sum_{r=0}^n \binom{n}{r}\\
&= \sum_{r=0}^{2t+1} \binom{2t+1}{r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=t+1}^{2t+1} \binom{2t+1}{r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=t+1}^{2t+1} \binom{2t+1}{(2t+1)-r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=0}^{t} \binom{2t+1}{r}\\
&= 2 \sum_{r=0}^{t} \binom{2t+1}{r}\\
&= 2 \left(\binom{2t+1}{0} + \binom{2t+1}{1} + \ldots + \binom{2t+1}{t}\right).
\end{align}$$
The fourth line is the critical step: $\binom{i}{j} = \binom{i}{i-j}$.
Once you've found an upper bound on $|C|$, you should try to find a binary $t$-error-correcting code of length $2t+1$ which meets that bound.
|
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|
Rolling dice probability by solving inequlity I was trying to solve a problem where I have to find the probability of the sum of $\mathcal 3$ rolls of a die being less than or equal to $\mathcal 9$.
In order to solve the problem I try first to find the number of non-negative integer solutions to the following inequality: $x_{1} + x_{2} + x_{3} \le 6$
From this it follows that by introducing the slack variable $x_{4}$ , I could rewrite the inequality as the following equation: $x_{1} + x_{2} + x_{3} + x_{4} = 6$ where $x_{4} = 6 - (x_{1} + x_{2} + x_{3})$
Therefore the answer would be $\binom{9}{3}$ solutions, where I have to remove 4 impossible solutions leaving me with $\frac{80}{6^3}$ as the probability.
I was wondering however, if the introduction of the slack variable $x_{4}$(forcing the equation to look like as if there were $\mathcal 4$ rolls of a die instead of $\mathcal 3$), could eventually result in a wrong probability? Is this method a reliable way for computing this probability?
|
An alternative solution (not a direct answer to your question):
Sums equal to $3$:
*
*We have $1$ permutation of $1,1,1$
Sums equal to $4$:
*
*We have $3$ permutations of $1,1,2$
Sums equal to $5$:
*
*We have $3$ permutations of $1,1,3$
*We have $3$ permutations of $1,2,2$
Sums equal to $6$:
*
*We have $3$ permutations of $1,1,4$
*We have $6$ permutations of $1,2,3$
*We have $1$ permutation of $2,2,2$
Sums equal to $7$:
*
*We have $3$ permutations of $1,1,5$
*We have $6$ permutations of $1,2,4$
*We have $3$ permutations of $1,3,3$
*We have $3$ permutations of $2,2,3$
Sums equal to $8$:
*
*We have $3$ permutations of $1,1,6$
*We have $6$ permutations of $1,2,5$
*We have $6$ permutations of $1,3,4$
*We have $3$ permutations of $2,2,4$
*We have $3$ permutations of $2,3,3$
Sums equal to $9$:
*
*We have $6$ permutations of $1,2,6$
*We have $6$ permutations of $1,3,5$
*We have $3$ permutations of $1,4,4$
*We have $3$ permutations of $2,2,5$
*We have $6$ permutations of $2,3,4$
*We have $1$ permutation of $3,3,3$
So the number of sums less than or equal to $9$ is $81$.
The total number of sums is $6^3=216$, hence the probability is $\frac{81}{216}$.
|
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|
how to find $a_{50}$ from a recursive term Given $a_{n+1}=a_n+2n+3,a_1=3$
How can I find $a_{50}$?
I can compute $a_2,a_3,...,a_{50}$
But it's a long way. Is there any smart technique to compute?
Thanks.
|
Other then methods above here is a general method which can work in many cases characteristic equation,but first you have to transform $a_{n+1}$ as combination of $a_k$'s without the constant terms and $n$ terms
$$a_{n+1}=a_n+2n+3\\a_n=a_{n-1}+2n+1\text{ I plugged n=n-1 into first equation}\\a_{n+1}-a_n=a_n-a_{n-1}+2\\a_{n+1}=2a_n-a_{n-1}+2\\a_n=2a_{n-1}-a_{n-2}+2\\a_{n+1}-a_n=2a_n-3a_{n-1}+a_{n-2}\\a_{n+1}-3a_n+3a_{n-1}-a_{n-2}=0$$
Now plug in $a_{n+1}=t^{n+1}$
$$t^{n+1}-3t^n+3t^{n-1}-t^{n-2}=0\\t^3-3t^2+3t-1=0\\(t-1)^3=0$$
Now since the roots of the polynomial are $t_1=t_2=t_3=1$
We have that $$a_n=k_1t_1^n+nk_2t_1^n+n^2k_3t_1^n=k_1+nk_2+n^2k_3$$
Now plugging in $n=1,2,3$ you'll get the coefficients
$$a_1=k_1+k_2+k_3=3\\a_2=k_1+2k_2+4k_3=8\\a_3=k_1+3k_2+9k_3=15\\k_2+3k_3=5\\2k_2+8k_3=12\\2k_2+6k_3=10\\k_3=1\\k_2=2\\k_1=0\\a_n=1n^2+2n+0=n(n+2)$$
|
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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
|
Note that $(1+x+x^2+x^3+x^4+x^5+x^6)$ has $z, z^2,z^3,z^4,z^5,z^6$ as roots, thus f(x) should have atmost one real root (MVT)
|
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|
$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to variables.
How to approach such question?
|
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca)$
therefore $$(ab+bc+ca)=((a+b+c)^2-1)/2$$ so the value is not fixed
|
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|
Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even
I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this without checking every combination of the parity of $a,b,c$?
|
$a$ and $a^2$ are either both even or both odd. So you don't change parity when you replace $a^2$ with $a$:
$$\begin{align}
ab(a+b)+bc(b+c)+ca(c+a)
&=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\\
&\equiv ab+ab+bc+bc+ac+ac\\
&= 2(ab+bc+ac)\\
\end{align}$$
which is even.
|
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|
Find coefficient of x in a generating function The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
I factored out some $x$'s, to get $x^3(1+x^2+x^3)x^4(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$and then combined the factored terms to get $x^7(1+x^2+x^4)(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$
Now I don't know what to do; usually it ends up factoring to $(1+x+x^2+...)$, but that doesn't appear to be the case here.
|
Multiplying the first two factors you find:
$x^7+ x^8+x^9+2x^{10} + $ other monomials of degree $n>10$ and, when you multiply such polynomial with the third factor, you see that the only monomial in $x^{10}$ has coefficient $2$ since all other terms have exponents $n=7,8,9$ or $n>10$.
Hint:
The OP has changed: so for this version the answer is:
Multiplying the first two factors you find:
$x^7+ x^8+x^9+3x^{10} + $ other monomials of degree $n>10$ and, when you multiply such polynomial with the third factor, you see that the only monomial in $x^{10}$ has coefficient $3$ since all other terms have exponents $n=7,8,9$ or $n>10$.
|
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|
Line integral along the curve $\gamma(t)=(4 \cos t, 4 \sin^2 t) $ Let us consider the vector field in the plane:
$$\vec{F}=\left(x \frac{e^{x^2+y^2} - e}{x^2+y^2},y\frac{e^{x^2+y^2} - e}{x^2+y^2}\right)$$
calculate the line integral along the curve defined by:
$$\gamma:
\begin{cases}
x=4 \cos t \\
y=4 \sin^2 t\\
\end{cases}
$$
with $t\in[0,\pi/2]$.
Any suggestions please?
|
Here is a direct evaluation using this link
$\begin{align}
dx&=-4\sin t \\
dy&=8\cos t \sin t \\
F&=\Big(\frac{2[\exp{(2\cos 4t +14)}-e] \cos t }{7+\cos 4t},\frac{2[\exp{(2\cos 4t +14})-e] \sin^2 t }{7+\cos 4t}\Big)
\end{align}$
Your integral will hence be
$\begin{align}I&=-\int_0^{\frac{\pi}{2}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\
&=-\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\
&=-\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt+\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\
&=0.
\end{align}$
where at last I have changed the variable for the second integral from $t$ to $\frac{\pi}{2}-t$.
|
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|
Problem while calculating a limit with tangens I'm having trouble calculating this limit:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x})$$
I've tried using the gact that $\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1$, but that obviously doesn't work here.
Neither does the l'Hospital's rule.
Could you give me some idea how to solve this?
Thanks.
|
apply l'Hospital's rule:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x}) = \lim _{x \rightarrow 0} (\frac{tanx - x}{2x^2 \tan x}) = \lim _{x \rightarrow 0} (\frac{\frac{1}{cos^2x} - 1}{4xtan x + \frac{2x^2}{cos^2x}})
= \lim _{x \rightarrow 0} (\frac{1 - cos^2x}{2x^2 + 4xsinxcosx}) = \lim _{x \rightarrow 0} (\frac{sin^2x}{2x^2 + 2xsin2x}) = \lim _{x \rightarrow 0} (\frac{sin2x}{4x+2sin2x+4xcos2x}) = \lim _{x \rightarrow 0} (\frac{2cos2x}{4 + 4cos2x + 4cos2x - 8xsin2x}) = \frac{2}{12} = \frac{1}{6} $$
|
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|
A chess player, X, plays a series of games against an opponent, Y A chess player $X$ plays a series of games against an opponent $Y$. For each game, the probability that $X$ wins is $p$, independently of the results of other games. If $X$ plays 4 games against $Y$, show that the probability that this series of games contains at least 2 consecutive wins by $X$ is $p^2(3-2p)$.
I know this can be done by listing all the possible ways and then summing the probabilities, but is there another way of doing this?
|
Ok, this is simpler ...
You want all cases with a "WW" in them. e.g. WWLL, WWWL, etc. Let "A" mean either win or lose Prob(A)=1 of course. The cases are:
WWAA: $p^2$
LWWA: $(1-p) \times p^2$
ALWW: $(1-p) \times p^2$
$= 3 p^2 -2 p^3$
$= p^2 (3 -2 p)$
Note - to avoid counting cases twice, I listed the cases by doing the WW moving from left to right, allowing anything to the right of the WW but no '2 consecutive wins' to the left of the WW.
For a larger number of games, this method is much better than the basic method, e.g. for 6 games, the answer would be:
WWAAAA: $p^2$
LWWAAA: $(1-p) \times p^2$
ALWWAA: $(1-p) \times p^2$
XXLWWA: $(1-P_2) \times (1-p) \times p^2$
XXXLWW: $(1-P_3) \times (1-p) \times p^2$
where $P_2 = p^2$ is the probability of having at least 2 consecutive wins in 2 games, and $P_3 = p^2 (2-p)$ is the probability of having at least 2 consecutive wins in 3 games.
$P(6) = 5 p^2 -4 p^3 -3 p^3 +4 p^5 -p^6$
The general answers are:
$P_2 = p^2$
$P_3 = 2 p^2 -p^3$
$P_4 = 3 p^2 -2 p^3$
$P_5 = 4 p^2 -3 p^3 -p^4 +p^5$
$P_6 = 5 p^2 -4 p^3 -3 p^4 +4 p^5 -p^6$
$P_7 = 6 p^2 -5 p^3 -6 p^4 +9 p^5 -3 p^6$
$P_8 = 7 p^2 -6 p^3 -10 p^4 +16 p^5 -5 p^6 -2 p^7 +p^8$
$P_9 = 8 p^2 -7 p^3 -15 p^4 +25 p^5 -6 p^6 -9 p^7 +6 p^8 -p^9$
$P_{10} = 9 p^2 -8 p^3 -21 p^4 +36 p^5 -5 p^6 -24 p^7 +18 p^8 -4 p^9$
There are some patterns to the coefficients of $P_n$ e.g. the $p^2$, $p^3$, $p^4$ coefficients, but I have not found a pattern for all of them.
|
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|
For what values of $a$, $\sum_{n=1}^{\infty}\frac{(\cos n)(\sin na)}{n}$ converges? For what values of $a$, $$\sum_{n=1}^{\infty}\dfrac{(\cos n)(\sin na)}{n}$$ converges? Every hint is appreciated.
I know that $(\cos n)(\sin na)=\dfrac{1}{2}(\sin (n+1)a+\sin (n-1)a)$.
|
Assuming $a$ is real, the series converges for all $a$. It follows from the fact that $\sum_{n = 1}^\infty \sin[n(a+1)]/n$ and $\sum_{n = 1}^\infty \sin[n(a-1)]/n$ converge for all $a$. Let's consider the latter series. If $a - 1$ is an integral multiple of $2\pi$, then each summand is $0$, so the series converges to $0$. If $a-1$ is not an integral multiple of $2\pi$, then the series converges by Dirichlet's test. To see this, note that for $N \ge 1$, \begin{align}\sum_{n = 1}^{N} \sin[n(a-1)] &= \csc\left(\frac{a-1}{2}\right) \sum_{n = 1}^{N} \sin[n(a-1)]\sin\left(\frac{a-1}{2}\right)\\
&= \csc\left(\frac{a-1}{2}\right) \sum_{n = 1}^{N-1} \left(\cos\left[\left(n-\frac{1}{2}\right)(a-1)\right] - \cos\left[\left(n + \frac{1}{2}\right)(a-1)\right]\right)\\
&= \csc\left(\frac{a-1}{2}\right) \left\{\cos\left(\frac{a-1}{2}\right) - \cos\left[\left(N + \frac{1}{2}\right)(a-1)\right]\right\}.
\end{align}
(Keep in mind that $\csc((a - 1)/2)$ is defined since $(a - 1)/2$ is not an integral multiple of $\pi$.) It follows that the sequence of partial sums of the series $\sum_{n = 1}^\infty \sin[n(a-1)]$ is bounded by $2|\csc((a-1)/2)|$. Furthermore, the sequence $\{\frac{1}{n}\}_{n = 1}^\infty$ decreases to $0$. Thus, by the Dirchlet test, the series $\sum_{n = 1}^\infty \sin[n(a-1)]/n$ converges. A similar argument shows that the series $\sum_{n = 1}^\infty \sin[n(a+1)]/n$ converges for all $a$.
|
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differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$
$(4x^4+4x^3+6x^2+6x+6x^4+12x^3-48x^2+3x^2+6x-24)$
$=10x^4+16x^3+39x^2+6x-18$
|
HINT : If $u,v,w$ are functions of $x$
Then $$\frac{d(uvw)}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}$$
|
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|
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me that:
$$\frac{x}{(x-1)^2} = \frac{1}{(x-1)} + \frac{1}{(x-1)^2}$$
How to intuitively think about this partial fraction expansion? I've seen some examples but suddenly these conter intuitive examples opo out and I get confused. I can check that his is true but I couldn't find this expansion by myself
Then:
$$\int \frac{x}{(x-1)^2} dx = \int \frac{1}{(x-1)} + \frac{1}{(x-1)^2}dx = \ln (x-1) + (x-1)^{-1}$$
Then:
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+\frac{x}{(x-1)^2}dx = \frac{x^2}{2} + 2x + 3[\ln(x-1)+(x-1)^{-1}]$$ but wolfram alpha gives another answer. What I did wrong?
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Your mistake is the partial fraction expansion. It should be $$\frac{x^3+2}{(x-1)^2}=x+2+\frac{3}{x-1}+\frac{3}{(x-1)^2}$$ Then the integral becomes $$x^2/2+2x+3\ln (x-1)-3(x-1)^{-1}$$
|
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What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$? Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come to the conclusion, using the pythagorean theorem, that the height of the trapezoid is $h=\sqrt { 1-\frac { x^{ 2 } }{ 4 } } $
The formula for the area of a trapezoid is $A=\frac { a+b }{ 2 } (h)$
I am basically solving for the $a$ here and so far I have:
$A=\frac { x+2 }{ 2 } (\sqrt { 1-\frac { x^{ 2 } }{ 4 } } )$
I simplified this in order to be able to easily take the derivative of it as such:
$$A=\frac { x+2 }{ 2 } (\sqrt { \frac { 4-x^{ 2 } }{ 4 } } )$$
$$A=\frac { x+2 }{ 2 } (\frac { \sqrt { 4-x^{ 2 } } }{ 2 } )$$
$$A=(\frac { 1 }{ 4 } )(x+2)(\sqrt { 4-x^{ 2 } } )$$
Then I took the derivative:
$$A'=\frac { 1 }{ 4 } [(1)(\sqrt { 4-x^{ 2 } } )+(x+2)((\frac { 1 }{ 2 } )(4-x^{ 2 })^{ -1/2 }(-2x)]$$
This all simplified to:
$$A'=\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } ]$$
Next, I set the derivative equal to zero in order to find the maximum point (I know that I can prove that it is actually a max point by taking the second derivative later)
$$\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } ]=0$$
$$\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } =0$$
$$\frac { 4-x^ 2-2x^ 2-4x }{ \sqrt { 4-x^ 2 } } =0$$
$$4-x^{ 2 }-2x^{ 2 }-4x=0$$
$$-3x^{ 2 }-4x+4=0$$
so I got:
$$x=-2\quad or\quad x=\frac { 2 }{ 3 }$$
$x=-2$ wouldn't make sense so I chose $x=\frac { 2 }{ 3 }$ and plugged it back into my original formula for the area of this trapezoid but my answer doesn't seem to match any of the multiple choice solutions. Where did I go wrong?
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When you went from $$\sqrt{4-x^2} - {{2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$
to $$ {{4 - x^2 - 2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$ you slipped a sign. The last term in the numerator should be $+4x$ instead of $-4x$.
Never mind. That error just reverses a previous erroneous sign. Your real error is in your first simplification of $A'$. You have factors of $1 \over 2$ and $-2x$. The $2$s should have cancelled, but you instead left out the $1 \over 2$.
|
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finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-10 x-7 x^2+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2 x-3 x^2+x^4}}$$.
But failed to solve $f'(x)=0$ for finding $f(x)_{max}$.
I would be glad for your help.
Thanks.
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I have notice that if $A(x^2,x),B(1,2),C(5,4)$, so $f(x)=AC-AB$.
Also $AC-AB\leq BC$. so the max of $f(x)$ will be $BC=\sqrt{20}$.
I checked and it's the right value by wolfram.
The problem is that i can't justify it.
For example if i take: $g(x)=\sqrt{(x-4)^2+(x-5)^2}-\sqrt{(x-2)^2+(x-1)^2}$
And choose: $A(x,x),B(1,2),C(5,4)$ it still be $\sqrt{20}$ but this function doesn't have a max value by wolfram.
Maybe someone can explain my mistake.
|
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|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
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For the last question, you have a triangle with side lengths $1,x,\sqrt{x^2+1}$ to have tangent $x$. That triangle has cosine $\frac{1}{x^2+1}$. Hence $$\cos \arctan x=\frac{1}{\sqrt{x^2+1}}$$
|
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|
Proof by induction and combinations I think I am stuck on this, I am not sure if I'm going down the correct path or not. I am trying to algebraically manipulate $p(k+1)$ so I can use $p(k)$ but I am unable to do so, so I am not sure if my math is bad or I am going about it the incorrect way.
15b.) Prove by induction that $(nCk) = \dfrac{(nC k - 1) \cdot (n - k +1)}{k}$
Starting step: prove $p(1)$
$$\begin{align}
p(1) & = {n \choose 1} = {n \choose 0} \cdot \frac{n + 0}{1} \\
& = \frac{n!}{1!(n-1)!} = \frac{n!}{0! (n!) \cdot (n)} \\
& = \left[\frac{n!}{1 \cdot (n-1)!}\right] = \left[\frac{n!}{1 \cdot (n)!}\right] \cdot (n) \\
& = \frac{n!}{(n-1)!} = \frac{n!}{n!} \cdot (n) \\
& = \frac{n!}{(n-1)!} = 1n \\
& = n = n
\end{align}$$
$p(k)$ is true: $$p(k) = \frac{n!}{[k! (n-k)!]} = \frac{n!}{[(k-1)!(n - k + 1)] \cdot [(n - k + 1) / k]}$$
Inductive step: show $p(k) \implies p(k+1)$
$$\begin{align}
p(k+1) & = {n \choose k + 1} = {n \choose k} \cdot \frac{n - k}{k+1} \\
& = \frac{n!}{(k+1)!(n - k + 1)!} = \frac{n!}{k! (n - k)!} \cdot \frac{n - k}{k+1} \\
& = \frac{n!}{(k+1)(k!)(n - k + 1)!} = \frac{n!}{k! (n - k)!} \cdot \frac{n - k}{k+1} \cdot k! (n - k)! \\
& = \frac{n! \cdot [k! (n - k)!]}{(k+1)(k!)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \\
& = n! \cdot \frac{(n - k)!}{(k+1)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \cdot (k+1) \\
& = n! \cdot (k+1) \cdot \frac{(n - k)!}{(k+1)(n - k + 1)!} = n! \cdot \frac{n - k}{k+1} \cdot (k+1) \\
& = n! \cdot \frac{(n - k)!}{(n - k + 1)!} = n! \cdot (n - k)
\end{align}$$
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You dont need to use mathematical induction on proving your claim. It will follow directly from the definition of the binomial coefficient.
Note that: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.
While $\binom{n}{k-1}=\frac{n!}{(k-1)!(n-(k-1))!}$ which reduces to $\frac{n!}{(k-1)!(n-k+1)!}$.
By multiplying $\frac{n-k+1}{k}$ in the last expression that is:
$\frac{n!}{(k-1)!(n-k+1)!}\cdot \frac{n-k+1}{k}$ we get
$\frac{n!}{k!(n-k)!}=\binom{n}{k}$.
If we use induction:
Suppose $\binom{n}{k-1}=\binom{n}{(k-1)-1}\cdot \frac{n-((k-1)+1)}{(k-1)}$.
Changing $k-1$ to $(k-1)+1$ will yield the answer above. So it actually holds by induction.
|
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|
Congruence modulo a power of two let $a \geq 3$. Let $2^a \mid c^2-1$.
By simple divisibility rules I get
$$c \equiv 1 \mod 2^a \quad \text{or} \quad c \equiv -1 \mod 2^a $$
or
$$c \equiv 1 \mod 2^{a-1},\quad c \equiv -1 \mod 2 $$
or
$$c \equiv -1 \mod 2^{a-1},\quad c \equiv 1 \mod 2 $$
Can you explain me how to obtain the following congruence:
$$
c \; \equiv \; \pm 1 + 2^{a-1} \mod 2^a
$$
??
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We want to solve the congruence $(c-1)(c+1)\equiv 0\pmod{2^a}$, where $a\ge 3$. Note that $c$ must be odd, and that $c-1$ and $c+1$ are two consecutive even numbers.
Since $c-1$ and $c+1$ are consecutive even numbers, one of them is congruent to $2$ modulo $4$, and therefore only has one $2$ to contribute to the product.
Case 1: [$c+1\equiv 2\pmod{4}$] Since $2^a$ divides $(c-1)(c+1)$, $2^{a-1}$ must divide $c-1$. So $c\equiv 1\pmod{2^{a-1}}$, and therefore $c=1+k2^{a-1}$ for some integer $k$. If $k$ is even, then $s\equiv 1\pmod{2^a}$. If $k$ is odd then $c\equiv 1+2^{a-1}\pmod{2^{a}}$.
It is clear that $c\equiv 1\pmod{2^a}$ and $c\equiv 1+2^{a-1}\pmod{2^a}$ are solutions.
Case 2: [$c-1\equiv 2\pmod{4}$] Then $2^{a-1}$ must divide $c+1$, and therefore $c=-1+k2^{a-1}$ for some integer $k$. If $k$ is even then $c\equiv -1+2^{a-1}\pmod{2^{a}}$, and if $k$ is odd then $c\equiv -1+2^{a-1}\pmod{2^{a}}$.
If $a\ge 3$, then the four solutions are distinct modulo $2^a$.
|
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Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$ Ok so I'm kind of struggling with this:
The question is:
"Use mathematical induction to prove that
1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"
Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2
Assuming P(k) is true for k≥1 gives:
1*3 + 2*4 + 3*5 + ... + k(k+2) ≥ (1/3)(k^3 + 5k)
And we want to show that P(k+1) is true, that is;
1*3 + 2*4 + 3*5 + ... + (k+1)(k+3) ≥ (1/3)((k+1)^3 + 5(k+1)) (inequality 1)
This is where I'm not sure if I'm thinking along the right lines or not. Surely, by P(k) and the rules for inequalities we get:
(1*3 + 2*4 + 3*5 + ... + k(k+2)) + (k+1)(k+3) ≥ (1/3)(k^3 + 5k) + (k+1)(k+3) (inequality 2)
but the right hand side in this inequality does not equal (1/3)((k+1)^3 + 5(k+1)) which is what I am trying to show. There's obviously a flaw in my thinking somewhere so any help would be greatly appreciated.
Thanks
Edit: As far as I understand it the right hand side of inequality 1 should equal the right hand side of inequality 2 but this isn't the case when expanded
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Here's the meat of the argument:
\begin{align}
1\cdot3+2\cdot 4+\cdots+(k+1)(k+3)&\geq \frac{k^3+5k}{3}+(k+1)(k+3)\tag{ind. hyp.}\\[1em]
&= \frac{k^3+5k+3(k^2+4k+3)}{3}\tag{simplify}\\[1em]
&= \frac{k^3+3k^2+17k+9}{3}\tag{simplify}\\[1em]
&\geq \frac{k^3+3k^2+8k+6}{3}\tag{since $k\geq 1$}\\[1em]
&= \frac{(k+1)^3+5(k+1)}{3}.
\end{align}
The last equality is what you want. The key is in realizing that since $k\geq 1$ you can obtain the desired expression on the right-hand side to show that
$$
1\cdot3+2\cdot 4+\cdots+(k+1)(k+3) \geq\frac{(k+1)^3+5(k+1)}{3}.
$$
|
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|
Solving the ODE $y+xy'=x^4 (y')^2$ I am trying to get to the solution which is $$y=c^2 +\frac{c}{x}$$
How would I go about solving this?
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$$y+x\frac{dy}{dx}=x^4\left(\frac{dy}{dx}\right)^2$$
Change of variable : $x=\frac{1}{X}$
$\frac{dy}{dx}=\frac{dy}{dX}\frac{dX}{dx}=-X^2\frac{dy}{dX}$
$$y-xX^2\frac{dy}{dX}=x^4\left(X^2\frac{dy}{dX}\right)^2$$
$$y-X\frac{dy}{dX}=\left(\frac{dy}{dX}\right)^2$$
On can see immediately that $y=aX+b$ is convenient :
$(aX+b)-aX=a^2$ imply $b=a^2$ hense $y=aX+a^2$
$$y=\frac{a}{x}+a^2$$
This is the expected result.
If we don't see the short-cut above,
$$y+\frac{X^2}{4}=\left(\frac{dy}{dX}+\frac{X}{2}\right)^2$$
$$y+\frac{X^2}{4}=\left(\frac{d}{dX}\left(y+\frac{X^2}{4}\right)\right)^2$$
Let $Y=y+\frac{X^2}{4}$
$$Y=\left(\frac{dY}{dX}\right)^2$$
leading to $Y=\left(\frac{X}{2}+c\right)^2=y+\frac{X^2}{4}$
$$y=cX+c^2=\frac{c}{x}+c^2$$
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Evaluate $\int\frac{x^3}{(x^2+1)^3}dx$ $\int\frac{x^3}{(x^2+1)^3}dx$ what is my $z$? I know that $x=z^n$ and $n$ is the LCD of the exponents, but my lcd here is just one? so my $z$ is gonna be $z=x^2+1$? but I saw the notes of my classmate its $z^2=x^2+1$ can someone explain this thoroughly please
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If you substitute $z^2=x^2+1$ , $2zdz=2xdx$ then $$\int\frac{x^3}{(x^2+1)^3}dx =\int\frac{z^2-1}{z^6}zdz = \int(\frac{1}{z^3}-\frac{1}{z^5})dz$$
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Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2) or 5 = 2(2) + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 3) = 2F(r + 1) + F(r)4) Then F(k + 4) = F(k + 2) + F(k + 3) = 5) 2F(k) + F(k - 1) + 2F(k + 1) + F(k) = 6) 2[F(k) + F(k + 1)] + [F(k - 1) + F(k)] = 7) 2F(k + 2) + F(k + 1)My first question here is how do I know how many values of n to test for? Here they chose two.My next question is how did they get from line 3 to line 4? I understand how the statement is correct but why is this chosen? I also understand that I need to prove it's true for all values of r because if I do that it implies that it is true for k + 1. Is it just to find a relation to F(r + 3) on line 3? If that was the case why not just have F(k + 3) = F(k + 2) + F(k + 1)?My final question about this is how did they get from line 4 to 5?The second equation I want to prove is:F(n + 6) = 4F(n + 3) + F(n) for n ≥ 1I'm able to prove n = 1 and n = 2 is true but I get stuck on going from what would be line 3 - 4 on this problem. As this is my problem for homework the answer is not in the back of the book.Now that I've gotten the help I just want to update this with the proof for my second equation (I haven't gotten the formatting down yet so bear with me):F(n + 6) = 4F(n + 3) + F(n)1) n = 1: F(7) = 4F(4) + F(1) or 13 = 12 + 1, true.2) n = 2: F(8) = 4F(5) + F(2) or 21 = 20 + 1, true.3) Assume for all r, 1 ≤ r ≤ k: F(r + 6) = 4F(r + 3) + F(r)4) Then F(k + 7) = 4F(k + 4) + F(k + 1) =5) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 4) + F(k + 1) =6) F(k + 4) + F(k + 4) + F(k + 4) + F(k + 3) + F(k + 2) F(k + 1) =7) F(k + 4) + F(k + 4) + F(k + 4) +F(k + 3) + F(k + 3) =8) F(k + 5) + F(k + 5) + F(k + 4) =9) F(k + 6) + F(k + 5) =10) F(k + 7)
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The reason for having two initial cases is that the recurrence defining the Fibonacci numbers defines each of them in terms of the two preceding Fibonacci numbers. The proof by induction uses the defining recurrence $F(n)=F(n-1)+F(n-2)$, and you can’t apply it unless you know something about two consecutive Fibonacci numbers.
Note that induction is not necessary: the first result follows directly from the definition of the Fibonacci numbers. Specifically,
$$\begin{align*}
F(n+3)&=\color{brown}{F(n+2)}+F(n+1)\\
&=\color{brown}{F(n+1)+F(n)}+F(n+1)\\
&=2F(n+1)+F(n)\;.
\end{align*}$$
You can use the first result to prove the second; here again no induction is needed. Start with the righthand side:
$$\begin{align*}
4F(n+3)+F(n)&=3F(n+3)+\color{brown}{F(n+3)}+F(n)\\
&=3F(n+3)+\color{brown}{F(n+2)+F(n+1)}+F(n)\\
&=2F(n+3)+\color{blue}{F(n+3)+F(n+2)}+\color{green}{F(n+1)+F(n)}\\
&=2F(n+3)+\color{blue}{F(n+4)}+\color{green}{F(n+2)}\\
&=F(n+4)+F(n+3)+\color{brown}{F(n+3)+F(n+2)}\\
&=F(n+4)+F(n+3)+\color{brown}{F(n+4)}\\
&=2F(n+4)+F(n+3)\;.
\end{align*}$$
Now apply the first result, $F(m+3)=2F(m+1)+F(m)$, with $m=n+4$.
You can also do it by induction, and again you’ll need two base cases to get the induction started. If you want to try this approach, I suggest that you model your work on the induction argument that crash gives.
|
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|
Using complex analysis to evaluate $\int_0^\infty\frac{(\ln x)^3}{1+x^2}d x$ Here is my attempt:
Let $R>1>r$ and $C$ be the closed curve in $\mathbb{C}$ consists of the following pieces:
$$C_1=\{Re^{it}: t\in(0,\pi)\},\quad C_2=[r,R],\quad C_3=\{re^{it}: t\in(0,\pi)\},\quad C_4=[-R,r]$$
all curves are oriented counterclockwise. It can be seen that $C$ is the boundary of the upper half of the annulus centred at $0$ with inner radius $r$ and outer radius $R$. Let $f(z)=\frac{(\text{Log } z)^3}{1+z^2},\quad Arg(z)\in\left(-\frac{\pi}{2},\frac{3\pi}{2}\right)$. Then $\int_Cf(z)d z=\sum_{i=1}^4\int_{C_i}f(z)d z$. Note that $f$ has a singularity $z=i$ inside $C$, thus by Cauchy's integral formula
$$\int_Cf(z)dz=2\pi i\cdot\frac{(\text{Log } i)^3}{i+i}=\pi\left(\ln|i|+\frac{\pi}{2}i\right)^3=-\frac{\pi^4}{8}i$$
Now consider $\int_{C_i}f(z)d z$:
$$\begin{aligned}&\left|\int_{C_1}f(z)d z\right|\leq\int_0^\pi\frac{(|\ln R|+|it|)^3}{|1-R^2e^{2it}|}|Re^{it}|dt\leq\int_0^\pi\frac{R(\ln R+t)^3}{R^2-1}|dt\leq \frac{\pi R(\ln R+\pi)^3}{R^2-1}
\end{aligned}$$
as $RHS\to 0$ as $R\to+\infty$, we have $\lim_{R\to\infty}\int_{C_1}f(z)d z=0$.
$$\int_{C_2}f(z)d z=\int_r^R\frac{(\ln x)^3}{1+x^2}d x\quad\Rightarrow\quad\lim_{r\to 0^+, R\to+\infty}\int_{C_2}f(z)d z=\int_0^\infty\frac{(\ln x)^3}{1+x^2}d x$$
$$\begin{aligned}&\left|\int_{C_3}f(z)d z\right|\leq\int_0^\pi\frac{(|\ln r|+|it|)^3}{|1+r^2e^{2it}|}|re^{it}|d t\leq\int_0^\pi\frac{r(\ln r+t)^3}{1-r^2}|d t\leq \frac{\pi r(\ln r+\pi)^3}{1-r^2}
\end{aligned}$$
as $RHS\to 0$ as $r\to0^+$, we have $\lim_{r\to 0^+}\int_{C_3}f(z)d z=0$
$$\begin{aligned}&\int_{C_4}f(z)d z=\int_{[-R,-r]}\frac{(\text{Log } z)^3}{1+z^2}d z=\int_{[-R,-r]}\frac{(\ln|z| +\pi i)^3}{1+z^2}d z=\int_r^R\frac{(\ln x +\pi i)^3}{1+x^2}d x\\
=&\int_r^R\frac{(\ln x)^3}{1+x^2}dx+3\pi i\int_r^R\frac{(\ln x)^2}{1+x^2}d x-3\pi^2\int_r^R\frac{\ln x}{1+x^2}d x-\pi^3 i\int_r^R\frac{1}{1+x^2}d x\end{aligned}$$
At this point I have to evaluate $\int_0^\infty\frac{(\ln x)^2}{1+x^2}d x$, which can be done by complex analysis again. However, this method is way too long and from my point of view, not the most efficient. Is there a shorter way to do this?
By the way, the answer is 0.
|
Using the contour I used here, we have that
$$
\int_0^{\infty}\frac{\log^3(x)}{1+x^2}dx = \int_{\Gamma}f(z)dz + \int_{-\infty}^{-\epsilon}f(z)dz + \int_{\gamma}f(z)dz + \int_{\epsilon}^{\infty}f(z)dz
\tag{1}
$$
where $f(z) = \frac{\log^3(z)}{z^2 + 1}$, $\Gamma$ is large semi circle, and $\gamma$ is small semi circle. As with the linked post, we will have the branch cut from $(-\infty, 0)$. By the estimation lemma, the first and third integrals of $(1)$ are zero. Now
$$
\log^3(z) = \log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert + i(3\arg(z)\log^2\lvert z\rvert - \arg^3(z))
$$
Additionally, we are taking the real Cauchy principal value of the integral. With that in mind and by the residue theorem, we have
\begin{align}
\int_{-\infty}^0\frac{\log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert}{z^2 + 1} dz + \int_0^{\infty}\frac{\log^3\lvert z\rvert - 3\arg^2(z)\log\lvert z\rvert}{z^2 + 1}dz &= \text{Re}\left(2\pi i\sum\text{Res}\right)\\
2\int_0^{\infty}\frac{\log^3(x)}{x^2 + 1}dz - 3\pi^2\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz & = 0\\
\int_0^{\infty}\frac{\log^3(x)}{x^2 + 1}dz&= 0\tag{2}
\end{align}
where $(2)$ occurs since $\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2 + 1}dz = 0$ and the residue was purely imaginary so the real principal value is zero. Using the same contour for $\frac{\log(z)}{z^2+1}$, we obtain
$$
\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2 + 1}dz = 2\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz + \int_0^{\infty}\frac{i\pi}{z^2+1}dz = \frac{i\pi^2}{2}\tag{3}
$$
Again, we were seeking the real Cauchy principal value so $(3)$ becomes
$$
\int_0^{\infty}\frac{\log\lvert z\rvert}{z^2+1}dz = 0
$$
|
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|
induction exercise - having struggles with $1^3 + 2^3 + ... + (n+1)^3 = [(n(n+1))/2]^2$ I'm trying to solve by induction that $1^3 + 2^3 + ... + (n+1)^3 = [(n(n+1))/2]^2$
However, I have a lot of trouble, and it must be said that I don't do a great deal of mathematics.
I keep getting $(k+1)^3 = (k+1)^2$ every time I try to solve it. Please point out where I am going wrong.
$[ (k(k+1))/2]^2 + (k+1)^3 = [ (k(k+1))/2 + (k+1) ]^2$
$(k^2(k+1)^2)/4 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^2$
Please help.
|
Please point out where I am going wrong.
$[ (k(k+1))/2]^2 + (k+1)^3 = [ (k(k+1))/2 + (k+1) ]^2$
$(k^2(k+1)^2)/4 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^2$
You did this on the RHS:
$$\left(\frac{k(k+1)}2+(k+1)\right)^2 = \frac{k^2(k+1)^2}4 + (k+1)^2.$$
This is not correct. You have $(a+b)^2=a^2+2ab+b^2$, not $(a+b)^2=a^2+b^2$.
So you should get
$$\left(\frac{k(k+1)}2+(k+1)\right)^2 = \frac{k^2(k+1)^2}4 + k(k+1)^2 + (k+1)^2.$$
You may notice that $k(k+1)^2 + (k+1)^2= k(k+1)^2 + 1(k+1)^2 = (k+1)(k+1)^2 = (k+1)^3$. So you get exactly the equality you wanted.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the right approach? Because when i try and work out the algebra i keep on hitting a wall.
\begin{align}
\left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\
&=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\
&=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2}
\end{align}
|
$\begin{aligned}& \prod_{2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln{\prod_{2\le k\le n}\bigg(1-\frac{1}{k^{2}}\bigg)}\bigg] = \exp\bigg[\sum_{2 \le k \le n} \ln{\bigg(\frac{k^2-1}{k^2} \bigg)}\bigg] = \exp(S_{n}). \\& \begin{aligned} & \begin{aligned} S_{n} & = \sum_{2 \le k \le n}\ln(k-1)+ \sum_{2 \le k \le n}\ln(k+1)-2 \sum_{2 \le k \le n}\ln(k) =\sum_{2 \le k+1 \le n}\ln(k)+ \sum_{2 \le k-1 \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) \\& = \sum_{1 \le k \le n-1}\ln(k)+ \sum_{3 \le k \le n+1}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln(1)-\ln(n)-\ln(2)+\ln(n+1)\\& +\sum_{2 \le k \le n}\ln(k)+ \sum_{2 \le k \le n}\ln(k)-2 \sum_{2 \le k \le n}\ln(k) = \ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg) .\end{aligned} \\& \begin{aligned} \therefore \prod_{ 2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln\bigg(\frac{1}{2}\bigg)+\ln\bigg(1+\frac{1}{n}\bigg)\bigg] = \frac{n+1}{2n}. \end{aligned} \end{aligned} \end{aligned} $
|
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|
Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
|
Hint
You have $$\frac{3 x^2+17 x}{x^3+3 x^2-6 x-8}=\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$$ So, write $$f(x)=3x^2+17x-\Big(A(x-2)(x+4)+B(x+1)(x+4)+C(x+1)(x-2)\Big)=0$$ Now $$f(2)=3\times 2^2+17 \times 2-B\times(2+1)\times(2+4)=46-18B=0$$ Repeat for $f(-1)$ and $f(-4)$; you just have one linear equation at the time.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you factorise $x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3$? I'm trying to factorise
$$ x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3 $$
into four linear factors. By plugging it into WolframAlpha I've learned that it's
$$-(x-y)(x-z)(y-z)(x+y+z)$$
My question is: what are the steps involved in factorising the expression? Is there a method I don't know about that I'd have access to with my limited maths?
Really appreciate any help!
|
Here's the way we did this when I was beginning high school: as it's a symmetric function of 3 variables, at some point we have to break the symmetry. All we meed is remarkable identities:
The expression can be rewritten as
\begin{align*}
(x^3-y^3)z&+(z^3-x^3)y+(y^3-z^3)x =(x^3-y^3)z+(z^3-y^3+y^3-x^3)y+(y^3-z^3)x\\
& =(x^3-y^3)(z-y)+(y^3-z^3)(x-y)\\
& =(x-y)(z-y)(x^2+xy+y^2)+(x-y)(y-z)(y^2+yz+z^2)\\
&= (x-y)(y-z)(z^2+yz-x^2-xy)\\
&= (x-y)(y-z)\bigl((z-x)(z+x)+y(z-x)\bigr)\\
&=(x-y)(y-z)(z-x)(x+y+z).
\end{align*}
|
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|
Expressing in terms of symmetric polynomials. How to express $$a^7+b^7+c^7$$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?
|
Hint: Expand $(a+b+c)^7=\sigma_1^7$ with Newton to find the terms you need to subtract from it to get $a^7+b^7+c^7.$ I think it's not too high of a power to do that, since its expansion has $\frac{8\cdot9}{2}=36$ terms, arguably not too many, but after all one can also compute $(a+b+c)^6$ by squaring $(a+b+c)^3$ and then multiply by $a+b+c=\sigma_1$.
For example, for the square we have
$$\require\color(a+b+c)^2=\sigma_1^2=a^2+b^2+c^2+2ab+2ac+2bc =a^2+b^2+c^2+2(ab+bc+ca) \iff \ \color\red{a^2+b^2+c^2=\sigma_1^2-2\sigma_2}$$
and for the cube
$$\require\color (a+b+c)^3=\sigma_1^3=a^3+b^3+c^3+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc=\\ a^3+b^3+c^3+3\left(ab(a+b)+bc(b+c)+ca(c+a)\right)+6abc=\\ a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-9abc+6abc \iff \\ \color\red{a^3+b^3+c^3 = \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3}.$$
|
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|
Describe the image of the set $\{z:|z|<1, Im(z)>0\}$ under the mapping $w =\frac{2z-i}{2+iz}$ Describe the image of the set $\{z:|z|<1, Im(z)>0\}$ under the
mapping $w =\frac{2z-i}{2+iz}$
First I need to find the inverse which is $z=\frac{2w+i}{2-iw}$.
Now let $w=u+iv$, we have
$$z=\frac{2w+i}{2-iw}=\frac{2u+2iv+i}{2+v-iu}$$
From this I get $x= \frac{3u}{(2+v)^2 +u^2}$ and $y=\frac{5v+2u^2 +2v^2+2}{(2+v)^2 +u^2 }$
Since $Im(z)>0$, $5v+2u^2 +2v^2+2>0$ and $|z|<1$ so $ \sqrt{x^2+y^2} <1$.
$$x^2 +y^2<1$$
$$(\frac{3u}{(2+v)^2 +u^2})^2 +(\frac{5v+2u^2 +2v^2+2}{(2+v)^2 +u^2 })^2 <1$$
$$3u^4+3v^4+20u^3+8v^3+6u^2v^2+8u^2v+4v^2-3v-12<0$$
Now I'm stuck, so I tried Mr. Blatter method and got$T(-1)=\frac{-2-i}{2+i}$, $T(0)=1$, $T(1)=\frac{2-i}{2+i}$, $T(i)=i$. So is this telling me that the image is the left side?
|
You have errors in your calculations of $T(-1), \space T(0)$
First verify that you can get $$T(-1)=-\frac{3}{5}-\frac{4}{5}i \quad T(0)=-\frac{i}{2}\quad T(1)=\frac{3}{5}-\frac{4}{5}i\quad T(i)=i$$
Next since you're looking for the mapping of the area between the real axis and the upper half unit disk, you first need to find the mapping of the real axis and the unit disk.
Starting with the unit disk, observe that $T(1) = \frac{3}{5}-\frac{4}{5}i, \space T(i) = i, \space T(-1)=-\frac{3}{5}-\frac{4}{5}i\space$ These are points on the unit circle so the unit circle is mapped to itself. We went counter-clockwise from $1$ to $i$ to $-1$ and got to the points $\frac{3}{5}-\frac{4}{5}i,\space i,\space -\frac{3}{5}-\frac{4}{5}i\space$ also counter-clockwise. So the interior is mapped to itself. Alternatively to this "direction" approach you can take one interior (or exterior) point and check where it's mapped to. We know that $T(0)=-\frac{i}{2}$ so an interior point is mapped to an interior point which means that the interior area is mapped to the interior area.
To check where the real axis is mapped to we look at $-1, 0, 1$ which are mapped to $-\frac{3}{5}-\frac{4}{5}i, \space -\frac{i}{2},\space \frac{3}{5}-\frac{4}{5}i$ respectively. So as we go over the real axis "from left to right" we get points on the circle in clock-wise direction. This means that points to the left of the real axis (the upper half plane) are mapped to points to the left of the circle (outside the circle).
So now you know that your image set is all the points inside the unit circle but outside the other circle.
To find that other circle plug in $z=t \in \mathbb{R}$ and represent the curve $h(t)$ with $x-y$ coordinates, take the derivative $h'(t)$ and check when the $y$ coordinate is $0$ which should be at $t=0, \infty$. Then you get $2$ points on the circle with distance $2r$. You should get the circle $|z+\frac{5}{4}i|=\frac{3}{4}$.
Update
For $z=t\in \mathbb{R}$ we have $w=\frac{2t-i}{2+it} = \frac{3t}{4+t^2}-i\frac{2+2t^2}{4+t^2}$ so the image curve is given by $h(t)=(\frac{3t}{4+t^2},-\frac{2+2t^2}{4+t^2})$
Now you can compute $h'(t)$ and check where the second coordinate is zero which will give you $2$ points where the tangent to the circle is parallel to the real axis meaning these points are at distance $2r$ from each other where $r$ is the radius.
|
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|
How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle. Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.
|
i will take the equilateral triangle base $ABC$ to have sides of length $2\sqrt3$ and the edges $AD= a, Bd = b, CD = c.$ set up the coordinate system so that $$A=(2,0,0),\, B=(-1, \sqrt 3, 0),\, C = (-1,-\sqrt 3, 0), D=(x,y,z) $$ we will derive a formula for $x,y,z$ in terms of $a,b,c.$
on equating the squares of length we have $$(x-2)^2 + y^2 + z^2=a^2,\,(x+1)^2+(y-\sqrt 3)^2 + z^2 = b^2,\, (x+1)^2+(y+\sqrt 3)^2 + z^2 = c^2 $$
from these we can derive $$6x - 2\sqrt 3 y = b^2 - a^2,\, 6x + 2\sqrt 3y = c^2 - a^2.$$
the solutions are $$x = \frac{c^2 + b^2 - 2a^2}{12}, \, y = \frac{c^2 - b^2}{4\sqrt3}, z = \pm\sqrt{a^2 - y^2 - (x-2)^2} . $$
|
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|
How to calculate $\int _0^{\frac{1}{3}} \frac{e^{-x^2}}{\sqrt{1-x^2}}dx$ can someone give me a hint on how to calculate this integral?
$\int _0^{\frac{1}{3}} \frac{e^{-x^2}}{\sqrt{1-x^2}}dx$
Thanks so much!
|
$\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^{\sin^{-1}\frac{1}{3}}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)$
$=\int_0^{\sin^{-1}\frac{1}{3}}e^\frac{\cos2x-1}{2}~dx$
$=e^{-\frac{1}{2}}\int_0^{2\sin^{-1}\frac{1}{3}}e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)$
$=\dfrac{e^{-\frac{1}{2}}}{2}\int_0^{2\sin^{-1}\frac{1}{3}}e^\frac{\cos x}{2}~dx$
Which can express in terms of Incomplete Bessel Functions
|
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|
Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method to use with this type of problem.
|
Hint $\ $ By CRT the square roots mod $5$ and $13$ lift to $4$ square roots mod $65$. There is an obvious root $\,{\rm mod}\ 65\!:\ x^2\equiv -1\equiv 64\,$ if $\,x\equiv \pm8\equiv \pm (2,5)\pmod {5,13}$. The other pair $\pm (-2,5),\,$ arise by multiplying the first pair by $\,(-1,1)\equiv 14,\,$ i.e. $\,\pm 8\cdot 14\equiv \pm18\pmod{65}.\ $ QED
|
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|
About the series $\sum_{n\geq 0}\frac{1}{(2n+1)^2+k}$ and the digamma function Let we provide a closed form for
$$ S_k = \sum_{n\geq 0}\frac{1}{(2n+1)^2+k} $$
for $k>0$ in terms of elementary functions. It is quite easy to check that $S_k$ can be computed in terms of the digamma function $\psi(x)$, but it is also true that:
$$ \int_{0}^{+\infty}\frac{\sin(mx)}{m}e^{-\sqrt{k}\,x}\,dx =\frac{1}{m^2+k}\tag{1} $$
and that, almost everywhere:
$$ \sum_{n\geq 0}\frac{\sin((2n+1) x)}{2n+1} = \frac{\pi}{4}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor}, \tag{2}$$
hence:
$$\begin{eqnarray*} S_k &=& \frac{\pi}{4}\int_{0}^{+\infty}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor} e^{-\sqrt{k}\,x}\,dx = \frac{\pi}{4}\sum_{n\geq 0}(-1)^n \int_{n\pi}^{(n+1)\pi}e^{-\sqrt{k}\,x}\,dx \\&=&\frac{\pi}{4\sqrt{k}}\sum_{n\geq 0}(-1)^n\left(e^{-n\pi\sqrt{k}}-e^{-(n+1)\pi\sqrt{k}}\right)\\&=&\frac{\pi\left(1-e^{-\pi\sqrt{k}}\right) }{4\sqrt{k}}\sum_{n\geq 0}(-1)^n e^{-n\pi\sqrt{k}}=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}.\tag{3} \end{eqnarray*}$$
Does this identity provide something interesting about special values of the digamma function?
|
Your result just confirms already known special values of the digamma function.
One may recall that the digamma function admits the following series representation
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{k=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant. By partial fraction decomposition we have
$$
\begin{align}
\frac{1}{(2n+1)^2+k} &= \frac{i}{2\sqrt{k}}\left(\frac{1}{(2n+1)+i\sqrt{k}}-\frac{1}{(2n+1)-i\sqrt{k}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac{1}{n+\frac{1-i\sqrt{k}}{2}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left[\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac1n\right)-\left(\frac{1}{n+\frac{1-i\sqrt{k}}{2}}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1-i\sqrt{k}}{2}+1\right)-\psi\left(\frac{1+i\sqrt{k}}{2}+1\right)\right)
$$ using
$$
\psi(x+1)=\psi(x)+\frac1x
$$ it gives the case $n=0$ to obtain
$$
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1}{2}-\frac{i\sqrt{k}}{2}\right)-\psi\left(\frac{1}{2}+\frac{i\sqrt{k}}{2}\right)\right)=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}
$$ where we have used the result (6.3.12):
$$
\Im \psi\left(\frac{1}{2}+iy\right)=\frac{\pi}{2}\tanh (\pi y).
$$
|
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|
What's wrong in this integral Where's the mistake in this solution? $$\int \tan^3x\sec^2xdx = \int \frac{\sin^3x}{\cos^5x}dx=\int\frac{\sin x(1-\cos²x)}{\cos^5x}dx=\int\frac{u^2-1}{u^5}du$$$$=\frac{1}{4u^4}-\frac{1}{2u^2}+C=\frac{1}{4\cos^4x}-\frac{1}{2\cos^2x}+C$$ for $ \cos x=u \to du=-\sin xdx$. I also tried doing $$\int \tan^3x\sec^2xdx = \int u^3du=\frac{u^4}{4}+C=\frac{\tan^4x}{4}+C$$ for $ tgx=u \to du=\sec^2xdx$
Please help!
|
\begin{align}
\frac{1}{4\cos^4x}-\frac{1}{2\cos^2x} +\frac{1}{4}
&= \frac{1-2\cos^2 x+\cos^4x}{4\cos^4 x}\\
&= \frac{(\cos^2x-1)^2}{4\cos^4x}\\
&= \frac{(-\sin^2x )^2}{4\cos^4x}\\
&= \frac{\tan^4x}{4}
\end{align}
|
{
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|
Find all solutions of the following linear congruence $2x+3y≡1\pmod 7$ I've never done one with two variables, I understand that $(2, 3, 7) = 1$ and $1|1$ so there are $1*7$ solutions, and $x=0,1,2,3,4,5,6$. I have no idea how to proceed from here, and any help would be greatly appreciated!
|
\begin{align}2x + 3y - 1 = 7k &\Rightarrow 3y = -2x + 1 - 7k = -3(x+2k) + x + 1 - k\\ &\Rightarrow y = -x - 2k + \dfrac{1+x-k}{3}
\end{align}
Since
\begin{align}
y \in \mathbb{Z} &\Rightarrow 3\mid (1+x-k)\\ &\Rightarrow 1+x-k = 3n \\&\Rightarrow x = k-1 +3n\\ &\Rightarrow y = -(k-1+3n) - 2k + n = -3k + 4n + 1$,
\end{align}
with $n,k \in \mathbb{Z}$.
|
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|
Bounds for $\log(1-x)$ I would like to show the following
$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequality on the left, the difference is $\frac{x^2}{2}-\left(\frac{x^3}{3}+\frac{x^4}{4} + \cdots\right)$. How do I show this is nonnegative?
|
$$\frac{d}{dx}\left(\log(1-x)+x+x^2\right) = 1+2x-\frac{1}{1-x}=\frac{x(1-2x)}{1-x} $$
is a non-negative function on $\left[0,\frac{1}{2}\right]$, hence the LHS-inequality follows.
|
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|
Putnam 2005 A1 Solution
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.
(For example, $23=9+8+6.)$
Suppose for $k = \{1, 2, 3... ,n-1\}$ this equality holds.
Let $n$ be an even integer. Since $\frac{n}{2} \le n -1$ if $n \ge 2$ then,
$$\frac{n}{2} = \sum 2^x 3^y \implies n = \sum 2^{x+1} 3^{y}$$
Proof hold for $P(n)$ from $P(n-1)$.
With odds, it gets trickier.
Suppose $n$ is a odd integer, NOT PRIME. So:
$$n = \{1, 9, 15, ...\}$$
Since, $\frac{n}{3} \le n-1 $ if $n \ge 3$ It follows,
$$\frac{n}{3} = \sum 2^x e^y \implies n = \sum 2^x 3^{y+1}$$
Suppose $n$ is PRIME.
$$n = \{3, 5, 7, 11, 13, 17, ... \}$$
There exists $t$ such that:
$$n < 3^{t-1} \implies \log_{3}(n) < t-1 = \sum 2^x 3^{y}$$
By strong induction,
$$\log_{3}(n) = \sum 2^x 3^y$$
Is there any thing I can do now ?I am stuck!
|
Your start is right, but odd numbers that are not prime are not all divisible by $3$.
Rather, you need to first show that if $n$ is divisible by $3$ or $2$, then you can proceed inductively. That is what you have (sort of) done already.
If $n$ is not divisible by $3$ or $2$, pick the largest, $k$, such that $n-2^k$ is positive and divisible by $3$. This will alway (why?) be either $k=\lfloor\log_2 n\rfloor$ or $k=\lfloor \log_2 n\rfloor-1$, which means that $4\cdot 2^k > n$.
Then proceed by induction using $\frac{n-2^k}{3}$. Since $\frac{n-2^k}{3}<2^k$, none of the terms are multiples of $2^k$. Multiply those terms by $3$ and then include $2^k$.
Lemma: Prove that if $n,m$ are both not divisible by $3$, then one of $n-m$ or $n-2m$ is divisible by $3$.
|
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|
$\sum_{n=1}^\infty\frac{\cos({2nt})}{n2^n} =$? for any $t\in\Bbb R$ Given that $\sum\limits_{n=1}^\infty\dfrac{z^{2n}}{n2^n} = -\log\left(1 - \frac{z^2}{2}\right)$ ,
could you calculate the sum in the title for every real $t$ ?
|
Since $$\cos(2nt) = \frac{e^{2nit} + e^{-2nit}}{2}$$ we have
\begin{align}\sum_{n = 1}^\infty \frac{\cos(2nt)}{n2^n} &= \frac{1}{2}\left(\sum_{n = 1}^\infty \frac{(e^{it})^{2n}}{n2^n} + \frac{(e^{-it})^{2n}}{n2^n}\right)\\
&= -\frac{1}{2}\left(\log\left(1 - \frac{e^{2it}}{2}\right) + \log\left(1 - \frac{e^{-2it}}{2}\right)\right)\\
&= -\frac{1}{2}\log\left|1 - \frac{e^{2it}}{2}\right|^2\\
&= -\frac{1}{2}\log\left(\frac{5}{4} - \cos(2t)\right).\\
\end{align}
|
{
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|
Is it generally true that $\arcsin \theta + \arccos \theta = \frac{\pi}{2}$?
Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)
How can one verify (1) when it is not generally true? We can rewrite (1) as
Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.
What about $\sin \frac{2\pi}{3}$ and $\cos \frac{\pi}{6}$? $\sin \frac{2\pi}{3} = \cos \frac{\pi}{6}$, but $\frac{2\pi}{3}+\frac{\pi}{6} \neq \frac{\pi}{2}$.We may try to verify (1) as:
Let $\arcsin \theta=u$ and $\arccos \theta=v$, then $\sin u=\cos v = \theta.$ We can write
$$\sin(u+v)=\sin u \cos v+ \cos u \sin v,$$
which becomes
$$\sin(u+v)=\theta^2+ \cos u \sin v.$$
Now, $\cos u=\pm \sqrt{1-\theta^2}$, and $\sin v=\pm \sqrt{1-\theta^2}$. For $\cos u= \sqrt{1-\theta^2}$ and $\sin v= \sqrt{1-\theta^2}$, we have
$$\sin(u+v)= \theta^2+(1-\theta^2)=1,$$
$$\therefore u+v=\arcsin \theta + \arccos \theta = \arcsin(1)=\frac{\pi}{2}.$$ But $\arcsin(1)$ is also equal to $\frac{\pi}{2}+2n\pi, n \in \mathbb{Z}.$ What about that? And what about the cases when $\cos u = -\sqrt{1-\theta^2}$ and $\sin u = \sqrt{1-\theta^2}$ and the other way? Then $\sin(u+v)=2\theta^2-1=1\ \mbox{iff}\ \theta =\pm 1.$
I don't know where I am making a mistake, what assumptions are wrong, or where my reasoning goes wrong. Please help.
|
here is a geometric argument to show that $$\sin^{-1}(b) + \cos^{-1}b = \pi/2 \tag 1$$ using the unit circle. i will make may argument for the case $0 \le b \le 1.$ the same should go through for $-1 \le b \le 0.$
we will interpret $\sin^{-1} b$ as the arc length $t$ between $-\pi/2$ and $\pi/2$ such that $\sin t$ is the $y$-coordinate of the terminal point $(a,b)$ on the unit circle. and interpret $\cos^{-1} a$ as the arc length $t$ between $0$ and $\pi$ such that $\cos t$ is the $x$-coordinate of the terminal point $(a,b)$ on the unit circle.
pick a point $P = (a,b), a \ge 0, b \ge 0$ on the unit circle. reflect this point $P$ on the line $y = x$ is $Q = (b, a)$ let $A = (1,0), B = (0,1).$ verify that the $$arc \,AP = arc\, BQ, arc\, AQ + arc \, BQ = \pi/2. \tag 2$$
now observe that $$\sin^{-1}b = arc \, AP, \cos^{-1}(b)= arc\, AQ$$ together with $(2)$ establishes $(1).$
|
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|
Application of Taylor's theorem Use Taylor's theorem to prove the estimate
$${|\sin(x)-x+x^3/6|} \le C|x|^5$$
for a suitable constant C. Hence prove that $$\lim_{x \to 0} \frac{\sin(x)-x}{x^3}$$ exists and determine its value.
I used Taylor's theorem with remainder in Lagrange form and got $$\left |\sin(x)-x+\frac{x^3}{6} \right| \le \frac{1}{5!}\left|x\right |^5$$
and then I am not sure what I should do next in order to get the inequality into the required form .
I am also not sure how I can use the inequality to find the limit as $x$ approaches zero.
Any help is appreciated, thanks in advance.
|
do integration by parts quite a few times staring with
$$\begin{align} \sin x &= \int_0^x d \sin t = \int_0^x \cos t\, d(t-x)\\
&= (t-x)\cos t\big|_0^x + \int_0^x (t-x) \sin t \, dt\\
&=x + \frac 12 (t-x)^2 \sin t\big|_0^x - \frac 12 \int_0^x (t-x)^2 \cos t\, dt\\
&=x - \frac 16 (t-x)^3 \cos t\big|_0^x - \frac 16\int_0^x (t-x)^3 \sin t\, dt\\
&=x - \frac 16 x^3- \frac 1{24}\sin t (t-x)^4\big|_0^x +\frac 1{24}\int_0^x (t-x)^4 \cos t\, dt\\
&=x - \frac 16 x^3 +\frac 1{24}\int_0^x (t-x)^4 \cos t\, dt\\
\end{align}$$
we can use the mean value theorem for integrals to find
$$\int_0^x (t-x)^4 \cos t\, dt = \cos c \int_0^x (t-x)^4 \,dt = \frac 15 x^5 \cos c$$
putting all these together we get $$\big|\sin x - x + \frac 16 x^3\big| \le \frac{1}{120} |x|^5$$
|
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|
Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?
Theorem. $\forall x\geq8, x$ can be represented by $5a + 3b$ where $a,b \in \mathbb{N}$.
Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$
Inductive step:
$n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\
b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\
c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\
\\
S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$
Basis stays true, because $8,9,10 \in S$
Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.
If $x \in a_n$ then $x+1 \in b_x$,
If $x \in b_x$ then $x+1 \in c_x$,
If $x \in c_x$ then $x+1 \in a_x$.
I can't prove that but it's obvious. What do you think about this?
|
I'm attempting to answer the question intuitively:
Every third number is a multiple of 3, and can definitely be written as a sum of 3s and 5s:
$3x$ + $5$ x $0$.
Now we need to deal with numbers $y$ for which $y mod 3 = 1$ and $y mod 3 = 2$
For $y mod 3 = 2$:
We know that $5 mod 3 = 2$, so 5 x 1 + 3$x$ should give us all numbers $y$ where $y mod 3 =2$
For $y mod 3 = 1$:
We know that $5$ x $2 mod 3 = 1$, so 5 x 2 + 3$x$ should give us all numbers $y$ where $y mod 3 =1$
The smallest case in which we can use this proof is 5 x 2 + 3 x 0 = 10
We can show it is possible for 8 and 9 as 5+3 =8 and 3 x 3 =9
Hence, we can show that is possible for all numbers greater than or equal to 8.
This is also proves that the maximum multiplier we would need for 5 is 2.
|
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|
Integrate $\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$
$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$
What I did was trig substitution:
$$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$
Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}}^{\frac 13 \sec\frac 23} \frac{81}{\sec^4 \theta} \, d\theta$$
But I got stuck here. What can I do next?
|
Substitute $y=3x$
$$I= \int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{1}{x^5\sqrt{9x^2-1}}dx = 81\int^2_{\sqrt2} \frac{1}{y^5\sqrt{y^2-1}}dy
$$
and then apply the reduction formula
$$\int \frac{1}{y^n\sqrt{y^2-1}}dy=I_n=\frac{\sqrt{y^2-1}}{(n-1)y^{n-1}}+\frac{n-2}{n-1}I_{n-2}
$$
to obtain
$$I=81\left( \frac{7\sqrt3-16}{64} +\frac38 \int^2_{\sqrt2} \frac{1}{y\sqrt{y^2-1}}dy\right)=\frac{81}{64}(7\sqrt3-16+2\pi)
$$
|
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|
How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$ How can I simplify as much as possible:
$$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$$
Also, first post here, looking forward to sticking around!
|
You could multiply by the conjugate of the denominator, then cancel.
\begin{align*}
\frac{a^2 + ab + b^2}{a + \sqrt{ab} + b} & = \frac{a^2 + ab + b^2}{a + b + \sqrt{ab}} \cdot \frac{a + b - \sqrt{ab}}{a + b - \sqrt{ab}}\\
& = \frac{(a^2 + ab + b^2)(a + b - \sqrt{ab})}{(a + b)^2 - ab}\\
& = \frac{(a^2 + ab + b^2)(a + b - \sqrt{ab})}{a^2 + 2ab + b^2 - ab}\\
& = \frac{(a^2 + ab + b^2)(a + b - \sqrt{ab})}{a^2 + ab + b^2}\\
& = a + b - \sqrt{ab}
\end{align*}
|
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|
Finding the sum of $\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}$
Find the sum of the series and for which values of $x$ does it converge:
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n} $$
My attempt:
$$\begin{align}&S_n=(x+1)^2+(x+1)^3/3+...+(x+1)^{n+1}/3^{n} \\
-(x+1)^2&S_n=-(x+1)^3/3-...-(x+1)^{n+2}/3^{n} \\
&S_n=\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2} \end{align}$$
And for $-4<x<2$ we have:
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2}=\frac{(x+1)^2}{1-(x+1)^2}$$
Which is the sum of the series for $-4<x<2$.
The sum looks a bit odd, is that alright?
Note: no integrals or Taylor or Zeta.
|
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}=(x+1)^2\sum_{n=0}^{\infty}\Big(\frac {x+1}{3}\Big)^n=(x+1)^2\frac{1}{1-\frac{x+1}{3}}=\frac{3(x+1)^2}{2-x}$$
|
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|
How to prove that $\frac{\log(1+a/x)}{\log(1+1/x)}$ is increasing, for $a\geq 1$? I need to prove that the sequence
$$
\frac{\log\big(1+\frac{a}{k}\big)}{\log\big(1+\frac{1}{k}\big)}
$$
is increasing for any $a\geq 1$, so that I thought of defining
$$
f(x)=\frac{\log\big(1+\frac{a}{x}\big)}{\log\big(1+\frac{1}{x}\big)}
$$
and prove that $f'(x)\geq 0$ for all $x\geq 1$. In view of some graphs I plotted, this function seems to be increasing. Moreover, it is easily seen from the expression that its limit will be precisely $a$. However, when trying to prove that $f'\geq 0$, I arrive to
$$
\frac{\log\big(1+\frac{a}{x}\big)}{1+\frac{1}{x}}-\frac{a\log\big(1+\frac{1}{x}\big)}{1+\frac{a}{x}} \overbrace{\geq}^?0.
$$
At that point I do not know how to proceed, so any hint would be kindly appreciated.
|
For later use, consider the function $\phi(x) = x \log x$, for which $\phi'(x) = 1 + \log x$ and $\phi''(x) = \frac{1}{x}$; in particular, $\phi$ is convex on $[1, \infty)$.
If
$$
f(x) = \frac{\log\bigl(1 + \frac{a}{x}\bigr)}{\log\bigl(1 + \frac{1}{x}\bigr)},
$$
then
$$
f'(x) = \frac{(x + a) \log\bigl(1 + \frac{a}{x}\bigr) - a(x + 1) \log\bigl(1 + \frac{1}{x}\bigr)}{(x + a)(x + 1)x \log^{2}\bigl(1 + \frac{1}{x}\bigr)}.
$$
The denominator is positive for $x \geq 1$, and the numerator is equal to
\begin{align*}
(x + a) &\log\bigl(1 + \tfrac{a}{x}\bigr) - a(x + 1) \log\bigl(1 + \tfrac{1}{x}\bigr) \\
&= (x + a) \bigl[\log(x + a) - \log x\bigr] - a(x + 1) \bigl[\log(x + 1) - \log x\bigr] \\
&= (x + a) \log(x + a) - a(x + 1) \log(x + 1) - \bigl[(x + a) - a(x + 1)\bigr]\log x \\
&= (x + a) \log(x + a) - a(x + 1) \log(x + 1) + (a - 1) x \log x \\
&= \phi(x + a) - a \phi(x + 1) + (a - 1) \phi(x).
\tag{1}
\end{align*}
Since $x + 1 = \frac{a - 1}{a} x + \frac{1}{a} (x + a)$, convexity of $\phi$ implies
$$
\phi(x + 1) \leq \tfrac{a - 1}{a} \phi(x) + \tfrac{1}{a} \phi(x + a),
$$
or
$$
a \phi(x + 1) \leq \phi(x + a) + (a - 1) \phi(x).
$$
That is, (1) is non-negative, so $f$ is non-decreasing (and strictly increasing if $a > 1$, since $\phi$ is strictly convex).
|
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|
The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
|
$$ \lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )=\lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )\frac{\sqrt{x^2 +x} + \sqrt{x^2 +1}}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=\lim _{x \to \infty } \frac{x-1}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=...=1/2$$
(if you need more I could finish it)
|
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|
When a function contains a sequence, and how to find the function's limit? Suppose $\lim \limits_{n \to \infty} a_n = 0$. Find the limit
$$\lim \limits_{n \to \infty} \left(1+a_n \frac{x}{n}\right)^n$$
It's kind intuitive that the answer is 1, but clearly I can't just say that the limits equal $\lim \limits_{n \to \infty} 1^n = 1$.
I feel like I should let $y =(1+a_n \frac{x}{n})^n $. Then take $log$ so that $log(y) = nlog(1+a_n \frac{x}{n})$ But L'hopital doesn't apply here either
|
Hint: Supose $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=0$. Note that $\lim_{n\to \infty} x\cdot a_n=0$,
$$
\left(
1+a_n\frac{x}{n}
\right)^{n}
=
\left(
1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}}
\right)^n
=
\left[
\left(
1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}}
\right)^{n\cdot\frac{1}{\color{red}{ x\cdot a_n}}}
\right]^{\color{red}{ x\cdot a_n}}
$$
and $
\lim_{n\to \infty}
\left(
1+\frac{1}{\color{blue}{n\cdot\frac{1}{ x\cdot a_n}}}
\right)^{\color{blue}{n\cdot\frac{1}{ x\cdot a_n}}}=e
$.
If $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=0$ then
$$
\lim_{n\to \infty}
\left[
\left(
1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}}
\right)^{n\cdot\frac{1}{\color{red}{ x\cdot a_n}}}
\right]^{\color{red}{ x\cdot a_n}}
=\ldots =e^0=1
$$
The case $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=L\in\mathbb{R}$ is trivial.
|
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|
Continued Fraction for Root 5 How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.
|
We can use the
$$x^2-5=0$$
$$x^2=5$$
$$x^2+x=5+x$$
$$x(x+1)=5+x$$
$$x=\frac{x+5}{x+1}$$
or
$$x=1+\frac{4}{1+x}$$
$$x=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
$$\sqrt{5}=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
|
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|
Find conditions on positive integers so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational
Find conditions on positive integers
$a, b, c$
so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational.
My solution:
if $ab$ is not the square of an integer,
then the expression is irrational.
I find it interesting
that $c$ does not come into this
at all.
My solution is modeled
(i.e., copied with modifications)
from dexter04's solution
to Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
.
Suppose $\sqrt{a}+\sqrt{b}+\sqrt{c} = r$
where $r$ is rational.
Then,
$(\sqrt{a}+\sqrt{b})^2
= (r-\sqrt{c})^2
\implies a+b+2\sqrt{ab}
= c+r^2-2r\sqrt{c}$.
So, $a+b-c-r^2+2\sqrt{ab} =-2r\sqrt{c}$.
Let $a+b-c-r^2 = k$,
which will be a rational number.
So,
$(k+2\sqrt{ab})^2 = k^2+ 4ab+4k\sqrt{ab} = 4cr^2$
or
$4k\sqrt{ab} = 4cr^2-k^2- 4ab$.
If $ab$ is not a square
of an integer,
then the LHS is irrational
while the RHS is rational.
Hence, we have a contradiction.
|
The answer is that if any of $a, b, c$ is not a square of an integer, then
$\sqrt{a} +\sqrt{b} +\sqrt{c}$ must be irrational.
The proof of the general case is not very easy. The paper
Square roots have no unexpected linear relationships by Qiaochu Yuan at
https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/
explains this non-trivial theorem.
|
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|
A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
|
$2ab \leq a^2+b^2$ so $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$. Now choose $a = \sqrt{x},b = \sqrt{y}$.
|
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|
Find rational points on $x^2 + y^2 = 3$ and on $x^2 + y^2 = 17$ $(a)$ Find all rational points on the circle $x^2 + y^2 = 3$, if there are any. If there is none, prove so.
$(b)$ Find all rational points on the circle $x^2 + y^2 = 17$, if there are any. If there is none, prove so.
I'm not sure how proceed with finding a general formula (if there is one)
I know that for $(a)$ there is no rational points but I don't know how to explain that there are none.
whereas for $(b)$ there are such points, $(1,4)$ for example. I think that we can find the intersection between the line $y=m(x-1)+4$ and $x^2 + y^2 = 17$
Any help is appreciated!
|
As shown in this answer, $n$ can be written as the sum of two squares if and only if, in the prime factorization of $n$, each prime that is $\equiv3\pmod4$ appears with even exponent.
If $x^z+y^2=3z^2$, then $3$ appears with odd exponent. Thus, there are no rational solutions of
$$
\left(\frac xz\right)^2+\left(\frac yz\right)^2=3\tag{1}
$$
As noted, $17=4^2+1^2$. Suppose that
$$
\left(\frac xz\right)^2+\left(\frac yz\right)^2=17\tag{2}
$$
then
$$
\begin{align}
1
&=\frac{x^2+y^2}{17z^2}\\
&=\frac{x+iy}{z(4+i)}\frac{x-iy}{z(4-i)}\tag{3}
\end{align}
$$
which means that
$$
\begin{align}
\frac{x+iy}{z(4+i)}\tag{4}
&=u+iv
\end{align}
$$
where $u,v\in\mathbb{Q}$ so that $u^2+v^2=1$.
Thus,
$$
\frac xz+i\,\frac yz=(4+i)\left(\frac ac+i\,\frac bc\right)\tag{5}
$$
where $a^2+b^2=c^2$ is a Pythagorean triple, all of which can be generated using the formula derived in this answer:
$$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\end{align}\tag{6}
$$
Using $(5)$ and $(6)$, we can compute all rational solutions of $(2)$.
Example
Using the Pythagorean triple $(3,4,5)$, we get
$$
\left(\frac35+i\,\frac45\right)(4+i)=\frac85+i\,\frac{19}5
$$
and
$$
\left(\frac35-i\,\frac45\right)(4+i)=\frac{16}5-i\,\frac{13}5
$$
Thus, we get
$$
\left(\frac85\right)^2+\left(\frac{19}5\right)^2=17
$$
and
$$
\left(\frac{13}5\right)^2+\left(\frac{16}5\right)^2=17
$$
|
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|
Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6
|
$$3k^2 + 3k + 12=3(k^2 + k +4)= 3(k(k+1)+4)$$
Can you see why $k(k+1)$ and $4$ are each divisible by $2$?
At least one of $k, k+1$ is even, as is $4$, hence $2$ divides $(k(k+1)+4)$, and with three as a factor of $\color{blue}{3}(k(k+1)+ 4)$, we have $$2\cdot 3 = 6\mid (3k^2 + 3k + 12).$$
|
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|
Differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.
*
*Show that the equation has $0$ as its singular regular point .
*Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$
*Show that if $a=n$, where $n \in \mathbb{N}$ then there is a polynomial solution of degree $n$.
*Let $L_n$ the polynomial $L_n(x)=e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x})$ (show that it is a polynomial), $n=1,2,3, \dots$. Show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$.
That's what I have tried:
*
*For $x \neq 0$ the differential equation can be written as: $y''+ \frac{1-x}{x}y'+ \frac{a}{x}y=0$.
$p(x)= \frac{1-x}{x}, q(x)= \frac{a}{x}$
The functions $x \cdot p(x)= 1-x, \ x^2q(x)=ax$ can be written as power series in a region of $0$.
Thus, $0$ is a singular regular point.
*We suppose that there is a solution of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n= \sum_{n=0}^{\infty} a_n x^{n+m}$.
Then $y'(x)= \sum_{n=0}^{\infty} a_n(n+m) x^{n+m-1} \Rightarrow -xy'(x)= \sum_{n=0}^{\infty} -a_n(n+m) x^{n+m}$ and $y''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-2} \Rightarrow xy''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-1}$
So we have
$$\sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m-1}+ \sum_{n=0}^{\infty} a_n (n+m) x^{n+m-1} + \sum_{n=1}^{\infty} -a_{n-1} (n+m-1) x^{n+m-1}+ \sum_{n=1}^{\infty} a a_{n-1} x^{n+m-1}=0 \\ \Rightarrow a_0 m (m-1) x^{m-1}+ a_0 m x^{m-1}+ \sum_{n=1}^{\infty} \left[ a_n (n+m) (n+m-1)+ a_n(n+m)-a_{n-1}(n+m-1)+ a a_{n-1} \right] x^{n+m-1}=0$$
It has to hold:
$$a_0 m^2=0 \overset{a_0 \neq 0}{ \Rightarrow } m=0$$
$$a_n (n+m) (n+m-1)+ a_n (n+m)- a_{n-1} (n+m-1)+ a a_{n-1}=0$$
For $m=0$: $a_{n} n (n-1)+ a_n n-a_{n-1} (n-1)+ a a_{n-1}=0 \Rightarrow a_n n^2+ a_{n-1} (a-n+1)=0 \Rightarrow a_n=- \frac{a_{n-1}(a-n+1)}{n^2}$
For $n=1: \ a_1=-aa_0$
For $n=2: \ a_2= \frac{aa_0(a-1)}{2^2} $
For $n=3: \ a_3=- \frac{aa_0(a-1) (a-2)}{2^2 3^2} $
For $n=4: \ a_4= \frac{aa_0(a-1)(a-2)(a-3)}{2^2 3^2 4^2} $
For $n=5: \ a_5= -\frac{aa_0(a-1)(a-2)(a-3)(a-4)}{2^2 3^2 4^2 5^2} $
We see that $a_n=(-1)^n a_0 \frac{\prod_{i=0}^{n-1} (a-i)}{\prod_{i=2}^n i^2}$
$$\frac{a_{n+1}}{a_n}=(-1) \frac{a-n}{(n+1)^2} \to 0$$
So the series $\sum_{n=0}^{\infty} a_n x^n$ converges and its radius of convergence is equal to $+\infty$.
Is it right so far? Could you give me a hint what we could do in order to answer the other two questions?
EDIT:
Do we differentiate the Leguerre polynomial as follows?
$$$$
$$\frac{d}{dx} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+\frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})$$
$$\frac{d^2}{dx^2} L_n(x)=e^x \frac{d^n}{dx^n} (x^n e^{-x})+ e^{x} \frac{d^{n+1}}{dx^{n+1}}(x^n e^{-x})+\frac{d^{n+2}}{dx^{n+2}}(x^n e^{-x})$$
EDIT: Substituting the above, we cannot show that $L_n$ satisfies the equation Laguerre if $a=n$.
Do you maybe know how else we could show this?
|
You have already done that third part, notice that if $a = n \in \mathbb{N}$ then $a_{i} = 0$ for all $i \ge n+1$. So you have a polynomial of degree $n$ as a solution.
For the last part, just some ideas - try differentiating the Leguerre polynomial and substituting it into the equation. Or, conversely, try to expand the derivative to show the coefficients of the power series are your $a_i$s?
|
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|
If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to find out relation but that will be long and tedious process , I want shorter approach to this problem . Btw I was given following options for this question
A) $a^2+c^2+ab=0$
B) $a^2-c^2+ab=0$
C) $a^2-c^2-ab=0$
D) $ap^2+bp+c=0$
Maybe we can relate something by looking at options?
|
You should have $(x^2+px+1)(\lambda x + D) = \lambda x^3 + (p\lambda + D) x^2 + (pD + \lambda) x + D$. thus $D = c$ and $\lambda = a$, and you need
$p \lambda + D = ap + c = 0$ and $pD + \lambda = cp + a = b$. Eliminating $p$ from these two gives you $0 = c (ap + c) - a(cp + a - b) = \ldots$.
|
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|
the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$
or, as another example if a = 0.01 .$$\frac{n (n + 0.01) (2 n + 0.01) }{6 \cdot 0.01} = 0.01^2 + 0.02^2 + \cdots + n^2 .$$
Now if we follow the same logic, I suppose if the difference between the closest numbers becomes smallest possible (a = 0.0...1), we will obtain
$$ \frac{n (n + 0.0..1) (2 n + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \ (0.0..1 ^ 2 + 0.0..2 ^ 2 .. + n ^ 2) * {0.0..1}$$ If we take that, a = 0.0...1 and, b = $\frac {n}{0.0...1}$ and we rearrange the formula we will get this:
$$\frac{n ^ 3}{3} = 1^2a^3 + 2^2a^3 + \dots+b^2 a^3$$
following the same logic we prove that the formula holds for each degree (it is easy to check), so that we could generalize, If we take that degree call m, we will get
$$\frac{n ^ m}{m} = 1^{m-1}a^m + 2^{m-1}a^m + \dots+b^{m-1} a^m$$
so the question is this expression can be interpreted as a Riemann sum, and why?
|
Indeed, if $\delta=\frac 1m$ then
$$ \frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2=\frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.$$
However, there is no such thing as "smalles possible" and from that point on you get some factros wrong.
What you really get in the limit as $m\to\infty$ is that
$$ \frac{n^3}{3}=\int_0^nx^2\,\mathrm dx.$$
This is because the last expression can be interpreted as a Riemann sum.
|
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|
Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
|
HINT:
$$[(x+1)^2-1]+(x+2)^3+(x+3)^4-1^4$$
$$=x(x+2)+(x+2)^3+[\{(x+3)^2-1\}\{(x+3)^2+1\}]$$
$$=x(x+2)+(x+2)^3+[(x+4)(x+2)\{(x+3)^2+1\}]$$
and
$$[(x+1)^2-3^2]+(x+2)^3+2^3+(x+3)^4-1^4$$
$$=(x+4)(x-2)+\{(x+2)+2\}[(x+2)^2-2(x+2)+2^2]+[(x+4)(x+2)\{(x+3)^2+1\}]$$
|
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|
Find all points on a surface which have a tangent plane parallel to given plane - is my method correct? The question given is to find all points on the surface given by $x^3 - y^3 - 2xy - z = 0$ which have a tangent plane which is parallel to $6x - 6y - z = 0$.
So, I found the two gradient vectors (which we defined as $f_{x}e_{1} + f_{y}e_{2} + f_{z}e_{3}$) and said that one must be a scalar multiple of the other in order to be parallel. Then, as $f_{x} = 6k$ and $f_{y} = -6k$, I was able to set $f_{x} = -f_{y}$ and by cancelling I solved to get $y = x - \frac{2}{3}$ and worked out that all points of the form $(x, x - \frac{2}{3}, z)$ appear to satisfy the conditions given. Then I substituted this into the surface equation to get a value for $z$ which simplified nicely to $\frac{8}{27}$. So, all points of the form $(x,x - \frac{2}{3}, \frac{8}{27})$ is the final answer I got. I have tried a couple of values for $x$ and the gradient vector does work out to be a scalar multiple of the plane's gradient.
I would really appreciate if somebody could quickly work out the answer and verify if I have found the correct answer, or tell me if I am doing it wrong?
Thanks! Helen
|
Given the surface
$$z = f(x,y) = {x^3} - {y^3} - 2xy$$
we have a parametrization:
$$\phi (x,y) = (x,y,{x^3} - {y^3} - 2xy)$$
with partials:
$$\begin{gathered}
{\partial _x}\phi = \left( {1,0,3{x^2} - 2y} \right) \hfill \\
{\partial _y}\phi = \left( {0,1, - 3{y^2} - 2x} \right) \hfill \\
\end{gathered} $$
and an normal vector:
$${\partial _x}\phi \times {\partial _y}\phi = \left( {2y - 3{x^2},2x + 3{y^2},1} \right)$$
Normal vector from the plane and normal vector must be parallel,
so we may set:
$$\left( {2y - 3{x^2},2x + 3{y^2},1} \right) = \left( { - 6,6,1} \right)$$
This leads to the system:
$$\begin{gathered}
2y - 3{x^2} = - 6 \hfill \\
2x + 3{y^2} = 6 \hfill \\
\end{gathered} $$
with four solutions:
$$\begin{gathered}
(x,y) = \left( {\frac{1}{3} - \sqrt {\frac{5}{3}} , - \frac{1}{3} - \sqrt {\frac{5}{3}} } \right) \hfill \\
(x,y) = \left( {\frac{1}{3} + \sqrt {\frac{5}{3}} , - \frac{1}{3} + \sqrt {\frac{5}{3}} } \right) \hfill \\
(x,y) = \left( {\frac{1}{3}\left( { - 1 - \sqrt {19} } \right),\frac{1}{3}\left( {1 + \sqrt {19} } \right)} \right) \hfill \\
(x,y) = \left( {\frac{1}{3}\left( { - 1 + \sqrt {19} } \right),\frac{1}{3}\left( {1 - \sqrt {19} } \right)} \right) \hfill \\
\end{gathered}$$
|
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|
Integral $\int\frac{dx}{(x^3-1)^2}$ Please help. I do not know what to do. You can just show the direction where to go and I continue. Here it is: $$\int\frac{dx}{(x^3-1)^2}$$
|
Let $$\displaystyle I = \int\frac{1}{(x^3-1)^2}dx = \frac{1}{3}\int\frac{1}{x^2}\cdot \frac{3x^2}{(x^3-1)^2}dx$$
Now Using Integration by parts, we get
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}-\frac{2}{3}\int \left[\frac{1}{x^3\cdot (x^3-1)}\right]dx$$
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}+\frac{2}{3}\int \left[\frac{1}{x^3}-\frac{1}{(x^3-1)}\right]dx$$
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}+\frac{2}{3}\int x^{-3}dx-\frac{2}{3}\int\frac{1}{(x-1)(x^2+x+1)}dx$$
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}-\frac{1}{3x^2}-\frac{2}{3}\int\frac{1}{(x-1)(x^2+x+1)}dx$$
Now Using partial fraction descomposition
$$\displaystyle \frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x-1}+\frac{Bx+c}{x^2+x+1}$$
Now camparing Coefficients, we get
$$\displaystyle A=\frac{1}{3}\;, B=-\frac{1}{3}\;,C=-\frac{2}{3}$$
So $$\displaystyle \int\frac{1}{(x-1)(x^2+x+1)} = \frac{1}{3}\int\frac{1}{x-1}-\frac{1}{3}\int\frac{x+2}{x^2+x+1}dx$$
$$\displaystyle =\frac{1}{3}\int\frac{1}{x-1}-\frac{1}{6}\int\frac{2x+1}{x^2+x+1}-\frac{1}{2}\int\frac{1}{x^2+x+1}dx$$
$$\displaystyle = \frac{1}{3}\ln |x-1|-\frac{1}{6}\ln|x^2+x+1|-\frac{1}{2}\int\frac{1}{x^2+x+1}dx$$
|
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta$, and I cannot see a way to obtain $\sqrt{x^2+y^2}$ in the above inequality to complete the proof. Am I mistaken in my process? Thank you.
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We have:
$$0\leq\frac{x^2y^2}{x^2+y^2}\leq \frac{x^2y^2}{y^2}=x^2$$
Since $f(x,y)$ is between $0$ and $x^2\rightarrow 0$, the function is arbitrarily close to zero, when the distance between $(x,y)$ and $(0,0)$ is sufficiently small. It follows that the limit of $f$ is $0$, just to limit definition.
|
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How many $5$ digit natural numbers have at least one $5$ or one $7$? There are $9 \cdot 10^4=90,000$ possible $5$ digit numbers.
Numbers with no $5$ is $8\cdot 9\cdot 9\cdot 9\cdot 9 = 52,488$.
Numbers with no $7$ is $52,488$.
Numbers with at least one $5$ is $90,000 - 52,488$
Similarly numbers with at least one $7$ is $90,000-52,488$
So answer is $2(37,512)=75,024$
But this answer is wrong. Please provide correct solution..
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Hint: Count the number of five digit-numbers that contain NO $5$'s and no $7$'s; that gives us $7\cdot 8^4$ such five-digit numbers.
The remainder ($90000 - 7\cdot 8^4$) will be the number of five digit numbers with at least one $5$ OR one $7$.
|
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Proof that $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$ $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$
my work:
I assumed $m = da$ , $n = db$ for $a,b \in \mathbb{Z}$.
Now, $2^m - 1$ = $2^{da} - 1$ = $(2^d)^a - 1$ = $x^a - 1$ where $x = 2^d$.
similarly
$2^n - 1$ = $x^b - 1$
Now,
using $x^a - 1 = (x - 1)(x^{a-1} + x^{a - 2} + .... + x + 1)$ and $x^b - 1 = (x - 1)(x^{b - 1} + x^{b - 2} + ..... + x + 1)$
clearly $(x - 1)$ is a factor common in both but how to show that
$(x^{a-1} + x^{a - 2} + .... + x + 1)$ and $(x^{b-1} + x^{b - 2} + .... + x + 1)$ can't have a common factor
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One way to deal with your particular step is to show that $\gcd(x^{a} -1, x^{b} - 1) = x - 1$, when $\gcd(a, b) = 1$. (Actually, a slight variation on the same argument would yield a proof of your original statement, but still.)
Assume $a \ge b \ge 0$, and proceed by induction on $b$. The case $b = 0$ being obvious, assume $b > 0$, divide $a$ by $b$ obtaining $a = q b + r$, with $0 \le r < b$, and note that
$$
x^{a} - 1 = x^{b q + r} - 1 = x^{b q} x^{r} - 1 \equiv x^{r} - 1 \pmod{x^{b} - 1}.
$$
Therefore $\gcd(x^{a} -1, x^{b} - 1) = \gcd(x^{b} -1, x^{r} - 1)$, which is $x - 1$ by the inductive hypothesis, as $1 = \gcd(a, b) = \gcd(b, r)$.
|
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$x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that $\sin \tfrac{2\pi x_1}{3}+\sin \tfrac{2\pi x_3}{3}=2\sin \tfrac{2\pi x_2}{3}$. Find $a$. $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that
$\sin \dfrac{2\pi x_1}{3}+\sin \dfrac{2\pi x_3}{3}=2\sin \dfrac{2\pi x_2}{3}$.
Find $a$
(Bulgari 1998)
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Using Prosthaphaeresis Formula,
$\sin\dfrac{2\pi x_1}3+\sin \dfrac{2\pi x_3}3=2\sin\dfrac{\pi(x_1+x_3)}3\cos\dfrac{\pi(x_1-x_3)}3$
As $x_1+x_2+x_3=3,$
$\sin\dfrac{\pi(x_1+x_3)}3=\sin\dfrac{\pi(3-x_2)}3=\sin\dfrac{\pi x_2}3$
So, we have $2\sin\dfrac{\pi x_2}3\left[\cos\dfrac{\pi(x_1-x_3)}3-2\cos\dfrac{\pi(x_2)}3\right]=0$
Now, $\cos\dfrac{\pi(x_2)}3=\cos\dfrac{\pi\{3-(x_1+x_3)\}}3=-\cos\dfrac{\pi(x_1+x_3)}3$
So, we have $2\sin\dfrac{\pi x_2}3\left[\cos\dfrac{\pi(x_1-x_3)}3+2\cos\dfrac{\pi(x_1+x_3)}3\right]=0$
As the product of two multiplicands is zero, at least one of them must be zero.
If $\sin\dfrac{\pi x_2}3=0,\dfrac{\pi x_2}3=n\pi\iff x_2=3n$ where $n$ is any integer
Else $\cos\dfrac{\pi(x_1-x_3)}3+2\cos\dfrac{\pi(x_1+x_3)}3=0$
$\iff3\cos\dfrac{\pi x_1}3\cos\dfrac{\pi x_3}3=\sin\dfrac{\pi x_1}3\sin\dfrac{\pi x_3}3$
$\iff\tan\dfrac{\pi x_1}3\tan\dfrac{\pi x_3}3=3$ which is not easily tractable
|
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Solving equation with rational exponent I have this equation: $$\mathrm{r} (x-1) = x^{8/9} - x^{1/9}$$
where $\mathrm{r}$ is a constant. Is there a general technique to solve such equations? Raising it to the 9nth power: $$\mathrm{r}^9 (x-1)^9 = (x^{8/9} - x^{1/9})^{9} \\ \mathrm{r}^9 (x-1)^9 = x (x^{7/9} - 1)^9\\$$ seems quite cumbersome.
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First make the substitution $y=x^{1/9}$ the equation becomes
$$r(y^9-1)=y^8-y$$
We can factor out from both sides $y-1$, giving us one solution $y=1$, and therefore the solution $x=1$. What remains is
$$r(y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=y(y^6+y^5+y^4+y^3+y^2+y+1)$$
or
$$ry^8+(r-1)\sum_{k=1}^{7}y^k+r=0$$
Observe that the coefficients of this polynomial are symmetric. Therefore we can use the substitution $z=y+\frac{1}{y}$. Observe that if we divide the polynomial above by $y^4$ we get
$$r\left(y^4+\frac{1}{y^4}\right)+(r-1)\sum_{k=1}^{3}\left(y^k+\frac{1}{y^k}\right)+(r-1)=0$$
Each $y^k+\frac{1}{y^k}$ can be written as a linear combination of powers of $\left(y+\frac{1}{y}\right)^n$, $n=0,1,2,...,k$. Therefore this can be written as a degree $4$ polynomial in $z=y+\frac{1}{y}$.
Tedious! But well... we get (using computations by Mark Bennet)
$$r(z^4-4z^2+2)+(r-1)(z^3-3z)+(r-1)(z^2-2)+(r-1)z+(r-1)=0$$
or
$$rz^4+(r-1)z^3-(3r+1)z^2+(2-2r)z+(r+1)=0$$
But check it yourself just in case.
Now, the degree $4$ polynomial equations we can solve in radicals.
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|
Solving the diophantine equation $p^2+n-3=6^n+n^6$
What are the pairs ($p,n$) of non-negative integers where $p$ is a prime number, such that $$p^2+n-3=6^n+n^6$$
How can I solve this diophantine equation?
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If $n \equiv 0,1\pmod3$, we have $3$ divides $6^n+n^6 - n+3$, except for $n=0$. Further, for $n=1$, we have $6^n+n^6 - n+3 = 9 = 3^2$. Hence, $n \equiv 2\pmod3$. However, if $n \equiv 2 \pmod3$, we have $$6^n+n^6 - n+3 \equiv 2\pmod3$$ and no square is $2 \pmod 3$. Hence, the solutions are $(n,p) = (0,2)$ and $(n,p) = (1,3)$.
|
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|
$z^n=(z+1)^n=1$, show that $n$ is divisible by $6$. we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two values $z=\operatorname{cis}(2 \pi /3)$ or $z=\operatorname{cis}(4 \pi/3)$. Then from $z^n=1=\operatorname{cis}(2 \pi k)$, $k$ integer I showed that $n$ should be divisible by $3$. I am left to show that $n$ should be divisible by $2$. I think if I figure out a way to show that $\operatorname{arg}(z+1)=\pi/3$ or $-\pi/3$ then I would be done, by using $(z+1)^n=\operatorname{cis}(2\pi k)$. but how can I do this? Any ideas are welcome! Thanks in advance!
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$z^n=1\implies z=\cos\tfrac{2\pi a}n+i\sin\tfrac{2\pi a}n$ where $a\equiv0,1,\cdots,n-1\pmod n$
$(z+1)^n=1\implies z+1=\cos\tfrac{2\pi b}n+i\sin\tfrac{2\pi b}n$ where $b\equiv0,1,\cdots,n-1\pmod n$
Equating the imaginary parts, $\sin\tfrac{2\pi a}n=\sin\tfrac{2\pi b}n$
$\implies\cos\tfrac{2\pi a}n=\pm\cos\tfrac{2\pi b}n$
Equating the real parts, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n-1$
Clearly, $\cos\tfrac{2\pi a}n=\cos\tfrac{2\pi b}n$ gives absurdity
$\implies\cos\tfrac{2\pi a}n=-\cos\tfrac{2\pi b}n$
Consequently, $-\cos\tfrac{2\pi b}n=\cos\tfrac{2\pi b}n-1\iff\cos\tfrac{2\pi b}n=\tfrac12=\cos\tfrac\pi3$
$\implies\tfrac{2\pi a}n=2m\pi\pm\tfrac\pi3=\tfrac\pi3(6m\pm1)$
$\iff\dfrac{n(6m\pm1)}6=a$ which is an integer
$\implies6|n(6m\pm1)$
But $(6m\pm1,6)=1$
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Solving a Diophantine equation: $p^n+144=m^2$ I found this Diophantine equation:
$$p^n+144=m^2$$
where $m$ and $n$ are integers and $p$ is a prime number. I solved it but I want to know if there exist other proofs through the use of rules of modular arithmetic.
This is my solution:
$$p^n+144=m^2$$
$$p^n=(m+12)(m-12)$$
With $(m+12)>(m-12)$ and $m-12$ is a divisor of $m+12$ because $m+12=p^x$ and $m-12=p^y$ with $n=x+y$ and $y<x$. Now if $m-12$ is a divisor of $m+12$, $\frac {m+12} {m-12} \ge 2$. If $\frac {m+12} {m-12} =2$, $m=36$ therefore $12<m<37$ and $24<m+12<49$. We can note that the only powers of prime numbers between values of $m+12$ are $25$, $27$ and $32$. By substitution we can find all possible values of $p$, $n$ and $m$.
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This gives us $m+12=p^a$ and $m-12 = p^b$. This means $p^a-p^b = 24$, i.e., $p^b(p^{a-b}-1) = 24 = 2^3 \cdot 3$. Note that $p^b$ and $p^{a-b}-1$ are of opposite parity.
*
*If $b=0$, we need $p^a-1 = 24 \implies p^a = 25 \implies p = 5,a=2$. Hence, $m=13$.
*If $b>0$, we need $p^b \mid 24$. Hence, $p=2$ or $p=3$.
*
*If $p=2$, we have $p^{a-b}-1$ to be odd. Hence, this means $p^{a-b}-1=3$ and $p^b=2^3$. This gives us $b=3$ and $a-b=2$, i.e., $a=5$. This gives us $m=20$.
*If $p=3$, we have $p^b=3$ and $p^{a-b}-1=2^3$. Hence, $b=1$ and $a-b=2$, i.e., $a=3$. This gives us $m=15$.
Hence, the solutions are ${\color{blue}{(m,p,n) = (13,5,2)}}$, ${\color{blue}{(m,p,n) = (20,2,8)}}$ and ${\color{blue}{(m,p,n) = (15,3,4)}}$.
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Easy inequality going wrong
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$
My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $x\leq-2, -5/4\leq x \leq -1$ and $-1\leq x\leq1$, x is less than or equal to zero.
But, this range i got is incorrect. Where did I go wrong then?
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Your only mistake is right at the end. Your polynomial changes sign at the points $-2,-\frac54, -1, 1, 5$, meaning that it is:
*
*Negative on $(-\infty, -2)$
*Positive on $(-2,-\frac54)$
*Negative on $(\frac54, -1)$
*Positive on $(-1,1)$
*Negative on $(1, 5)$
*Positive on $(5,\infty)$
As for the precise values in which the sign changes, you must be careful, because $$(x-5)(4x+5)(x-1)(x+1)(x+2)$$ may be defined on the whole real line, but the fraction $$\frac{(4x+5)(x-5)}{(x+1)(x-1)(x+2)}$$
is NOT defined on $x=1,-1,-2$, so the inequality cannot hold for these three numbers. It does hold for $x=-\frac54$ and $x=5$, however, so your whole answer is
$$x<-2\text{ or }-\frac54 \leq x < -1\text{ or } 1<x\leq 5$$
|
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Show that $2^{105} + 3^{105}$ is divisible by $7$ I know that $$\frac{(ak \pm 1)^n}{a}$$ gives remainder $a - 1$ is n is odd or $1$ is n is even.
So, I wrote $ 2^{105} + 3^{105}$ as $8^{35} + 27^{35}$ and then as $(7\cdot 1+1)^{35} + (7\cdot 4-1)^{35}$, which on division should give remainder of $6$ for each term and total remainder of 5 (12/7).
But, as per question, it should be divisible by 7, so remainder should be zero not 5. Where did I go wrong?
[note: i don't know binomial theorem or number theory.]
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You can write $2^{105} = 2^{6 \cdot 17}\cdot 8$ and $3^{105} = 3^{6 \cdot 17} \cdot 3^3$ and you can use Fermat's little theorem.
$2^{6 \cdot 17} \equiv 1 \pmod 7$, $8 \equiv 1 \pmod 7$ therefore $2^{2015} \equiv 1 \pmod 7$ while $3^{6 \cdot 17} \equiv 1 \pmod 7$ and $3^3 \equiv 6 \pmod 7$ therefore $3^{105} \equiv 6 \pmod 7$. $2^{105} + 3^{105}\equiv 1 + 6 \equiv 7 \equiv 0 \pmod7.$ We demonstrated that the number is divisible for $7.$
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Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx $
but I am not sure how this was accomplished. Any ideas? I am more interested in the method than the answer.
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Since $x^2+x-2 = (x-1)(x+2)$, we can use partial fraction decomposition:
$$\frac{x-2}{x^2+x-2} = \frac{A}{x-1} + \frac{B}{x+2}.$$
Then $x-2 = A(x+2) + B(x-1)$.
Equating the coefficients of each power of $x$, we have that $1=A+B$, and $-2=2A-B$.
Solving for $A$ and $B$ decomposes our original rational expression into two simple fractions.
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Identity in Ramanujan style Is it possible to represent
$$ \sqrt[3] {7\sqrt[3]{20}-1} =\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$ with rational $A,\,B,$ and $C?$
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Yes.
$$ \sqrt[3] {-1+7\sqrt[3]{20}} =\sqrt[3]{\frac{16}{9}}+\sqrt[3]{\frac{100}{9}}-\sqrt[3]{\frac{5}{9}}\tag0$$
Solution: More generally, given the three roots $x_i$ of any cubic equation,
$$x^3+ax^2+bx+c=0\tag1$$
then sums involving the cube roots of the $x_i$ can be given in the simple form,
$$(u+x_1)^{1/3}+(u+x_2)^{1/3}+(u+x_3)^{1/3} = \big(w+3\,\sqrt[\color{blue}6]{d}\big)^{1/3}$$
where $u,w$ are the constants,
$$u = \frac{ab-9c+\sqrt{d}}{2(a^2-3b)}\tag2$$
$$w = -\frac{(2a^3-9ab+27c)+9\sqrt{d}}{2(a^2-3b)}\tag3$$
and $d$ is,
$$d = \tfrac{1}{27}\Bigl(4(a^2-3b)^3-(2a^3-9ab+27c)^2\Bigr)\tag4$$
Example: For your question, we have,
$$w=-1, \quad d =\frac{7^6\cdot20^2}{3^6}$$
and subbing these into $(3),(4)$, and using $Mathematica$ to simplify, we get $b,c$ for arbitrary $a$ as,
$$b= \tfrac{1}{9}(-343+3a^2)$$
$$c= \tfrac{1}{27}(-2058-343a+a^3)$$
Substituting $b,c,d$ into $(2)$ and $(1)$, we find that,
$$u=\tfrac{1}{9}(37+3a)$$
and $(1)$ factors as,
$$ (7 + a + 3 x) (14 + a + 3 x) (-21 + a + 3 x) = 0\tag5$$
giving $x_1, x_2, x_3$. The expression $u+x_1$ simplifies as just $\frac{16}{9}$ and similarly for $x_2, x_3$, thus resolving into the numerical relation $(0)$ given above.
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How to prove this inequality? $(a+b+c=1)$ Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold:
$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$, but I don'think that's a good idea because $abc$ can be positive and negative.
I have also tried substituting $a^3+b^3+c^3 \geq 3abc$, but that isn't strong enough.
Any help/hints please??
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Firstly, multiply the right-hand side by $a+b+c$, so that both sides are now degree-zero. Then you want to prove
$$2a^3+2b^3+2c^3\geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$$
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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.