Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A triangular inequality including squares of sides Show that for any triangle ABC, the following inequality is true
$$a^2 + b^2 + c^2 > \sqrt{3} \max\{|a^2-b^2|,|b^2-c^2|,|c^2-a^2|\}$$
where $a,b,c$ are the sides of the triangle
| Let $c^2-a^2$ be the maximal term. By triangle inequality, we have $b^2>a^2-2ac+c^2$. Substituting in the equation of the OP, we get
$$a^2 + a^2 -2ac+ c^2 + c^2 > \sqrt{3} (c^2-a^2)$$
which leads to
$$ \frac{1}{2}[(2+\sqrt{3})a^2+(2-\sqrt{3})c^2]>ac $$
This last inequality is always true because it corresponds to an AM-GM inequality.
| {
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Verifying $\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$
Verify: $$\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$$
My solution:
$$
\begin{align}\sec^2x+\tan^2x&=\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\cdot\frac{1-\sin^2x}{1-\sin^2x}\\
&=\frac{1-\sin^4x}{\cos^2x\cdot\cos^2x}\\
&=\frac{1-\sin^4x}{\cos^4x}\\
&=\frac{1}{\cos^4x}-\frac{\sin^4x}{\cos^4x}\\
&=\sec^4x-\sin^4x\sec^4x\\
&=\sec^4x(1-\sin^4x)\\
\end{align}$$
Is it incorrect to multiply in $1-\sin^2x$ like in the fourth equality?
| No it's absolutely correct
You may be worried about What if (Sin^2)x= 1
But it can not be achieved in the domain of the equality otherwise (tan^2)x and (sec^2)x at (sin^2)x=1 will not be defined
| {
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Example 3.53 in Baby Rudin Here's Example 3.53 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.
Consider the convergent series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots$$ and one of its rearrangements $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots$$ in which two positive terms are always followed by one negative. If $s$ is the sum of the original series, then $$s < 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.$$ Since $$\frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} = \frac{8k-4}{(4k-1)(4k-3)} - \frac{1}{2k} = \frac{2k(8k-4) - (4k-1)(4k-3)}{2k(4k-1)(4k-3)} = \frac{8k-3}{2k(4k-1)(4k-3)} > 0$$ for $k \geq 1$, we see that $$s^\prime_3 < s^\prime_6 < s^\prime_9 < \ldots,$$ where $s^\prime_n$ is the $n$th partial sum of the series after the rearrangement. Hence $$\lim_{n\to\infty}\sup s^\prime_n > s^\prime_3 = \frac{5}{6},$$ so that the rearranged series certainly does not converge to $s$.
Now here's my question:
How to determine, using the machinery developed by Rudin upto this point in the book, if the new (or rearranged) series converges at all? Rudin leaves it to the reader to check that the new series does converge. How to prove this convergence?
I would like to have answers that use only the results that Rudin has discussed so far in the book.
| We can show that the series converges, and find its sum, as follows:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get
$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.
Adding this to the original series gives
$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.
| {
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Minimum of $ay+az+bz+bx+cx+cy$ with $ab+bc+ca=xy+yz+zx=1$ Let $a,b,c,x,y,z\in\mathbb{R}^+$, and $ab+bc+ca=xy+yz+zx=1$. What is the minimum value of $ay+az+bz+bx+cx+cy$?
When $a=b=c=x=y=z=\dfrac{1}{\sqrt{3}}$, the desired value is $2$.
When $a=b=x=y\rightarrow 1$ and $c=z\rightarrow 0$, the desired value also approaches $2$, so it seems likely that $2$ is the minimum.
| For any $a,b,c,x,y,z$, we have
$$ (a+b)(x+y)(ay+az+bz+bx+cx+cy) \\
= (xy+yz+zx)(a+b)^2 + (ab+bc+ca)(x+y)^2 + (bx-ay)^2 $$
So, in our situation with $xy+yz+zx = ab+bc+ca = 1$,
\begin{align*}
(a+b)(x+y)(ay+az+bz+bx+cx+cy)
&= (a+b)^2 + (x+y)^2 + (bx-ay)^2 \\
&\ge (a+b)^2 + (x+y)^2 \\
&\ge 2(a+b)(x+y)
\end{align*}
by AM/GM. Since our numbers are all positive, dividing by $(a+b)(x+y)$ yields the desired inequality. The equality case is that $a+b=x+y$ and $ay=bx$, which is equivalent to $a=x$ and $b=y$ (in which case also $c=z$).
I'm not too happy with the symmetry breaking in this argument, but there it is.
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Find the linear fractional transformation that maps the circles |z-1/4| = 1/4 and |z|=1 onto two concentric circles centered at w=0? I am very close to the solution I think. Since the circles cross the real axis, I want to find mappings from $z \to w$ such that $1/2 \to m$ and $0 \to -m$ and $-1 \to n$ and $1 \to -n$. Using
$$ w = \frac{az+b}{cz+d}$$
we get
$$ \frac{a+2b}{c+2d} = -\frac{b}{d}$$
and
$$ \frac{b-a}{d-c} = - \frac{a+b}{c+d}$$
but where do we go from here? I am a little lost in the simplifying algebra here to solve for the $a,b,c,d$ in the transformation.
| Starting from your ansatz, we can reduce the degrees of freedom by positing that we want to map $1 \mapsto 1$ and $-1\mapsto -1$. If your ansatz works at all (it does, but one needs to know a bit about Möbius transformations to know that a priori), we can achieve that by afterwards dividing by $n$.
That gives us the relations
\begin{gather}
a+b = c+d\\
a-b = d-c\\
(a+2b)d = - (c+2d)b
\end{gather}
for the desired transformation. From the first two, we obtain $a = d$ and $b = c$, so inserting that into the third, we get
$$(a+2b)a = - (b+2a)b,$$
and that becomes
$$a^2 + 4ab + b^2 = 0,$$
or
$$(a+2b)^2 = 3b^2.$$
We can normalise $b = 1$ (since $b = 0$ would lead to $a = 0$ and that doesn't yield a Möbius transformation), then $(a+2)^2 = 3$ gives us $a = -2 \pm \sqrt{3}$.
Let's pick $a = -2-\sqrt{3}$. So we look at
$$T(z) = \frac{(-2-\sqrt{3})z + 1}{z -(2+\sqrt{3})}.$$
That gives us
$$\begin{aligned}
T(1) &= \frac{-1-\sqrt{3}}{-1-\sqrt{3}} = 1; & T(-1) &= \frac{3+\sqrt{3}}{-3-\sqrt{3}} = -1;\\ T(0) &= \frac{1}{-(2+\sqrt{3})} = -(2-\sqrt{3}); & T\left(\tfrac{1}{2}\right) &= \frac{-\sqrt{3}}{1-2(2+\sqrt{3})} = \frac{-\sqrt{3}}{-\sqrt{3}(2+\sqrt{3})} = 2-\sqrt{3};
\end{aligned}$$
and the four points are mapped as desired. Since $T(\mathbb{R}\cup \{\infty\} = \mathbb{R} \cup \{\infty\}$, and the original two circles intersected the real line at right angles, the two image circles also do that, hence their centres are real, and since the intersections of the image circles with the real line are symmetric with respect to $0$, the centre of both is $0$.
| {
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sum of three inverse matrices The following Searle identity computes the sum of two inverses:
$A^{-1}+B^{-1} = A^{-1}(A+B)B^{-1}$.
Is there any generalisation of this for the sum of three inverses?
$A^{-1}+B^{-1}+C^{-1} = ? $
| Well, we could say that if $A+B$ is invertible,
$$
(A^{-1} + B^{-1})^{-1} = B(A+B)^{-1}A
$$
so that
$$
A^{-1} + B^{-1} + C^{-1} = \\
(A^{-1} + B^{-1}) + C^{-1} = \\
(A^{-1} + B^{-1})^{-1}((A^{-1} + B^{-1})^{-1} + C)C^{-1} =\\
B(A+B)^{-1}A(B(A+B)^{-1}A + C)C^{-1}
$$
| {
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Proof by Induction Divisibility. $6^n-5n+4$ is divisible by 5 for all positive integers $n$.
$n >=1$
Prove By Induction
My attempt is as follows:
$n=1$
$6^1-5(1) +4$
$=5$, Therefore 5 is divisible by 5 so $n=1$ is true
Assume its true for $n=k$
consider $n=k+1$
$6^k-5k+4=5.x$
I am stuck here would appreciate some assistance.
| Assume $6^n+5n+4$ is multiple of $5$.
Let us consider $$6^{n+1}-5(n+1)+4=(5+1)6^n-5n-5+4=5(6^n-1)+(6^n-5n+4)$$
The first summand is multiple of $5$ and the second is multiple of $5$ by inductive assumtion.
| {
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Finding the length of the side of the equilateral triangle
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
| Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^{\pi i/6}(a+2i) = X $$
so:
$$\Im \left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\cdot\left(a+2i\right)\right]=3, $$
or:
$$ 1+\frac{\sqrt{3}}{2}a = 3 $$
so $a = \frac{4}{\sqrt{3}}$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = \frac{16}{3}+4 = \frac{28}{3} $$
so the side length is $2\sqrt{\frac{7}{3}}$.
| {
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Beautifully looking little geometry/trigonometry problem Given triangle ABC, a,b,c as its sides, p is a half perimeter, such that $\dfrac{p-a}{11}=\dfrac{p-b}{12}=\dfrac{p-c}{13}$.
We need to find $(\tan\dfrac{A}{2})^2$
(A)$\dfrac{143}{432}$ B)$\dfrac{13}{33}$ C)$\dfrac{11}{39}$ D)$\dfrac{12}{37}$
I've tried to substitute the area of the triangle as $S=(1/2)(ab\sin C)=(1/2)(ac\sin B)=(1/2)(bc\sin A)$ and I know that $(\tan A/2)^2=(1-\cos A)/(1+\cos A)$ And some other formulas, but how is $\tan A/2$ derived from the given?
Sorry, the problem's given was confused, p is half-perimeter.
| Let: $$x=\dfrac{p-a}{11}=\dfrac{p-b}{12}=\dfrac{p-c}{13}$$
Thus, $11x+12x+13x=3p-(a+b+c)$, and so $36x=p$.
Denoting $r$ as inradius and $\Delta$ as area, we that:
$$r=\frac{\Delta}{s}=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{p}=\sqrt{\frac{11\cdot 12\cdot 13\cdot x^3}{36x}}$$
And:
$$\tan^2\frac{A}2=\frac{r^2}{(p-a)^2}=\frac{11\cdot 12\cdot 13\cdot x^2}{36\cdot (11x)^2}= { ...}$$
As to why $\tan\frac{A}2=\frac{r}{p-a}$, look at this diagram, and ask if you have any further doubts :)
| {
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Calculating semi axes from given tilted ellipse equation Hopefully no duplicate of Ellipse $3x^2-x+6xy-3y+5y^2=0$: what are the semi-major and semi-minor axes, displacement of centre, and angle of incline? (see below)
Let the following equation $$x^2 - \frac{1}{2}xy + y^2 - 11x - y = 18$$ be the ellipse in question. Plotting it on Webgraphing.com yields some nice values, e.g. the center at $(6, 2)$, but also the major/minor axis, which is what I am interested in. Their length is $\approx9.24$ and $\approx 7.16$ respectively.
Calculating the center is no problem, but I never get the axis length right. If anyone could spot the mistake in the following calculation, that would be very helpful!
Let $A = 1 (x^2)$, $B = -\frac{1}{2} (xy)$, $C = 1(y^2)$ and $D = 18$.
The matrix
$$
\left(\begin{matrix} A - \lambda & \frac{B}{2} \\ \frac{B}{2} & C - \lambda \end{matrix}\right) =
\left(\begin{matrix} 1 - \lambda & -\frac{1}{4} \\ -\frac{1}{4} & 1 - \lambda
\end{matrix}\right)
$$ yields the eigenvalues $\lambda_{1,2} = 1 \pm \frac{1}{4}$ and the corresponding eigenvector matrix
$$
\frac{1}{\sqrt{2}}\left(\begin{matrix} 1 & -1 \\ 1 & 1 \end{matrix}\right)
$$
It is obvious from the above image that the vectors point in the right direction. But when attempting to calculate their lengths, I never get the right result. I somewhere read that the absolute value in the equation ($D$) had to be normalised somehow, and the eigenvalues don't have the right proportion, so I took their root. This results in the lengths
$$l_1 = 2 * \sqrt{\lambda_1} * \sqrt{D} = 2 * \sqrt{\frac{5}{4}} * \sqrt{18} \approx 9.48 \neq 9.24$$
$$l_2 = 2 * \sqrt{\lambda_2} * \sqrt{D} = 2 * \sqrt{\frac{3}{4}} * \sqrt{18} \approx 7.34 \neq 7.16$$
that do not yield the correct result when printed (the red dots were the basis for the equation):
For other ellipses, the error is much bigger. Can anyone point out where I am wrong in the calculation?
Concerning the thread linked above (that answers the question for another example), the method described there does not work for me (I may be applying it wrong):
$$\sqrt{\frac{35}{\lambda_2}} = \sqrt{\frac{35}{1 - \frac{1}{4}}} \approx 6.83 \neq 9.24$$
| The equation of the equation centred at the origin is
$$(x+6)^2-\frac12 (x+6)(y+2)+(y+2)^2-11(x+6)-(y+2)=18\\
x^2+y^2-\frac12 xy=34$$
The quadratic coefficient matrix is
$$\left(\begin{array}{cc}1 & -\tfrac14 \\ -\tfrac 14 & 1 \end{array}\right)$$
The resulting characteristic equation is
$$\left|\begin{array}{cc}1-\lambda & -\tfrac14 \\ -\tfrac 14 & 1-\lambda \end{array}\right|=0\\
16\lambda^2-32\lambda+15=0\\(4\lambda-3)(4\lambda-5)=0\\\lambda=\frac34,\frac54\qquad \text{(eigenvalues)}$$
The equation of the "untilted" ellipse is
$$\tfrac34 x^2 +\tfrac54 y^2=34\\
\frac{x^2}{\left(2\sqrt{\tfrac{34}3}\right)^2}
+\frac{y^2}{\left(2\sqrt{\tfrac{34}5}\right)^2}=1$$
Hence semi-major and semi-minor axes are
$$2\sqrt{\frac{34}3}, 2\sqrt{\frac{34}5}$$
respectively.
It can easily be shown that the corresponding eigenvectors are $(m,m)$ and $(-m,m)$ respectively, thus the directions of the major and minor axes are $\pi/4$ and $-\pi/4$ respectively. $\blacksquare$
| {
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Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$
Calculate the definite integral
$$
I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx
$$
given that $a>b>0$.
My Attempt:
If we replace $x$ by $C$, then
$$
I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC
$$
Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives
$$
\begin{align}
\cos C &= \frac{a^2+b^2-c^2}{2ab}\\
a^2+b^2-2ab\cos C &= c^2
\end{align}
$$
From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to
$$
\begin{align}
I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC
\end{align}
$$
Is my process right? If not, how can I calculate the above integral?
| Let $p=a^2+b^2,q=2ab$ so that $p^2-q^2=(a^2-b^2)^2$ and the integrand can be rewritten as $$\frac{\sin^2x}{p-q\cos x} = \frac{1}{q}\left(\cos x+\frac{p}{q}+\frac{1}{q}\cdot\frac{q^2-p^2}{p-q\cos x} \right) $$ and thus the desired integral is equal to $$0+\frac{p\pi}{q^2}+\frac{q^2-p^2}{q^2}\int_{0}^{\pi}\frac{dx}{p-q\cos x} $$ The integral above is famously equal to $\pi/\sqrt{p^2-q^2}$ and hence the value of the integral in question is equal to $$\frac{\pi} {q^2}\cdot(p-\sqrt{p^2-q^2})=\frac{\pi}{2a^2}$$
| {
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Which expansion of $e$ is more accurate? We have two forms of $e^x$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$$
and
$$e^x=\frac{1}{\displaystyle 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+....}$$
The second form comes from $e^x=1/e^{-x}$
Which one is more accurate if I want to find any value of $e^x$
| The accuracy of any Taylor expansion of a given order is given by the next order of the expansion. For this very case, the answer is that it depends on how far out you want to take the expansions. For example,
$$\frac1{1-x} = 1+x+x^2+O(x^3)$$
This has twice the error as the expansion $1+x$, whose next term is $x^2/2!$. However,
$$\begin{align}\frac1{1-x+x^2/2} &= 1+\left (x-\frac{x^2}{2} \right ) + \left (x-\frac{x^2}{2} \right )^2+ \left (x-\frac{x^2}{2} \right )^3 + \cdots \\&= 1+x+\frac{x^2}{2} + O(x^4)\end{align}$$
Thus, for the second-order expansion, the reciprocal has a smaller error. However, for the cubic expansion, the expansion is
$$1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12} + O(x^5) $$
whereas the fourth order term in the direct, Taylor expansion of the exponential is $x^4/24$. Thus, the direct Taylor expansion is more accurate out to fourth order.
There is a systematic way to prove that the accuracy alternates with order oddness or evenness by considering the error in the expansion of $e^x \cdot e^{-x}=1$.
ADDENDUM
The proof is actually not all that hard. Consider the finite approximations to the equation $e^x \cdot e^{-x}=1$:
$$\left (\sum_{k=0}^m \frac{x^k}{k!} \right ) \left (\sum_{k=0}^m (-1)^k \frac{x^k}{k!} \right ) $$
It may be easily shown that the coefficients of $x$, $x^2$, ..., $x^m$ are zero. We now consider the first error term, i.e., the coefficient of $x^{m+1}$:
$$\frac{x^1}{1!} \frac{(-1)^m x^m}{m!} + \frac{x^2}{2!} \frac{(-1)^{m-1} x^{m-1}}{(m-1)!} +\cdots + \frac{x^m}{m!} \frac{(-1)^1 x^1}{1!} = \sum_{k=1}^m \frac{(-1)^{m-k+1}}{k! (m-k+1)!} x^{m+1}$$
Rewrite this sum, sans the $x^{m+1}$ term, as
$$\sum_{k=0}^{m-1} \frac{(-1)^{m-k}}{(k+1)! (m-k)!} = (-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1}$$
We can evaluate this sum by realizing that
$$(1-x)^m = \sum_{k=0}^m (-1)^k \binom{m}{k} x^k $$
so that
$$\sum_{k=0}^m (-1)^k \binom{m}{k} \frac1{k+1} = \int_0^1 dx \, (1-x)^m = \frac1{m+1}$$
Therefore, by subtracting off the last term in the sum, we find that
$$(-1)^m \sum_{k=0}^{m-1} \frac{(-1)^k}{k! (m-k)!} \frac1{k+1} = - \frac{1-(-1)^m}{(m+1)!}$$
Therefore, we now may say that
$$\frac1{\displaystyle \sum_{k=0}^m (-1)^k \frac{x^k}{k!} } =
\sum_{k=0}^m \frac{x^k}{k!} + \frac{1-(-1)^m}{(m+1)!} x^{m+1} +
O(x^{m+2})$$
For even values of $m$, the next term error is zero, which is smaller than that for the direct Taylor series, which has error $x^{m+1}/(m+1)!$. On the other hand, for odd values, the error is double that of the direct Taylor series. This agrees with the above examples.
ADDENUDUM II
You may use the above analysis to determine whether you can get better accuracy using
$$e^x = \frac{e^{x/2}}{e^{-x/2}} $$
(Generally speaking, rational approximations to functions that agree with Taylor series in certain limits are called Pade approximates, and may be very useful in some situations.)
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Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$, the x-axis and the lines $x = 1$ and $x = k$ is 16. Find $k$. Question: Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$,
the $x$-axis and the lines $x = 1$ and $x = k$ is $16$. Find $k$.
I tried but found no solution.
What I did:
$$\int_1^k (3x^2-12x+9) \,dx = \big[x^3-6x+9x \big]_1^k = k^3-6k^2+9k-4 =16$$
$$\Rightarrow k^3-6k^2+9k-20 = 0$$
Then after simplifying I got:
$$(k-5)(k^2-k+4) = 0$$
This means $k = 5$, because $(k^2-k+4)$ has no real roots, but this is wrong. The reason is after drawing the graph, the area under the curve, from $x=1$ to $x=5$ is:
$$-\int_1^3 (3x^2-12x+9) \, dx + \int_3^5 (3x^2-12x+9) \, dx = 24 \ne 16$$
Can anyone help? What mistake did I make?
| As I set the question, but do not have the reputation to comment, I will say what the question should have been, and answer it. I will also provide an answer to the question you attempted, and point out your flaw.
There was a typo on the sheet: the 1 should have been a -1 (this was corrected at some point though).
Therefore, the question goes as follows:
$$
\begin{align*}
16&=\int\limits_{-1}^k3x^2-12x+9\mathrm{d}x &\text{NOTE: Do not omit the dx (as in the OP). You'll drop a mark.}\\
&=\left[x^3-6x^2+9x\right]_{-1}^k\\
&=k^3-6k^2+9k-(-1-6-9)\\
\Rightarrow 0&=k^3-6k^2+9k\\
\Rightarrow k&=0\text{ or }k=3
\end{align*}
$$
We discard the $k=3$ solution so the answer is $k=0$. I will leave you to work out why we discard $k=3$ rather than $k=0$...
If we take the lower limit to be $x=1$, then note that the area between $x=1$ and $x=3$ (the root we care about) is $4$. Hence, $k>3$. Therefore, the equation you should have been solving is the following.
$$
\begin{align*}
16&=-\int\limits_{-1}^33x^2-12x+9\mathrm{d}x+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x\\
&=4+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x
\end{align*}
$$
This gives you the equation $(k-3)^2k=12$. The only real root is $k\approx4.6129$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using the ratio test for series I need to show whether $\displaystyle\sum_{n=1}^{\infty}{\frac{2^n+3^n}{4^n-5^n}}$ converges or diverges using the ratio test.
So far I have $\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}+3^{n+1}}{4^{n+1}-5^{n+1}} . \dfrac{4^n-5^n}{2^n+3^n}$
I know I could maybe use division by $5^n$ at some point but I am not sure how I could simplify from here in order to obtain an expression that I could easily show has a limit $<1$ for convergence or $>1$ for divergence.
The only tests I can utilise at the moment are ratio and comparison, along with the use of geometric series. I would easily be able to solve this if not for the minus sign in the denominator by using comparison with geometric series for example.
| You can also use $\displaystyle\frac{|a_{n+1}|}{|a_n|}=\frac{2^{n+1}+3^{n+1}}{5^{n+1}-4^{n+1}} \cdot \frac{5^n-4^n}{2^n+3^n}=\frac{2^{n+1}+3^{n+1}}{2^n+3^n}\cdot\frac{5^n-4^n}{5^{n+1}-4^{n+1}}$
$=\displaystyle\frac{2(\frac{2}{3})^n+3}{(\frac{2}{3})^n+1}\cdot\frac{1-(\frac{4}{5})^n}{5-4(\frac{4}{5})^n}\to \frac{3}{1}\cdot\frac{1}{5}=\frac{3}{5}$
| {
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"answer_count": 2,
"answer_id": 1
} |
about a difficult and weird Probability question Let W be the random variable that counts the number of tails before one gets r heads for a coin whose probability of heads is θ. Without using moment generating function, show that the mean and variance for W are [r(1-θ)]/θ and [r(1-θ)]/θ^2
Please help me with the complete proof. I have some ideas, but I can't write it in a formal way.
Thanks!
| Let's apply the binomial theorem to see that
$$
(1-x)^{-n-1} = \sum_{k\geq 0} (-1^k) \binom{-n-1}{k} x^k
$$
We can negate the upper index in the binomial coefficient, incorporating the $(-1)^k$, to get
$$
\frac{1}{(1-x)^{(n+1)}} = \sum_{k\geq 0} \binom{n+k}{k} x^k = \sum_{k\geq 0} \binom{n+k}{n} x^k
$$
and if we multiply that by $z^n$ and note that binomial coefficients with integer top less than the integer bottom are zero, we get our key identity
$$
\frac{x^n}{(1-x)^{(n+1)}} = \sum_{k\geq 0} \binom{k}{n} x^k
$$
Consider $k$ to be the total number of tosses until the $r$-th head turns up on the $k$-th toss. (Obviously $k \geq r$.) The probability of that happening is
$$f(k;r) = \binom{k-1}{r-1} \theta^{r-1} (1-\theta)^{k-r} \theta $$ which we can read off as " $r-1$ out of the first $k-1$ tosses came up heads; the remaining $k-r$ were tails; and then the last toss was the $r$-th head."
$$f(k;r) = \left( \frac{\theta}{1-\theta} \right)^r \binom{k-1}{r-1} (1-\theta)^k $$
Now let's calculate the expectation of the total number of tosses (not just the tails):
$$E(k;r) = \sum_{k\geq r} k \left( \frac{\theta}{1-\theta} \right)^r \binom{k-1}{r-1} (1-\theta)^k = \left( \frac{\theta}{1-\theta} \right)^r \sum_{k\geq r} k \binom{k-1}{r-1} (1-\theta)^k
$$
But it is easy to see that $k \binom{k-1}{r-1} = r \binom{k}{r}$ by incorporating the $k$ factor into the numerator and then seeing how to ket to a binomial coefficient by multiplying the denominator (by $r$).
$$E(k;r) = = \left( \frac{\theta}{1-\theta} \right)^r r \sum_{k\geq r} \binom{k}{r} (1-\theta)^k
$$
upon which we apply our key identity (with $x = (1-\theta)$)to get
$$E(k;r) = \left( \frac{\theta}{1-\theta} \right)^r r \frac{(1-\theta)^r}{\theta^{r+1}} = \frac{r}{\theta}
$$
The mean number of tails is then
$$\frac{r}{\theta} - r = r \frac{1-\theta}{\theta}$$ as was to be shown.
The variance of the number of tails is the same as the variance of the total number of tosses, as these differ by a fixed constant $r$. This in turn is
$E(k^2;r) - \left( E(k;r) \right)^2$.
$$ E(k^2;r) = \left( \frac{\theta}{1-\theta} \right)^r \sum_{k\geq r} k^2 \binom{k-1}{r-1} (1-\theta)^k = \left( \frac{\theta}{1-\theta} \right)^r \left[ \sum_{k\geq r} (k+1)k \binom{k-1}{r-1} (1-\theta)^k - \sum_{k\geq r} k \binom{k-1}{r-1} (1-\theta)^k \right]
$$
$$= \left( \frac{\theta}{1-\theta} \right)^r \left[ r(r+1) \sum_{k\geq r} \binom{k+1}{r+1}(1-\theta)^k - r \sum_{k\geq r}\binom{k}{r} (1-\theta)^k \right] =
\left( \frac{\theta}{1-\theta} \right)^r \left[ r(r+1) \sum_{m > r} \binom{m}{r+1}(1-\theta)^{m-1} - r \frac{(1-\theta)^r}{\theta^{r+1}} \right]=\left( \frac{\theta}{1-\theta} \right)^r \left[ r(r+1) \frac{1}{1-\theta} \sum_{m > r} \binom{m}{r+1}(1-\theta)^m \right]- \frac{r}{\theta} $$
$$= \left( \frac{\theta}{1-\theta} \right)^r r(r+1) \frac{1}{1-\theta} \frac{(1-\theta)^{r+1}}{\theta^{r+2}} - \frac{r}{\theta} = r\frac{r+1 - \theta}{\theta^2}
$$
From which
$$\sigma^2 = E(k^2;r) - \left( E(k;r) \right)^2= r \frac{1-\theta}{\theta^2}$$
which was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find closed formula for $S(n,n-4)$. Find closed formula for $S(n,n-4)$.
These are stirling numbers of 2nd kind.
My attempt: I uses the recurrence relation $$S(n,k)=S(n-1,k-1)+kS(n-1,k)$$
$S(n,n-4)=S(n-1,n-5)+(n-4)S(n-1,n-4)=S(n-2,n-6)+(n-5)S(n-2,n-5)+(n-4)\{S(n-2,n-5)+(n-4)S(n-2,n-4)\}$
But it seems not helpful for me.How I can find closed formula for $S(n,n-4)$?
| Here is a solution by generating functions for verification purpose
until something simpler appears.
Observe that there cannot be a set of size at least six because that
leaves $n-6$ items which can form at most $n-6$ sets for a total of
$n-5$ sets. Similarly for a set of length seven and so on.
The species of set partitions with sets of length at most five is
is
$$\mathfrak{P}
(\mathcal{A}_1 \mathfrak{P}_{=1}(\mathcal{Z})
+ \mathcal{A}_2 \mathfrak{P}_{=2}(\mathcal{Z})
+ \cdots
+ \mathcal{A}_5 \mathfrak{P}_{=5}(\mathcal{Z})).$$
This gives the generating function
$$G(z) = \exp\left(\sum_{q=1}^5 a_q \frac{z^q}{q!}\right)
= \prod_{q=1}^5 \exp\left( a_q \frac{z^q}{q!} \right).$$
Now there are several cases.
First case.
A five-set and $n-5$ fixed points.
This gives the generating function
$$\frac{1}{(n-5)!} \left(\frac{z}{1!}\right)^{n-5}
\exp\left(a_2 \frac{z^2}{2!}\right)
\exp\left(a_3 \frac{z^3}{3!}\right)
\exp\left(a_4 \frac{z^4}{4!}\right)
\frac{1}{1} \left(\frac{z^5}{5!}\right)^1.$$
We can drop the contributions in $a_2, a_3, a_4$ as no such sets
appear because there aren't any items left over, for an answer of
$$\frac{1}{(n-5)!} \left(\frac{z}{1!}\right)^{n-5}
\left(\frac{z^5}{5!}\right)^1.$$
Second case.
A four-set, a two-set and $n-6$ fixed points, for a
contribution of
$$\frac{1}{(n-6)!} \left(\frac{z}{1!}\right)^{n-6}
\left(\frac{z^2}{2!}\right)^1
\left(\frac{z^4}{4!}\right)^1.$$
Third case. Two three-sets and $n-6$ fixed points, for a
contribution of
$$\frac{1}{(n-6)!}\left(\frac{z}{1!}\right)^{n-6}
\frac{1}{2} \left(\frac{z^3}{3!}\right)^2.$$
Fourth case. A three-set, two two-sets and $n-7$ fixed
points for a contribution of
$$\frac{1}{(n-7)!} \left(\frac{z}{1!}\right)^{n-7}
\frac{1}{2} \left(\frac{z^2}{2!}\right)^2
\left(\frac{z^3}{3!}\right)^1.$$
Fifth case. Four two-sets and $n-8$ fixed
points for a contribution of
$$\frac{1}{(n-8)!}\left(\frac{z}{1!}\right)^{n-8}
\frac{1}{24} \left(\frac{z^2}{2!}\right)^4.$$
Adding the contributions from these generating functions we obtain
$$\frac{z^n}{(n-5)!} \times
\left(\frac{1}{120}
+ \frac{n-5}{48}
+ \frac{n-5}{72}
+ \frac{(n-5)(n-6)}{48}
+ \frac{(n-5)(n-6)(n-7)}{384} \right)
\\ = \frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right).$$
Performing coefficient extraction on this we obtain the answer
$$n! [z^n]
\frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right)
\\ = \frac{n!}{(n-5)!}
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right).$$
This is very similar to the computation at this
MSE link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
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How to prove the matrix inequality $\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}+\det{(B)}|$ Let $A,B\in M_{2}(R)$ be two square matrices so that
$AB=BA$, prove the inequality
$$\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}|$$
I only know prove this
$$\sqrt{a^2-ab+b^2}+\sqrt{a^2+ab+b^2}\ge |a+b|,a,b\in R$$
because
$$\sqrt{x}+\sqrt{y}\ge\sqrt{x+y}$$
so
$$LHS\ge\sqrt{2(a^2+b^2)}\ge (a+b)$$
But for matrices I can't prove it. Thank you
| As user 1551 points out, it suffices to prove the inequality for $AB = BA$ where, $A,B$ are invertible. Since, commuting matrices preserve each others eigenspaces, they are simultaneous diagonalisable, $\exists \, U \in U_2(\mathbb{R})$, such that $U^{-1}AU = D[x_1,y_1] = D_1$ and $U^{-1}BU = D[x_2,y_2] = D_2$.
Thus, $\displaystyle \sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}| \iff \sqrt{\det{(D_1^2-D_1D_2+D_2^2)}}+\sqrt{\det{(D_1^2+D_1D_2+D_2^2)}}\ge |\det{(D_1)}
+\det{(D_2)}|$
(since, $|\det(U)| = 1$)
That is,
$\displaystyle \sqrt{(x_1^2-x_1x_2+x_2^2)(y_1^2-y_1y_2+y_2^2)}+\sqrt{(x_1^2+x_1x_2+x_2^2)(y_1^2+y_1y_2+y_2^2)} \ge |x_1y_1+x_2y_2| \iff \left(\frac{3}{4}(x_1-x_2)^2+\frac{1}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{3}{4}(y_1-y_2)^2+\frac{1}{4}(y_1+y_2)^2\right)^{1/2} + \left(\frac{1}{4}(x_1-x_2)^2+\frac{3}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{1}{4}(y_1-y_2)^2+\frac{3}{4}(y_1+y_2)^2\right)^{1/2} \ge_{C.S.} \frac{3}{4}|x_1-x_2||y_1-y_2|+\frac{1}{4}|x_1+x_2||y_1+y_2|+\frac{1}{4}|x_1-x_2||y_1-y_2|+\frac{3}{4}|x_1+x_2||y_1+y_2| \ge |x_1y_1+x_2y_2|$ (by triangle inequality).
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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Why four roots to this equation: $(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$
$$(7x+1)^{1 \over 3}+(8+x-x^2)^{1 \over 3}+(x^2-8x-1)^{1 \over 3}=2$$
I figured the roots are $0$, $1$, $-1$, and $9$. But why?
| Why does this equation has $4$ roots? The short answer is it is an accident.
For similar problem like
$$(7x-6)^{1/3}+(8+x-x^2)^{1/3}+(x^2-8x-1)^{1/3}=1
$$
You can get $5$ instead of $4$ solutions.
Define $a,b,c$ like Blue's answer.
$$\begin{cases}
a^3 &= 7x + 1 \\
b^3 &= 8 + x - x^2 \\
c^3 &= x^2 - 8 x - 1
\end{cases}
$$
The key of the whole thing is $a^3 + b^3 + c^3 = 8$ which is independent of $x$
and $8 = 2^3$.
If $a + b + c = 2$, we will have
$$0 = (a+b+c)^3 - (a^3+b^3+c^3) = 3(a+b)(b+c)(c+a) = 3(2-c)(2-a)(2-b)$$
This implies one of $a, b, c$ is equal to $2$ and hence one of $a^3, b^3, c^3$ is $8$.
*
*$a^3 = 8 \implies x = 1$,
*$b^3 = 8 \implies x - x^2 = 0 \iff x(x-1) = 0 \implies x = 0 \text{ or } 1$.
*$c^3 = 8 \implies x^2 - 8x = 9 \iff (x-9)(x+1) = 0 \implies x = -1\text{ or } 9$.
As you can see, the first case is a linear equation, it give you one possible solution.
The remaining two cases are quadratic equation, each of them given you two possible solutions. This means in principle, you can have $5$ possible solutions.
However, one of the solution $x = 1$ has been duplicated, so you are left with $4$ distinct possible solutions.
It remains to check the above 4 possible choice of $x$ are indeed solutions of the
original problem. Instead of plugging in the values of $x$ to check it is a solution, let me
use $x = 0$ as an example to illustrate the general pattern. In that case $b^3 = 8$ and we have
$$a^3 + b^3 + c^3 = 8 \implies a^3 + c^3 = (a+c)(a^2 - ac + c^2) = 0$$
Notice $a, c$ are real numbers, this means
$a^2 - ac + c^2 = (a-\frac{c}{2})^2 + \frac34 c^2 \ne 0$ whenever $(a,c) \ne 0$.
This implies $ a+c = 0$ and hence
$$(a + b + c)^3 = a^3 + b^3 + c^3 = 8 \implies a + b + c = 2$$
As a result, $x = 0$ is indeed a solution. By a similar argument, the other $3$ choices of $x$ are also solutions of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047284",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Finding the derivative of $f(x) = \frac{8}{\sqrt{x -2}}$ using first principles. How would you go about determining the derivative of ( $f(x) = \frac{8}{\sqrt{x -2}}$ ) using the limit definition of the derivative
(i.e. $\lim\limits_{h\to 0} = \frac{f(x+h) - f(x)}{h}$) as opposed to just applying the chain rule. So I'm thinking this counts as an algebra question but can't find too many examples dealing with simplification of polynomial expressions with fractional exponents.
I got as far as,
$\frac {1}{h} (\frac{8}{\sqrt{x+h-2}} - \frac{8}{\sqrt{x-2}})$
In case I'm not using correct terminology or being unclear, what I mean is, can one algebraically eliminate the $h$ from the denominator in the above expression to take the value of the limit?
| Hint: For $x>2$ and any non-zero $h$ such that $x+h>2$:
$$\frac {1}{h} \left(\frac{8}{\sqrt{x+h-2}} - \frac{8}{\sqrt{x-2}}\right)$$
$$ = \frac {1}{h} \frac{8(\sqrt{x-2} - \sqrt{x+h-2})}{\sqrt{x+h-2}\sqrt{x-2}} $$
$$ = \frac {1}{h} \frac{8[(x-2) - (x+h-2)]}{\sqrt{x+h-2}\sqrt{x-2}(\sqrt{x-2} + \sqrt{x+h-2})} $$
$$ = -\frac{8}{\sqrt{x+h-2}\sqrt{x-2}(\sqrt{x-2} + \sqrt{x+h-2})}. $$
When $h$ approaches $0$, the last fraction approaches:
$$ -\frac{4}{(x-2)\sqrt{x-2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove inquality involving factorial This is part of an analysis problem I'm working on.
Show that:
$ \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + ... < \frac{3}{(n+1)!}$
After some algebra I got
$ \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + ... < 2$
Now I'm stuck.
| If $n \geq 0$, you have
$$ \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + \cdots$$
$$ \leq
\frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \cdots $$
$$ < \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots = 1 < 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Partial Fractions Variables $$\int\frac{x^3+6x^2+3x+16}{x^3+4x}\,dx$$
Eventually one solves that a variable (I used $C$) ${}= -x$.
By variables I mean the decomposition yields $A + Bx+C$.
Therefore $C = -1$.
I think that $C$ should equal $-x$. How is what I think not possible?
| Partial fraction decomposition here only works because it is proven that if $P(x)$ is a polynomial, an expression
\begin{equation}
\frac{P(x)}{(x)(ax^{2}+b)}
\end{equation}
...can be put in the form:
\begin{equation}
\frac{A}{x}+\frac{Bx+C}{ax^{2}+b}
\end{equation}
...IF A, B, and C are constants. Otherwise, all bets are off.
It is possible that it works with C=-x in this case by coincidence, which you can check, but that's not a partial fraction decomposition; a partial fraction decomposition is one in which A, B, and C are constant.
I'll do the problem here:
First, factor the denominator so that it contains only one term:
\begin{equation}
x^{3}+4x=x(x^{2}+4)
\end{equation}
The goal is to find an A, B, and C (constants) such that:
\begin{equation}
\frac{x^{3}+6x^{2}+3x+16}{x(x^{2}+4)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+4}
\end{equation}
Multiplying both sides by $x(x^{2}+4)$:
\begin{equation}
x^{3}+6x^{2}+3x+16=A(x^{2}+4)+(Bx+C)x=A(x^{2}+4)+Bx^{2}+Cx
\end{equation}
Plugging in x=0 will cancel everything on the right except the A term. The left will become 16:
\begin{equation}
16=A(4) \implies A=4
\end{equation}
Unfortunately, we can't use that strategy to find B and C, because there is no number which will make the A term 0. So I'll use a different strategy.
Working on the right hand side, I'll plug in A=4 because we know that now, and order everything by the power:
\begin{equation}
4(x^{2}+4)+Bx^{2}+Cx=4x^{2}+16+Bx^{2}+Cx=Bx^{2}+4x^{2}+Cx+16
\end{equation}
Coming back to the whole equation (having simplified the right side), we have:
\begin{equation}
x^{3}+6x^{2}+3x+16=Bx^{2}+4x^{2}+Cx+16
\end{equation}
This is the interesting part: The only way for these two to be equal for all x is if all the $x^{2}$ coefficients on the left sum to the $x^{2}$ coefficients on the right. So...
\begin{equation}
6=B+4 \implies B=2
\end{equation}
Now, same with the x coefficients:
\begin{equation}
3=C
\end{equation}
In conclusion,
\begin{equation}
\frac{x^{3}+6x^{2}+3x+16}{x(x^{2}+4)}=\frac{4}{x}+\frac{2x+3}{x^{2}+4}
\end{equation}
This new function is easier to integrate. Split the second fraction into two fractions:
\begin{equation}
=\int \frac{4}{x}+\frac{2x}{x^{2}+4}+\frac{3}{x^{2}+4}dx
\end{equation}
With $u=x^{2}+4$, and dividing by 4 on the top and bottom of the 3rd fraction, this equals
\begin{equation}
4\int \frac{1}{x}dx+\int \frac{du}{u}+\int \frac{3/4}{(x/2)^{2}+1}dx
\end{equation}
Then the first two terms are just natural logs (see below). The rightmost term is an arctan.
\begin{equation}
\int \frac{1}{x}dx=ln(x)+C
\end{equation}
\begin{equation}
\int \frac{1}{x^{2}+1}dx=arctan(x)+C
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Evaluating $\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}$
Show that series $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}$$ converges by simplifying its sequence of partial sums and find its sum.
I don't have much detail but this all I have:
$$\begin{align}\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}
&=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{6}{m(m+1)(m+2)}\\
&=\lim_{m\to\infty}\frac{3(m^2+3m)}{2(m+1)(m+2)}\\
&=\frac{3}{2}\\
\end{align}$$
I know, I don't have much but any help will do with the detail or is it right.
| Since
\begin{gather*}
\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right),
\end{gather*}
we have, by telescoping summation,
\begin{align*}
\sum_{n=1}^N\frac{1}{n(n+1)(n+2)}&=\frac{1}{2}\sum_{n=1}^N\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=1}^N\frac{1}{(n+1)(n+2)}\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{m=2}^{N+1}\frac{1}{m(m+1)}\quad \quad (m=n+1)\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n(n+1)}\\
&=\frac{1}{2}\cdot\frac{1}{2}+\sum_{n=2}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N}\frac{1}{n(n+1)}-\frac{1}{2}\cdot\frac{1}{(N+1)(N+2)}\\
&=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{(N+1)(N+2)}\right)\to \frac{1}{4}, \qquad \text{as } N\to\infty,
\end{align*}
from which we can infer that
\begin{gather*}
\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}=6\cdot\frac{1}{4}=3/2.
\end{gather*}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then, we have $x = 2$, $y=2$, $z=3$. Putting this all into the inequality, we get:
$(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$
$(2a+2b+3c)^2 \leq (a^2+b^2+c^2)(2^2+2^2+3^2)$
$(26+a)^2 \leq (52)(17)$
$(26+a)^2 \leq 884$
$26 + a \leq \sqrt{884}$
$a \leq 2\sqrt{221} - 26 \approx 3.73$
However, the listed answer gives $a \leq \frac{26}{7} \approx 3.71$. What am I doing wrong?
| You can solve $a + 2b + 3c = 26$ for $b$ and plug into $a^2 + b^2 + c^2 = 52$ which gives:
$$13c^2+6ac-156c+5a^2-52a+676=208 \tag{A}$$
Which is a conic section (ellipse) and the maximum can be found by taking a derivative wrt $c$:
$$26c + 6a + 6c\frac{da}{dc} - 156 + 10a\frac{da}{dc} - 52\frac{da}{dc} = 0$$
and setting $\frac{da}{dc} = 0$ :
$$26c + 6a - 156 = 0 \tag{B}$$
Solving (B) for $c$ and plugging into (A) gives:
$$a = \frac{27}6$$
as the max.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $3x\equiv 4\pmod 7$ I'm trying to learn about linear congruences of the form ax = b(mod m). In my book, it's written that if $\gcd(a, m) = 1$ then there must exist an integer $a'$ which is an inverse of $a \pmod{m}$. I'm trying to solve this example:
$$3x \equiv 4 \pmod 7$$
First I noticed $\gcd(3, 7) = 1$.
Therefore, there must exist an integer which is the multiplicative inverse of $3 \pmod 7$.
According to Bezout's Theorem, if $\gcd(a, m) = 1$ then there are integers $s$ and $t$ such that $sa+tm=1$
where $s$ is the multiplicative inverse of $a\pmod{m}$.
Using that theorem:
$\begin{align}7 = 3\cdot2 +1\\7 - 3\cdot2 = 1 \\-2\cdot3 + 7 = 1\end{align}$
$s=-2$ in the above equation so $-2$ is the inverse of $3 \pmod{7}$.
The book says that the next step to solve $3x \equiv 4 \pmod{7}$ is to multiply $-2$ on both sides.
By doing that I get:
$\begin{align}-2\cdot3x \equiv -2\cdot4 \pmod 7\\-6x\equiv -8 \pmod 7\end{align}$
What should I do after that?
I am working on this problem for hours.
Thanks :)
| $$3x\equiv4\pmod{7}\\-6x\equiv -8\pmod{7}\\-6x\equiv-1-7\\-6x\equiv-1\pmod{7}\\(7-6)x\equiv-1\equiv6\pmod{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
What can I conclude from the fact that I got a zeroes row?
Does this help solving the problem?
| In this particular case it'd perhaps be easier to understand what's going on performing column elementary operations:
$$\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}\stackrel{C_2-C_1\,,\,\,C_3-2C_1\,,\,C_4+C_1}\longrightarrow\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 1 & 3 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}\stackrel{C_3-\frac12C_2\,,\,C_4-\frac32C_2}\longrightarrow$$
$$\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}$$
The above means that
$$\begin{align}&C_3-2C_1-\frac12(C_2-C_1)=0\implies &C_3=\frac32C_1+\frac12C_2\\
&C_4+C_1-\frac32(C_2-C_1)=0\implies &C_4=-\frac52C_1+\frac32C_2\end{align}$$
and the above means that any vector $\;(x_\;x_2\;x_3\;x_4)^t\;$in the image of $\;A\;$ has to fulfill the same relations as shown, meaning:
$$\begin{align}&x_3=\;\;\;\,\frac32x_1+\frac12x_2\\&x_4=-\frac52x_1+\frac34x_2\end{align}$$
Now, does you right side matrix (formed with three row (column) vectors) fulfill this?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Finding $\tan \pi/8$ from $\sqrt{1+i}$. I want to prove that $\tan \pi/8 = \sqrt{2} - 1$ using $\sqrt{1+i}$ in some way.
Write: $\sqrt{1+i} = a+bi$, and let's find $a$ and $b$. We have: $$1+i = a^2+b^2 + 2abi,$$ so $a^2+b^2 = 1$ and $ab = 1/2$. This already seems wrong to me, because we know that $|1+i| = \sqrt{2} \implies |\sqrt{1+i}| = \sqrt[4]{2},$ regardless of what root we choose. Let's pretend everything is ok. Substituition gives $$a^2 + \frac{1}{4a^2} = 1 \implies a^4 - a^2 + \frac{1}{4} = 0 \implies a^2 = \frac{1}{2} \implies a = \pm \frac{\sqrt{2}}{2},$$ and $$b = \pm\frac{1}{2\frac{\sqrt{2}}{2}} = \pm\frac{\sqrt{2}}{2} = a.$$
And writing $1+i = re^{i((\pi/4)+2k\pi)}$, we get the roots are given by $\sqrt[4]{2}e^{i((\pi/8)+k\pi)}$.
Now, $z = x+iy \implies \tan \arg(x+iy) = y/x$ would give us $\tan \pi/8 = 1$!
Can someone give me a light here? It must be simple, but I'm failing to make sense of this. Thanks.
Yes, I know better ways to find $\tan \pi/8$. I want to do it this way.
| I've got it figured out. Silly mistake, as always. I should have $$1+i = a^2 - b^2 + 2abi,$$ as pointed by Omnomnomnom in the comments. Then: $$\begin{cases} a^2 - b^2 = 1 \\ ab = 1/2\end{cases} \implies a^2 - \frac{1}{4a^2} = 1 \implies a^4 - a^2 - \frac{1}{4} = 1,$$ so: $$a^2 = \frac{1\pm \sqrt{2}}{2} \implies a = \pm \sqrt{\frac{1+\sqrt{2}}{2}},$$ since $a$ must be real. Then: $$b = \pm\frac{1}{2\sqrt{\frac{1+\sqrt{2}}{2}}} \implies b = \pm\frac{1}{\sqrt{2}\sqrt{\sqrt{2}+1}}.$$
Finally, $$\tan \pi/8 = \frac{b}{a} = \pm\frac{1}{\sqrt{2}\sqrt{\sqrt{2}+1}}\frac{\pm \sqrt{2}}{\sqrt{\sqrt{2}+1}} = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\sin\theta +\sin3\theta =0$
Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$
Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta }{ 2 } \cos\frac { \theta -3\theta }{ 2 }\\
0&=2\sin\frac { 4\theta }{ 2 } \cos\frac { -2\theta }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta }{ 2 }\end{align}$$
| $$\sin(\theta) + \sin(3\theta) = 0$$
Using sum-to-product,
$$2\sin\left(\frac{\theta+3\theta}{2}\right)\cos\left(\frac{\theta - 3\theta}{2}\right) = 0$$
$$2\sin(2\theta)\cos(-\theta) = 0$$
$$2\sin(\theta+\theta)\cos(-\theta)= 0$$
Using either the double angle or sum-difference formulas,
$$4\sin(\theta)\cos(\theta)\cos(-\theta) = 0$$
By the definition of cosine and getting rid of the 4,
$$\sin(\theta)\cos^2(\theta) = 0$$
Using co-function identities,
$$\tag{1}\sin(\theta)\sin^2(\frac{\pi}{2} - \theta) = 0 $$
We know that $\sin$ is zero at $k\pi$ for $k \in \mathbb{Z}$ so,
$$\theta = k\pi$$ in which case the left-hand $\sin$ in (1) is 0 or, $$\theta = \frac{\pi}{2} + k\pi$$ in which case the right-hand $\sin$ in (1) is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Use of Poincare Lemma in solving $\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$ UPDATED
You are given the following statement of the Poincaré Lemma: If $\Phi_t$ is a one-parameter family of diffeomorphisms on $\mathbb R^n$ (not necessarily a subgroup) and $X_t$ the vector field defined by
$$X_t \circ \Phi_t = \frac{\partial}{\partial t}\,\Phi_t,$$
and if $\beta$ is a closed $k$-form on $\mathbb R^n$ such that
$$\Phi_1^* \beta = \beta, \quad \lim_{\epsilon \to 0} \Phi_\epsilon^* \beta = 0,$$
then $\beta = d\alpha$, where
$$\alpha=\int_0^1 \Phi_t^* i_{X_t}\beta\,dt.$$
Let $U = \mathbb{R}^3 \setminus \{(0,0,z) \}$ (i.e. $\mathbb{R}^3$ with the $z$-axis removed ) and consider $\beta$ on $U$ given by
$$\beta = \frac{x \,dy \wedge dz + y \,dz \wedge dx + z \,dx \wedge dy}{(x^2+y^2+z^2)^{3/2}}$$
One can show that $d\beta=0$.
Let $\Phi_t(x,y,z)=(x,y,tz)$. One can show that $\Phi_1^*\beta=\beta$, $\lim_{\epsilon \rightarrow 0} \Phi_{\epsilon}^*\beta=0$.
So use the Poincaré Lemma to find a vector field $\textbf{A}(\textbf{r})$ on $U$ such that
$$\nabla \times \textbf{A}(\textbf{r})=\frac{\textbf{r}}{r^3}$$
I believe that you just need to compute $\displaystyle \alpha=\int_0^1 \Phi_t^* i_{X_t}\beta \,dt$ as $\beta = d\alpha$.
$$
\begin{align}
\hat{\mathbb{X}}_t
&= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \hat{\Phi}_t^{-1} \\
&= \left(\frac{\partial}{\partial t}\hat{\Phi}_t\right) \left(x,y,\frac{z}{t}\right) \\
&=\left(0,0,z/t\right)
\end{align}$$
Now $\Phi_t \,dx =dx$, $\Phi_t\, dy =dy$ and $\Phi_t\, dz = tdz$.
So
$$
\begin{align}
i_{\hat{X}_t}\beta
&= \frac{x}{r^3}i_{\hat{X}_t}(dy \wedge dz)+\frac{y}{r^3}i_{\hat{X}_t}(dz \wedge dx)+\frac{z}{r^3}i_{\hat{X}_t}(dx \wedge dy) \\
&= \frac{x}{r^3}\left(\frac{-z}{t}dy\right)+\frac{y}{r^3}\left(\frac{z}{t}dx\right) \\
&= \frac{-zxt}{r^3}dy+\frac{zty}{r^3}dx
\end{align}
$$
So
$$
\begin{align}\Phi_t^*i_{\hat{X}_t}\beta
&= \Phi_t^* \left[\frac{-zxt}{r^3}dy+\frac{zty}{r^3}dx \right] \\
&= \frac{-(tz)xt}{(x^2+y^2+t^2z^2)^{3/2}}dy+\frac{(tz)ty}{(x^2+y^2+t^2z^2)^{3/2}}dx \\
&= \frac{-xzt^2}{(x^2+y^2+t^2z^2)^{3/2}}dy+\frac{yzt^2}{(x^2+y^2+t^2z^2)^{3/2}}dx
\end{align}
$$
I am not sure that this is correct as it doesnt look very integrable.
| There's a mistake in the last line in the calculation of $i_{\hat{X}_t}\beta$. The $t$'s in the numerators should be in the denominators:
$$i_{\hat{X}_t}\beta = \frac{-zx}{\color{red}{t}r^3}\, dy + \frac{zy}{\color{red}{t}r^3}\, dx = \frac{z}{\color{red}{t}r^3}(y\, dx - x\, dy).$$
Then
$$\Phi_t^*i_{\hat{X}_t}\beta = \frac{z}{(x^2 + y^2 + t^2 z^2)^{3/2}}(y\, dx - x\, dy).$$
Consequently, $$\alpha = I(x,y,z)(y\, dx - x\, dy),$$
where
$$I(x,y,z) = \int_0^1 \frac{z\, dt}{(x^2 + y^2 + t^2 z^2)^{3/2}}.$$
To evaluate $I(x,y,z)$, let $tz = \sqrt{x^2 + y^2}\tan\theta$. Then $z\, dt = \sqrt{x^2 + y^2}\sec^2\theta\, d\theta$, and thus
\begin{align}I(x,y,z) &= \int_0^{\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)} \frac{\sqrt{x^2+y^2}\sec^2\theta\, d\theta}{(x^2 + y^2)\sqrt{x^2 + y^2} \sec^3\theta}\\
&= \frac{1}{x^2 + y^2} \int_0^{\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)} \cos\theta\, d\theta\\
&= \frac{1}{x^2+y^2}\sin\left(\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)\\
&= \frac{z}{r(x^2+y^2)}.\end{align}
Therefore,
$$\alpha = \frac{yz}{r(x^2 + y^2)}\, dx - \frac{xz}{r(x^2 + y^2)}\, dy$$
and
$$\mathbf{A}(\mathbf{r}) = \left\langle \frac{yz}{r(x^2 + y^2)}, -\frac{xz}{r(x^2 + y^2)}, 0\right\rangle.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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How to find maximum value?
Let $R$ be the region $z \leq 1$. Compute the maximum value of $|z^2 + z + 2|$ in
$R$ and find out the point where this function reaches this value in $R$.
I let $$z = \cos \theta + i \sin \theta$$ Then $$|z^2 + z +2|^2 = (\cos 2 \theta + 2 \cos \theta)^2 + (\sin 2 \theta + 2 \sin \theta)^2$$
$$= \cos^2 2\theta + 4\cos^2 \theta+ 4 + 4\cos\theta \cos 2\theta+ 4\cos2\theta+8\cos\theta+\sin^2 2\theta+ 4\sin^2\theta+4\sin\theta\sin2\theta$$
I am having trouble simplifying this and going further to get $$8(\cos \theta + 3/4)^2 + 1/2$$ That's the answer(maximum value $5$, occurs at $\theta=0$) that is given. I would be grateful for any help to solve this question.
| By the maximum modulus principle, the maximum must occur on the boundary. By the triangle inequality, $|z^2+z+2|\leq|z|^2+|z|+2\leq4$. We can check $1^2+1+2=4$, so the function attains its maximum when $z=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$
Prove that $$\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$
I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$
And maybe taking it's derivative but it didn't work out well.
Any ideas?
| For a geometric series.
$$\sum_{n=0}^{\infty} (-1)^n(x^n) = \frac{1}{1+x}$$
Substitute $x \rightarrow x^3$
$$\sum_{n=0}^{\infty} (-1)^n (x^{3n}) = \frac{1}{1+x^3}$$
Integrate the sides.
$$\sum_{n=0}^{\infty} (-1)^n\frac{x^{3n + 1}}{3n + 1} = \int \frac{dx}{1+x^3}$$
The hard part is the integration. Then let $x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\ln \big( b + \sqrt{b^2 + c^2 + x^2}\,\big)$? I am looking to demonstrate the following result. Any ideas are much appreciated.
$$
\begin{align}\int \ln \left( b + \sqrt{b^2 + c^2 + x^2}\right) dx = &\;x \ln \left( b + \sqrt{b^2 +c^2 +x^2}\right) +b \ln \left(2x + 2\sqrt{ b^2 +c^2 +x^2} \right)\\&\; - c \arctan \left(\frac{ b x} { c \sqrt{b^2 +c^2 +x^2}}\right) + c \arctan \left(\frac{x}{c}\right) -x \end{align}$$
| From the first term at the right let's try integrations by parts :
\begin{align}
I(b,c)&:=\int \ln \left( b + \sqrt{b^2 + c^2 + x^2}\right) dx-x \ln \left( b + \sqrt{b^2 +c^2 +x^2}\right) \\
&= -\int x\frac{1}{b + \sqrt{b^2 +c^2 +x^2}}\frac{2\,x}{2\sqrt{b^2 +c^2 +x^2}}\,dx\\
&= \int \frac{{b - \sqrt{b^2 +c^2 +x^2}}}{c^2 +x^2}\frac{x^2}{\sqrt{b^2 +c^2 +x^2}}\,dx\\
&= b\int \frac{c^2+x^2-c^2}{(c^2 +x^2)\sqrt{b^2 +c^2 +x^2}}\,dx-\int\frac{x^2}{c^2 +x^2}\,dx\\
&= b\int \frac{dx}{\sqrt{b^2 +c^2 +x^2}}-\int \frac{b\,c^2}{(c^2 +x^2)\sqrt{b^2 +c^2 +x^2}}\,dx+\int\frac{c^2-(c^2+x^2)}{c^2 +x^2}\,dx\\
&= b\,\ln \left(2x + 2\sqrt{ b^2 +c^2 +x^2} \right) - c\,\arctan \frac{ b x} { c \sqrt{b^2 +c^2 +x^2}} + c\, \arctan \frac{x}{c} -x\\
\end{align}
The three last ones should be more direct !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$ Find the set of complex numbers $z$ which satisfy
$$\left\lvert\frac{z-3}{z+3}\right\rvert=2\text.$$
I need help on that one. Thank you.
| Putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$\left|\frac{z-3}{z+3}\right|=2\iff |(x-3)+iy|^2=4|(x+3)+iy|^2\iff$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2\iff$$
$$x^2+10x+y^2=-9\iff (x-5)^2+y^2=16$$
Thus, the set of complex numbers fulfilling the above condition is a circle of radius four.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
mechanics piston problem involving rotational motion.
The above figure shows a piston driving a crank OP pivoted at the end $O$.
The piston slides in a straight cylinder and the crank is made to rotate with constant angular velocity $ \omega $. Find the distance $OQ$ in terms of the lengths $b,c$ and the angle $\theta$. Show that, when $b/c$ is small, $OQ$ is given approximately by $OQ = c + b\cos(\theta)-\frac{b^2}{2c}\sin^2(\theta)$
I have sketched a little diagram that goes as follows;
$ \cos \theta = \frac{x}{b} $ this implies that $ b\cos\theta = x$
$ \sin \theta = \frac{h}{b} $ this implies that $ b\sin\theta = h$
now $ c^2 = h^2 + y^2 $ so $ c^2 - h^2 = y^2 $
now i am letting the length $OQ = z$.
$z = x + y = b\cos\theta + \sqrt{c^2 - b^2(\sin\theta)^2}$
Now I know I can manipulate this more. but I feel as though i am getting further and further away. I may have made a mistake, but it is basic trig?
| The key thing to remember is that the linking rods are rigid and must maintain their length throughout.
Put the origin at O. The point P is instantaneously $b\cos\theta, b\sin\theta$ which automatically satisfies $|OP|=b$. If $Q=(z,0)$, one must have $|PQ|=c$. This gives you
$$
(z-b\cos\theta)^2 + (b\sin\theta)^2 = c^2 \\
z^2 - 2bz \cos\theta + b^2-c^2 = 0
$$
Solving this quadratic for $z$ you have
$$
z = b\cos\theta \pm \sqrt{c^2-b^2\sin^2\theta} \\
= b\cos\theta \pm c\sqrt{1-\frac{b^2}{c^2}\sin^2\theta}
$$
You have to take the positive root since the length is b+c at $\theta=0$.
For small $b/c$, you can expand the square root term in a Taylor series as $(1+x)^{(1/2)} \approx 1 + \frac{x}{2}$, giving you
$$
z \approx b\cos\theta + c - \frac{b^2}{2c} \sin^2\theta
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to show that $3^x+4^x=5^x$ has only one solution? How to show that $3^x+4^x=5^x$ has only one solution? Thanks in advice.
| Define
$$f(x)=\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x.$$
It is clearly monotonic decreasing. Note also that
$$f(2)=1,$$
so that
$$x=2$$ is the only real solution to $f(x)=1$.
More generally, for the equation
$$3^x+4^x+5^x=6^x,$$
define
$$f(x)=\left(\dfrac{3}{6}\right)^x+\left(\dfrac{4}{6}\right)^x+\left(\dfrac{5}{6}\right)^x.$$
This function is also decreasing, and as Euler first noted,
$$3^3+4^3+5^3=6^3,$$
so that
$$f(3)=1,$$
and $x=3$ is the only real solution to $f(x)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$ \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x $ Here is a problem in calculus shared by a friend. Compute $$ \lim_{n \to \infty} \displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x. $$ It is not trivial to find the partial sum. Can we pull the limit inside to get $$ \displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{\infty} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x? $$Still, this is hard.
|
Lemma: $ \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 kx \, \mathrm{d}x = \frac {\pi}{4} $, for any integer $k$.
Proof. First note that $ \sin^2 (kx) = \frac {1 - \cos (2kx)}{2} $, so our integral is $$ \begin {align*} \frac {1}{2} \cdot \displaystyle\int_{0}^{\frac {\pi}{2}} \left( 1 - \cos (2kx) \right) \, \mathrm{d}x &= \frac {1}{2} \cdot \left[ x - \frac {\sin (2kx)}{2k} \right]_{0}^{\frac {\pi}{2}} \\&= \frac {1}{2} \cdot \left( \frac {\pi}{2} - \frac {\sin \left( \pi k x \right)}{2k} \right) \\&= \frac {1}{2} \cdot \left( \frac {\pi}{2} - 0 \right) \\&= \frac {\pi}{4}, \end {align*} $$as desired. $\Box$
$$ \begin {align*} \displaystyle\int_{0}^{\frac{\pi}{2}} \displaystyle\sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x &= \displaystyle\sum_{k=1}^{n} \displaystyle\int_{0}^{\frac{\pi}{2}} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x \\&= \displaystyle\sum_{k=1}^{n} \left( \frac {1}{k^2} \cdot \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 kx \, \mathrm{d}x \right) \\&= \frac {\pi}{4} \cdot \displaystyle\sum_{k=1}^{n} \frac {1}{k^2}. \end {align*} $$Taking the limit, our answer is $$ \begin {align*} \frac {\pi}{4} \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac {1}{k^2} &= \frac {\pi}{4} \cdot \displaystyle\sum_{k=1}^{\infty} \frac {1}{k^2} \\&= \dfrac {\pi}{4} \cdot \dfrac {\pi^2}{6} \\&= \boxed {\dfrac {\pi^3}{24}}. \end {align*} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Product of numbers in any two cells sharing a side is $2$ In a $3\times 3$ square, every cell has a positive number. The product of numbers in any two cells sharing a side is exactly $2$. What is the minimum sum of all the numbers?
We may color the cells in chessboard fashion, and write $a$ in black and $b$ in white cells, where $ab=2$. The sum of all the numbers is $5a+4b\geq 2\sqrt{20ab}=4\sqrt{10}$. Inequality holds when $5a=4b$ and $ab=2$. But how to show that it suffices to consider this configuration? (It makes sense since it's the extreme configuration, but I'm not sure how to prove it.)
| The point is once you fix a cell number, all the neighbours (and hence, all cells) are fixed.
Suppose then the center element is $a$. Then the four neighbours must be $b=2/a$. And then the corners must be $a$. Hence the total sum is $5a + 8/a$. You must find $a$ that minimizes this.
You can do that by analysis, or using AM-GM: $\frac{1+x^2}{2}\ge x$ with equality at $x=1$
$$5a + 8/a=\frac{8}{a} \left(\frac{5}{8}a^2 + 1\right)=\frac{8}{a}\left(\left(\sqrt{\frac{5}{8}}a\right)^2 + 1\right)\ge \frac{8}{a} 2 \,\sqrt{\frac{5}{8}}a= 4\sqrt{10}$$
And the minimum happens at $\sqrt{\frac{5}{8}}a=1 \implies a=\sqrt{\frac{8}{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer $m$ such that $2^m\equiv\pm 1\pmod{2n+1}$ Let $n$ be a positive integer. Does there always exist a positive integer $m\leq n$ such that $2^m\equiv\pm 1\pmod{2n+1}$?
It is true that $2^{\phi({2n+1})}\equiv 1\pmod{2n+1}$. If $2n+1$ is prime, the statement follows by taking $m=\phi(2n+1)/2$. But if $2n+1$ is composite, then it might be possible that $2^{\phi({2n+1})/2}$ is not $\pm 1\pmod{2n+1}$.
| Suppose that there exist positive integers $m_1<m_2\le n$ such that $2^{m_1}\equiv 2^{m_2}(\operatorname{mod} 2n+1)$. Then $2^{m_2-m_1}(2^{m_1}-1)$ is divisible by $2n+1$. Since $\operatorname{GCD}(2^{m_2-m_1},2n+1)=1$, we see that $2^{m_1}-1$ is divisible by $2n+1$. Now consider numbers $2^0,2^2,\dots, 2^n$. If $2^{m}\equiv 1 (\operatorname{mod} 2n+1)$ then the claim is proved. So assume the converse. Then residues $2^0,2^2,\dots, 2^n$ are mutually different modulo $2n+1$. Then residues $-2^0,-2^2,\dots, -2^n$ are mutually different modulo $2n+1$ too. Since each of these two set of residues contains $n+1$ element, and $2(n+1)>2n+1,$ by pigeonhole principle, there exist integers $0\le m_1<m_2\le n$ such that $-2^{m_1}\equiv 2^{m_2}(\operatorname{mod} 2n+1)$. Then $2^{m_2-m_1}(2^{m_1}+1)$ is divisible by $2n+1$. Since $\operatorname{GCD}(2^{m_2-m_1},2n+1)=1$, we see that $2^{m_1}+1$ is divisible by $2n+1$.
| {
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"url": "https://math.stackexchange.com/questions/1076277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality.such as Nesbitt Let $a,b,c >0 $ , prove that:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \leq \dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}$$
| Solution 1:
\begin{align*}&\sum\dfrac{a^2}{b^2+c^2}-\sum\dfrac{a}{b+c}=\sum_{cyc}\dfrac{ab(a-b)+ac(a-c)}{(b^2+c^2)(b+c)}\\
&=\sum_{cyc}\dfrac{ab}{(b^2+c^2)(b+c)}(a-b)+\dfrac{ac}{(b^2+c^2)(b+c)}(a-c)\\
&=\sum_{cyc}\left[\dfrac{ab}{(b^2+c^2)(b+c)}-\dfrac{ab}{(c+a)(c^2+a^2)}\right](a-b)\\
&=\sum_{cyc}\dfrac{(a+c)(a^2+c^2)-(b+c)(b^2+c^2)}{(a+c)(b+c)(a^2+c^2)(b^2+c^2)}(a-b)ab\\
&=\sum_{cyc}\dfrac{(a-b)(a^2+b^2+c^2+ab+bc+ac)}{(a+c)(b+c)(a^2+c^2)(b^2+c^2)}(a-b)ab\\
&=\sum_{cyc}\dfrac{ab(a-b)^2(a^2+b^2+c^2+ab+bc+ac)}{(a+c)(b+c)(b^2+c^2)(a^2+c^2)}\\
&\ge 0
\end{align*}
Solution 2:
let $$f(x)=\dfrac{a^x}{b^x+c^x}+\dfrac{b^x}{c^x+a^x}+\dfrac{c^x}{a^x+b^x}$$
then we have
\begin{align*}
f'(x)&=\sum_{cyc}\dfrac{a^x(b^x+c^x)\ln{a}-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}\\
&=\sum_{cyc}\dfrac{a^xb^x(a^x-b^x)(\ln{a}-\ln{b})(2c^x+a^x+b^x)}{(a^x+b^x)^2(b^x+c^x)^2}\\
&\ge 0
\end{align*}
$$f(2)\ge f(1)$$
By done!
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$
Prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$.
Let $\theta$ be a root of $f(x)$. Compute $\frac{1}{\theta}$ and $(2 + \theta^2)^{-1} $ in $\mathbb Q[\theta ]$.
\begin{array}{l}
f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\
\Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\
\Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\
\end{array}
\begin{array}{l}
\frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\
\Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\
\Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\
\Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\
\end{array}
But I can't show that $f$ is irreducible.
| Another idea is that if $f(x)$ is factorable over $\Bbb Q[x]$ it must have at least one rational zero. However, by the rational zero theorem, $\pm 1$ are the only possible rational zeros, but neither one is a zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\sum ( \cos \sqrt[3]{n^3 + \sqrt n + 7} - \cos \sqrt[3]{n^3 - 2\sqrt n + 3})$ I have some problem with this example: $$\displaystyle \sum_{n=2}^{\infty}\Bigg(\cos\Big(\sqrt[3]{n^3+\sqrt{n}+7}\Big) -\cos\Big(\sqrt[3]{n^3-2\sqrt{n}+3}\Big)\Bigg)$$
the only idea that crossed my mind is to use that $\cos x-\cos y=-2\sin\big({\frac{x+y}{2}}\big)\sin\big({\frac{x-y}{2}}\big)$ but later I don't know what to do with sines how to compare them or what else I can do with them ?
| Methodology
*
*Write $\cos x - \cos y = -2 \sin \frac{x+y}{2}\sin \frac{x-y}{2}$.
*Disregard the $-2$ and the $\sin \frac{x+y}{2}$, as $\lvert \sin \alpha \rvert \leq 1$
*Instead, focus on the $\sin \frac{x-y}{2}$ term. Intuitively, as $x \approx y$ when $n$ is large, we should expect this to get small, and therefore for this sine term to get small.
*Expand $x-y$ in power series to see that $x-y = \frac{1}{n^{3/2}} + \frac{4}{3n^2} + \dots$
*As $\sin x \approx x$ when $x$ is small, conclude that $\sin \frac{x-y}{2}$ behaves like $\frac{1}{2n^{3/2}}$
*Conclude that your series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can prove that $\sum_{n=1}^{\infty }\frac{\zeta (2n)}{4^{n-1}}(1-\frac{1}{4^n})=\frac{\pi }{2}$
$$\zeta (2)(1-\frac{1}{4})+\frac{\zeta (4)}{4}(1-\frac{1}{4^2})+\frac{\zeta (6)}{4^2}(1-\frac{1}{4^3})+...=\frac{\pi }{2}$$
The WolframAlph couldn't recognize the closed-form which is $\pi/2$ when I gave the series,
so I used the WolframAlph again to compute many terms of infinite series.
I think that the WolfarmAlph cannot say the value is $\pi/2$,So we need to prove it.
| We have:
$$\zeta(2n)\left(1-\frac{1}{4^n}\right)=\sum_{k=0}^{+\infty}\frac{1}{(2k+1)^{2n}}\tag{1}$$
hence:
$$\sum_{n=1}^{+\infty}\frac{\zeta(2n)}{4^{n-1}}\left(1-\frac{1}{4^n}\right)=4\sum_{n=1}^{+\infty}\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^{2n}}=4\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}\tag{2}$$
and the result follows from:
$$\sum_{k=0}^{+\infty}\frac{1}{(4k+2)^2-1}=\frac{1}{2}\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=\frac{1}{2}\arctan 1 = \color{red}{\frac{\pi}{8}}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
Solve $2(x+y)+xy=x^2+y^2$ where $x,y \in \mathbb{Z}$ Solve the equation: $$2(x+y)+xy=x^2+y^2$$
How should I go about solving this? Any guidance appreciated.
Thanks!
| Apply Cauchy-Schwarz inequality twice to both left and right sides of the equation:
$2(x+y) + \dfrac{(x+y)^2}{4} \geq LHS = RHS \geq \dfrac{(x+y)^2}{2} \Rightarrow 2(x+y) + \dfrac{(x+y)^2}{4} \geq \dfrac{(x+y)^2}{2} \Rightarrow 2(x+y) - \dfrac{(x+y)^2}{4} \geq 0 \Rightarrow (x+y)(8-(x+y)) \geq 0 \Rightarrow 8 \geq x+y \geq 0$. So there are a total of $9$ cases to consider. I will do a few ones and you do the rest.
Case 1: $x+y = 0 \Rightarrow x = -y \Rightarrow 2\cdot 0 + xy = x^2+y^2 \Rightarrow -x^2 = 2x^2 \Rightarrow 3x^2 = 0 \Rightarrow x = 0 = y$.
Case 2: $x+y = 1 \Rightarrow 2\cdot 1 + xy = (x+y)^2 - 2xy = 1 - 2xy \Rightarrow 3xy = -1 \Rightarrow$ no solution for this case.
etc...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $[2x]+[2y] \geq [x]+[y]+[x+y]$ Prove that $[2x]+[2y] \geq [x]+[y]+[x+y]$ whenever $x$ and $y$ are real numbers.
The $[]$ symbol is the greatest integer or floor function.
I have proved this fact by cases, but I stumbled upon what I believe to be another way to prove the above inequality, and I was wondering if my sequence of statements are legitimate.
I make use of two lemmas that I have proved
Lemma 1. If $x$ is a real number and m is an integer, then $[x+m] = [x]+m$.
Lemma 2. $\displaystyle [x]+\left[x+\frac{1}{2} \right] = [2x]$.
My proof begins with an "obvious" statement
$$x+y \leq x+y+1$$
I then take the floor of the inequality to get
$$[x+y] \leq [x+y]+1$$ (1)
which is true in virtue of lemma 1.
Furthermore, if I add the following statements
$$[x+1/2] \leq x + 1/2$$
$$[y+1/2] \leq y + 1/2$$
I procure
$$\left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2}\right] \leq x+y+1$$
which by definition of the floor function renders the equation
$$[x+y]+1 = \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$ (2)
Substituting (2) for (1), I have
$$[x+y] \leq \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right]$$
I then add $[x]$ and $[y]$ to the above inequality to produce
$$[x]+[y]+[x+y] \leq [x]+\left[x+\frac{1}{2} \right]+[y]+\left[y+\frac{1}{2} \right]$$
And in virtue of Lemma 2, the right hand side of the inequality becomes
$$[x]+[y]+[x+y] \leq [2x]+[2y].$$
I personally don't see anything wrong, except maybe for the implication made to establish (2).
Solving these kinds of problems is solely for personal gratification, so I will greatly appreciate feedback.
Thanx.
| Let $x=m+\theta_1$ and $y=n+\theta_2$ where $n,m\in\Bbb{Z}$ and $0\le\theta_1,\theta_2<1$
Case I: $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12$
Now \begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor\\
&=2m+2n\\
\end{align} since $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12\implies 0\le2\theta_1<1$ and $0\le2\theta_2<1$
and \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n)\\
&=2m+2n,
\end{align} since $0\le\theta_1<\dfrac12$ and $0\le\theta_2<\dfrac12\implies 0\le\theta_1+\theta_2<1$
$\color{red}{So\quad \lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor=\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
Case II: $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1$
\begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor=2m+(2n+1),
\end{align} since $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1\implies 0\le2\theta_1<1$ and $1\le2\theta_2<2$
and since $0\le\theta_1<\dfrac12$ and $\dfrac12\le\theta_2<1\implies \dfrac12\le\theta_1+\theta_2<\dfrac32$ \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n)=2m+2n, \text{if}\quad\dfrac12\le\theta_1+\theta_2<1\\
&=2m+2n+1, \text{if}\quad 1\le\theta_1+\theta_2<\dfrac32
\end{align}
$\color{red}{So \quad\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor\le\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
Case III: $\dfrac12\le\theta_1<1$ and $0\le\theta_2<\dfrac12$. $\color{blue}{\textbf{[Same as Case II]}}$
Case IV: $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1$
\begin{align}
\lfloor 2x\rfloor +\lfloor 2y\rfloor&=\lfloor 2m+2\theta_1\rfloor+\lfloor 2n+2\theta_2\rfloor=(2m+1)+(2n+1)=2m+2n+2,
\end{align} since $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1\implies 1\le2\theta_1<2$ and $1\le2\theta_2<2$
and since $\dfrac12\le\theta_1<1$ and $\dfrac12\le\theta_2<1\implies 1\le\theta_1+\theta_2<2$ \begin{align}
\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor &=\lfloor m+\theta_1\rfloor+\lfloor n+\theta_2\rfloor+\lfloor m+n+\theta_1+\theta_2\rfloor\\
&=m+n+(m+n+1)=2m+2n+1\\
\end{align}
$\color{red}{So \quad\lfloor x\rfloor +\lfloor y\rfloor+\lfloor x+y\rfloor\le\lfloor 2x\rfloor +\lfloor 2y\rfloor}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it.
| As $x\to\infty$ $$\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}=\displaystyle \frac{4}{\sqrt{4+\dfrac{5}{x}} + \sqrt{4+\dfrac{1}{x}}}\to1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Stars & Bars Question: Identical Balls in Distinct Boxes I am terrible at combinatorics so any and all help would be appreciated.
20 identical balls are put into 10 distinct boxes so that at most 3 boxes are empty. In how many ways can this be done?
Thanks
| We can split this into $4$ cases based on the number of empty boxes, and use stars and bars for each case.
Case 1: No empty boxes.
We place $1$ ball in each box, and we imagine $9$ dividers and the remaining $10$ balls, so the number of possible ways to arrange the $19$ objects is $$\binom{19}{9}$$
Case 2: One empty box.
There are $10$ ways to choose the empty box. We place $1$ ball in each of $9$ boxes, and imagine $8$ dividers with the remaining $11$ balls, so the number of ways to arrange the $19$ objects is $$10\binom{19}{8}$$
Case 3: Two empty boxes.
There are $\binom{10}{2}=45$ ways to choose which boxes are empty. We place $1$ ball in each of $8$ boxes, and imagine $7$ dividers with the remaining $12$ balls, which gives $$45\binom{19}{7}$$
Case 4: Three empty boxes.
There are $\binom{10}{3}=120$ ways to choose which boxes are empty. We place $1$ ball in each of $7$ boxes, and imagine $6$ dividers with the remaining $13$ balls, which gives $$120\binom{19}{6}$$
Our final answer is
$$120\binom{19}{6}+45\binom{19}{7}+10\binom{19}{8}+\binom{19}{9}=\boxed{6371498}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed form for series $\sum_{m=1}^{N}m^n\binom{N}{m}$ How can we calculate the series
$$
I_N(n)=\sum_{m=1}^{N}m^n\binom{N}{m}?
$$
with $n,N$ are integers.
The first three ones are
$$
I_N(1)=N2^{N-1}; I_N(2)=N(N+1)2^{N-2}; I_N(3)=N^2(N+3)2^{N-3}
$$
| Suppose we are trying to evaluate
$$I_N(n) = \sum_{m=1}^N {N\choose m} m^n.$$
Observe that
$$m^n = \sum_{q=0}^m {n\brace q} \frac{m!}{(m-q)!}.$$
Note also that ${n\brace 0} = 0$ so that we get for the sum
$$\sum_{m=0}^N {N\choose m}
\sum_{q=0}^m {n\brace q} \frac{m!}{(m-q)!}.$$
Re-write this as
$$\sum_{q=0}^N {n\brace q}
\sum_{m=q}^N {N\choose m} \frac{m!}{(m-q)!}$$
or
$$\sum_{q=0}^N {n\brace q} \times q! \times
\sum_{m=q}^N {N\choose m} {m\choose q}.$$
We now see by inspection (i.e. considering subsets of size $q$ of $N$
elements) that the inner sum can be simplified to give
$$\sum_{q=0}^N {n\brace q} \times q! \times
{N\choose q} 2^{N-q}.$$
Now it remains to show how to compute the Stirling numbers for fixed
$q.$ Recall the marked species of set partitions
$$\mathfrak{P}(\mathcal{U}(\mathfrak{P}_{\ge 1}(\mathcal{Z})))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1))$$
and hence
$${n\brace q} = n! [z^n] \frac{(\exp(z)-1)^q}{q!}.$$
Suppose we wanted to compute $I_N(3).$
We get
$${n\brace 1} = n! [z^n] \frac{\exp(z)-1}{1!}
= n! \frac{1}{n!} = 1$$
and
$${n\brace 2} = n! [z^n] \frac{(\exp(z)-1)^2}{2!}
= \frac{n!}{2!} \times
\left(\frac{2^n}{n!}-2\frac{1}{n!}\right)
= 2^{n-1} - 1.$$
and finally
$${n\brace 3} = n! [z^n] \frac{(\exp(z)-1)^3}{3!}
= \frac{n!}{3!} \times
\left(\frac{3^n}{n!} - 3\frac{2^n}{n!}
+ 3\frac{1}{n!}\right)
\\ = \frac{1}{6} 3^n
- \frac{1}{2} 2^n + \frac{1}{2}.$$
This gives for $I_N(3)$ the expression
$${N\choose 1} 2^{N-1}
+ (2^2-1) \times 2 \times {N\choose 2} 2^{N-2}
+ \left(\frac{1}{6} 3^3 - \frac{1}{2} \times 2^3 + \frac{1}{2}\right)
\times 6 \times {N\choose 3} 2^{N-3}.$$
This is
$$2^{N-3} \times
\left(4N + 6N(N-1)+ N(N-1)(N-2)\right).$$
which simplifies to
$$I_N(3) = N^2\times (N+3)\times 2^{N-3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Trigonalise a matrix
Let $\mathbf{A}=\begin{bmatrix}
0 & 1 & 1 \\
-1 & 1 & 1 \\
-1 & 1 & 2
\end{bmatrix}$
*
*Trigonalise a matrix
Could someone trigonalise this matrix in details way, i wanna see the difference between yours and the french one
to start here's $P_{1}(\lambda)=(\lambda -1)^3$
note that some authors in book's frnech use :
$$T = \left( \begin{array}{ccc} \lambda_{1} & 1 & 0 \\ 0 & \lambda_{2} & 1 \\ 0 & 0 & \lambda_{3} \\ \end{array} \right)$$
and other use $$T = \left( \begin{array}{ccc} \lambda_{1} & a & b \\ 0 & \lambda_{2} & c \\ 0 & 0 & \lambda_{3} \\ \end{array} \right)$$
im wondering wich way that you using in usa english's books
| From @Ian's comment below, the author appears to be saying to "make triangular by similarity transformations", which gives something of the form (or like you added to the comment).
$$T = \left( \begin{array}{ccc} \lambda_{1} & a & b \\ 0 & \lambda_{2} & c \\ 0 & 0 & \lambda_{3} \\ \end{array} \right)$$
We try two things: $1.)~$ Diagonalize when possible, and if that fails, $2.)~$ Find the Jordan Normal Form.
For this specific problem, we have:
$$\mathbf{A}=\begin{bmatrix}
0 & 1 & 1 \\
-1 & 1 & 1 \\
-1 & 1 & 2
\end{bmatrix}$$
We find the characteristic polynomial and eigenvalues using $|A - \lambda I| = 0$, resulting in:
$$-\lambda ^3+3 \lambda ^2-3 \lambda +1 = 0 \implies -(-1 + \lambda)^3 = 0 \implies \lambda_{1, 2, 3} = 1$$
Next, we want to find three linearly independent eigenvectors using $[A- \lambda I]v_i = 0$, but this is not always possible due to algebraic and geometric difference (deficient matrices), so we have to resort to generalized eigenvectors.
So, we have $[A - \lambda_1 I]v_1 = [A -I]v_1 = 0$ as:
$$\left(
\begin{array}{ccc}
-1 & 1 & 1 \\
-1 & 0 & 1 \\
-1 & 1 & 1 \\
\end{array}
\right)v_1 = 0$$
If we put that in row-reduced-echelon-form (RREF), we arrive at:
$$\left(
\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right) v_1 = 0$$
This only provides one linearly independent eigenvector as:
$$v_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$
Now, we need to find two more generalized eigenvectors and there are many approaches to that. One approach is to try $[A - \lambda I]v_2 = v_1$, yielding an augmented RREF of:
$$\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
This yields $a = c, b = 1$, so choose $c = 0$, thus:
$$v_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$
Repeating this process, we set up and solve $[A - I]v_3 = v_2$, yielding an augmented RREF of:
$$\begin{pmatrix} 1 & 0 & -1 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$
This yields $a = -1 + c, b = -1$, so choose $c = 0$, thus:
$$v_3 = \begin{pmatrix} -1 \\ -1 \\ 0 \end{pmatrix}$$
We can now form $P = (v_1~|~v_2~|~v_3)$ as:
$$P = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 1 & 0 & 0 \end{pmatrix}$$
This yields the Jordan Normal Form:
$$T = PAP^{-1} = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)$$
Note: there are various ways to solve these problems.
*
*You want to look for references with the Jordan Normal Form.
*You can sometimes infer $T$ (see references below).
Some sources for generalized eigenvectors and the Jordan Normal Form are:
*
*Wiki
*Notes 1
*Notes 2
*Jordan Normal Form and Chaining
*Book - Abstract Algebra by Dummit and Foote
*Books - many books on linear algebra
| {
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"url": "https://math.stackexchange.com/questions/1083516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Derivative of absolute value of $|x^5|$
Differentiate $|x^5|$.
I know the formula for the derivative of absolute value but I can't seem to apply it to get $5x|x^3|$.
| Note: $x^5$ can be simplified to $x^4\cdot|x|$
Apply the product rule: $\frac{\text{d}}{\text{d}x}(x^4\cdot|x|)$:
$$\frac{\text{d}}{\text{d}x}(x^4\cdot|x|)=$$
$$\frac{\text{d}}{\text{d}x}(x^4)\cdot|x|+x^4\cdot\frac{\text{d}}{\text{d}x}(|x|)=$$
$$4x^3\cdot|x|+x^4\cdot\frac{x}{|x|}=$$
$$4x^3\cdot|x|+\frac{x^5}{|x|}=$$
Since $|x^2|=x^2$
$$4x^3\cdot|x|+\frac{x^3\cdot|x^2|}{|x|}=$$
$$4x^3\cdot|x|+x^3\cdot|x|=$$
$$5x^3\cdot|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve this equation $\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$
Solve this equation
$$\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$$
This equation has solutions $x=\pm 1$ but I cannot find a solution.
Do you think using an inequality will is the way forward?
| $$\frac{1}{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}=\frac{7x^2+1}{4x^3+4x}$$
$$\sqrt{x^2-1}=\frac{4x^4-3x^2-1}{4x^3+4x}=\frac{(4x^2+1)\cdot(x-1)\cdot(x+1)}{4x\cdot(x^2+1)}$$
$$\sqrt{x^2-1}\geq0,\space so\space\space x \in [-1,0] \cup \{1\}$$
$$x^2-1=(x-1)\cdot(x+1)=\frac{(4x^2+1)^2\cdot(x-1)^2\cdot(x+1)^2}{16x^2\cdot(x^2+1)^2}$$
$$(x-1)\cdot(x+1)\cdot(\frac{(4x^2+1)^2\cdot(x-1)\cdot(x+1)}{16x^2\cdot(x^2+1)^2}-1)=0$$
$$\rightarrow x \in \{-1,1\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrals of Trigonometry functions Show that
$$\int^\pi_0 \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx =\int^\pi_0 \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\,dx$$
I tried to use to check if
$$\int^\pi_0\left(\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}} - \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}+\sqrt{1-\cos x}}\right)\,dx=0$$
but it didn't turn out well.
Please help.
| Your idea actually works; you just have to pursue it. Abbreviating $\cos x$ to $c$ (and $\sin x$ to $s$), we have
$${\sqrt{1+c}-\sqrt{1-c}\over\sqrt{1+c}+\sqrt{1-c}}={(\sqrt{1+c}-\sqrt{1-c})^2\over(1+c)-(1-c)}={1+c-2\sqrt{1-c^2}+1-c\over2c}={1-s\over c}={c\over1+s}$$
(NB: $\sqrt{1-c^2}=s$ has the correct sign, since $s=\sin x\ge0$ for $0\le x\le\pi$.) The substitution $u=1+\sin x$, $du=\cos x\ dx$ gives
$$\int_0^\pi{\sqrt{1+\cos x}-\sqrt{1-\cos x}\over\sqrt{1+\cos x}+\sqrt{1-\cos x}}dx=\int_0^\pi{\cos x\ dx\over1+\sin x}=\int_1^1{du\over u}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Find Minimum Value of: $P=x^2+y^2+2xy+2-\frac{1}{xy}$ Given: $x,y>0$ and $x^2y-x+xy^2-y-3xy=0$
Find Minimum Value of:
$P=x^2+y^2+2xy+2-\frac{1}{xy}$
I found $\min P =\frac{71}{4}$ at $x=y=2$ but I cant prove that
Could some one help me ?
| here is another way which is slightly different from Emil's.
$a=x+y,b=xy \implies 4b\le a^2 \iff 0<4a=4b(a-3) \le a^2(a-3) \iff 4\le a^2-3a \iff (a+1)(a-4)\ge 0 \iff a \ge 4, \\ P=a^2+\dfrac{3}{a}+1 $
note $f(t)=t^2+\dfrac{3}{t} =t(t-3)+3(t+\dfrac{1}{t})$ is mono increase function when $t>3$ because :
$t(t-3)$ is mono increasing function when $t>3$,
$t+\dfrac{1}{t}$ is mono increasing function when $t>1$
(indeed $f(t) \ge 3(t^2*(\dfrac{3}{2t})^2)^{\frac{1}{3}}=3(\dfrac{3}{2})^{\frac{2}{3}}, f(t) $is mono increasing function when $t \ge (\dfrac{3}{2})^{\frac{1}{3}}$)
$P_{min}=P(a=4)=4^2+\dfrac{3}{4}+1$
| {
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"url": "https://math.stackexchange.com/questions/1087317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum $\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$ A series is given as follows
$$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$
Can you give me hints to get started finding its value? Thanks.
| Consider the binomial expansion of $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$
Multiplying the series term by term by $2$
On comparing, $nx=\frac{1}{3}$ and $\frac{n(n-1)}{2}x^2=\frac{5}{3.12}$
Solving for $n,x$ we get $n=\frac{-2}{3},x=\frac{-1}{2}$
So the sum becomes $\{(\frac{1}{2})^{\frac{-2}{3}}-1\}=2^\frac{2}{3}-1=4^\frac{1}{3}-1$
since $2$ has been multiplied before divide the result by $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Deriving inverse matrix formula If matrix $A$ is given with dimensions $2 \times2 $ then,
A is invertible if, and only, if $ad - bc \neq 0$:
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
How can this be derived?
I just need hints, I am not good at properties etc.. of matrices, please help!
| 2*2 Matrices inverse proof
As $A \times A^{-1}=I$
$\begin{bmatrix} x_{11} &x_{12} \\ x_{21}&x_{22} \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$
$ax_{11}+cx_{12} = 1$
$ax_{21}+cx_{22} = 0$
$bx_{11}+dx_{12} = 0$
$bx_{21}+dx_{22} = 1$
$b(ax_{11}+cx_{12}) = abx_{11}+bcx_{12}=b$
$a(bx_{11}+dx_{12}) = abx_{11}+adx_{12}=0$
$(abx_{11}+adx_{12})-(abx_{11}+bcx_{12})=-b$
$x_{12}(ad-bc)=-b$
Row 1 Column 2 : $x_{12} =$ $-b\over{ad-bc}$
$b(ax_{21}+cx_{22}) = abx_{21}+bcx_{22}=0$
$a(bx_{21}+dx_{22}) = abx_{21}+adx_{22}=a$
$(abx_{21}+adx_{22})-(abx_{21}+bcx_{22})=a$
$x_{22}(ad-bc)=a$
Row 2 Column 2 : $x_{22} = $$a \over ad-bc$
$d(ax_{11}+cx_{12}) = adx_{11}+cdx_{12}=d$
$c(bx_{11}+dx_{12}) = bcx_{11}+cdx_{12}=0$
$(adx_{11}+cdx_{12})-(bcx_{11}+cdx_{12})=d$
$x_{11}(ad-bc)=d$
Row 1 Column 1 : $x_{11} = $$d\over ad-bc$
$d(ax_{21}+cx_{22}) = adx_{21}+cdx_{22}=0$
$c(bx_{21}+dx_{22}) = bcx_{21}+cdx_{22}=c$
$(adx_{21}+cdx_{22})-(bcx_{21}+cdx_{22})=-c$
$x_{21}(ad-bc)=-c$
Row 2 Column 1 : $x_{21} = $$-c \over ad-bc$
As $x_{11}=$$d \over ad-bc$, $x_{12}=$$-b \over ad-bc$, $x_{21}$$=-c \over ad-bc $and $x_{22}=$$a \over ad-bc$
$$
A^{-1}= \begin{bmatrix} \frac{d}{detA} & \frac{-b}{detA}\\ \frac{-c}{detA}& \frac{a}{detA} \end{bmatrix} =
\frac{1}{detA}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}
$$
My first reply was unreadable due to lack of formatting, hope this new revision could help!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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How to compute the following integral $\int{\frac{x^2+d^2}{\sqrt{(x^4+b^2x^2+c^2)^3}}\mathrm dx}\,$? For some time, I have been struggling with the following integral:
$$\int{\frac{x^2+d^2}{\sqrt{\left(x^4+b^2x^2+c^2\right)^3}}\mathrm dx}\;,$$
where $b^4-4c^2>0$.
I did my best, but I am still far away from solving this problem. Does anyone have a clue how to compute it?
I would be very grateful for any hint.
| Hint for first step: This just describes how the initial four parameter problem can simplified to a three parameter one. I find that in solving these types of multi-parameters problems, this should almost always be your first step.
The integral in question may be written as a definite integral with variable lower limit of integration:
$$\mathcal{I}{\left(b,c,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x,$$
where $b,c,d,u\in\mathrm{R}$ with $b^4-4c^2>0$. For simplicity, we shall assume that the three parameters $b,c,d$ are positive.
Given $c>0$, we can rescale the other three parameters and the integration variable by $\sqrt{c}$, i.e.,
$$\begin{cases}
x=y\sqrt{c},\\
\frac{b}{\sqrt{c}}=:\sqrt{2}\,\beta,\\
\frac{d}{\sqrt{c}}=:\delta,\\
\frac{u}{\sqrt{c}}=:\eta,\\
\end{cases}$$
and in the process reduce the original four parameter integral to an integral with one fewer free parameters:
$$\begin{align}
\mathcal{I}{\left(b,c,d,u\right)}
&=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x\\
&=\int_{u/\sqrt{c}}^{\infty}\frac{cy^2+d^2}{(c^2y^4+b^2cy^2+c^2)^{3/2}}\sqrt{c}\,\mathrm{d}y\\
&=c^{-3/2}\int_{\eta}^{\infty}\frac{y^2+\delta^2}{(y^4+2\beta^2y^2+1)^{3/2}}\,\mathrm{d}y\\
&=c^{-3/2}\mathcal{I}{\left(\sqrt{2}\,\beta,1,\delta,\eta\right)}.\\
\end{align}$$
That is, it is sufficient to solve,
$$\tilde{\mathcal{I}}{\left(b,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+2b^2x^2+1)^{3/2}}\,\mathrm{d}x;~~\text{where }b>1.$$
More substantial hint:
Given two real parameters $u,b\in\mathbb{R}$ such that $0\le u$ and $1<b$, define the two-variable function $f{\left(u,b\right)}$ by the definite integral,
$$f{\left(u,b\right)}:=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}.$$
We'll find it convenient to additionally define the auxiliary parameters,
$$\begin{cases}
\varphi:=\arcsin{\left(\frac{1-u^2}{u^2+1}\right)};\\
\kappa:=\sqrt{\frac{b^2-1}{1+b^2}}.\\
\end{cases}$$
Note in particular that, for $0\le u$, we have
$$-1<\frac{1-u^2}{u^2+1}\le 1,$$
and for $1<b$, we have
$$0<\frac{b^2-1}{1+b^2}<1;$$
which guarantee that our parameters $\varphi$ and $\kappa$ are well-defined real numbers.
The integral $f{\left(u,b\right)}$ can be reduced to elliptic integrals of the first kind with two substitutions, $x=\frac{\sqrt{1-y}}{\sqrt{y}}\implies y=\frac{1}{x^2+1}$ and $y=\frac{z+1}{2}\implies z=2y-1$:
$$\begin{align}
f{\left(u,b\right)}
&=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}\\
&=\int_{\frac{1}{u^2+1}}^{0}\frac{\left(-\frac{y^{-3/2}}{2\sqrt{1-y}}\right)\,\mathrm{d}y}{\sqrt{\left(\frac{1-y}{y}\right)^2+2b^2\left(\frac{1-y}{y}\right)+1}}\\
&=\frac12\int_{0}^{\frac{1}{u^2+1}}\frac{\mathrm{d}y}{\sqrt{y(1-y)}\sqrt{1+2(b^2-1)y(1-y)}}\\
&=\frac{\sqrt{2}}{2\sqrt{1+b^2}}\int_{-1}^{\frac{1-u^2}{u^2+1}}\frac{\mathrm{d}z}{\sqrt{1-z^2}\sqrt{1-\left(\frac{b^2-1}{1+b^2}\right)z^2}}\\
&=\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}\\
&=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{0}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\
&=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\
&=\frac{K{\left(\kappa\right)}+F{\left(\varphi,\kappa\right)}}{\sqrt{2}\sqrt{1+b^2}}.\\
\end{align}$$
Now try to find a way to express $\mathcal{I}$ as a combination of derivatives of $f$.
| {
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Find integer solution to the given system of equation. find all positive integers $a,b,c$ such that
$abc=24$
$ab+bc+ca=38$
If particular values are given then we can easily find the solution but I am searching for some short general method. Is there any sufficient condition for the constant values to have integer solution?
| $ab+bc+ca=38 \Rightarrow a^2(b+c)+24=38a$. From the function $f(a)=a^2(b+c)-38a+24$ we get $38^2 \ge 4\cdot 24\cdot (b+c) \Rightarrow 15 \ge (b+c)$. You have now 13 values for b+c but not for every value will $38^2-4\cdot 24\cdot (b+c)$ be a perfect square and we also need $a_{1,2}=\frac{38 \pm \sqrt{38^2-4\cdot 24\cdot (b+c)}}{2(b+c)}$ to be an integer. We always get a and (b+c), but b and c can be calculated using $bc=\frac{24}a$. You will get (1,2,12) and his permutations as the only solution.
Note: we see from the solution that $a+b+c=15$ is always constant, so proving that would give us an even more simple solution bcs then we know that a,b,c are the solutions to the equasion $x^3-15x^2+38x-24=(x-1)(x-2)(x-12)=0$
This is as much as I know, if there is a faster solution that can covert them all please let me know.
| {
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Find the set of all $\alpha$ such that Matrix A is invertible and calculate the inverse for all $\alpha$ $A=\begin{pmatrix}
0 & 1 & -1 & 2\\
2 & -1 & 3 & 0 \\
\alpha & 0 & 1& 0 \\
3 & -1 &4 & 0
\end{pmatrix}$
I know that a Matrix is invertible only if $det(A)\not=0$
$det(A)=0\begin{vmatrix}
-1 & 3 & 0 \\
0 & 1 & 0 \\
-1 & 4 & 0 \\
\end{vmatrix}-1\begin{vmatrix}
2 & 3 & 0 \\
\alpha & 1 & 0 \\
3 & 4 & 0 \\
\end{vmatrix}-1\begin{vmatrix}
2 & -1 & 0 \\
\alpha & 0 & 0 \\
3 & -1 & 0 \\
\end{vmatrix}-2\begin{vmatrix}
2 & -1 & 3 \\
\alpha & 0 & 1 \\
3 & -1 & 4 \\
\end{vmatrix}$
$=-[2\begin{vmatrix}
1 & 0\\
4 & 0\\
\end{vmatrix}-3\begin{vmatrix}
\alpha & 0\\
3 & 0\\
\end{vmatrix}]-[2\begin{vmatrix}
0 & 0\\
-1 & 0\\
\end{vmatrix}+\begin{vmatrix}
\alpha & 0\\
3 & 0\\
\end{vmatrix}]-2[2\begin{vmatrix}
0 & 1\\
-1 & 4\\
\end{vmatrix}+\begin{vmatrix}
\alpha & 1\\
3 & 4\\
\end{vmatrix}+3\begin{vmatrix}
\alpha & 0\\
3 & -1\\
\end{vmatrix}]$
$=-2(2+4\alpha-3-3\alpha)=-4-8\alpha+6+6\alpha=-2\alpha+2$
$-2\alpha+2=0 \iff \alpha=1$
So the matrix is invertible for all $\alpha \in \Bbb R| \alpha\not=0$
But how do I calculate the inverse of A for ALL other $\alpha$ without spending the rest of my life on this question? Am I missing something here?
Thanks in advance?
| The determinant of $A$ is given by $\det(A)=2(1-a)$, and the inverse, for all $a\neq 1$, is given by
$$
A^{-1}=
\frac{1}{\det(A)}\begin{pmatrix} 0 & -2 & -2 & 2 \cr
0 & 2(4a-3) & 2 & 2(2-3a) \cr
0 & 2a & 2 & -2a \cr
1-a & 3(1-a) & 0 & 2(a-1)
\end{pmatrix}.
$$
You don't need to spend the rest of your life for its computation. For example, the Gauss elimination with $A$ in parallel with the identity matrix, gives you the result fairly easy. Also, it is useful to get familiar with a computer algebra system "early in life". Almost all people I know do no longer compute inverse matrices of size $4$ by hand.
| {
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Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$
If $x>0$, $y>0$, $z>0$ and $xyz = 1$ then
$$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$
I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution,
$\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$
So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$
Squaring again does not lead anywhere nice. Is the last inequality true ? If not is there a better way of showing the result ?
| To use Lagrange multipliers, we can define
$$L = \sqrt2(x+y+z) -(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}) - \lambda(xyz-1)$$
Setting the partial derivatives to zero, e.g. $\dfrac{\partial L}{\partial x} = \sqrt2-\dfrac{x}{\sqrt{x^2+1}} - \lambda yz = 0 $ easily gives using the constraint,
$$\lambda = \sqrt2 x - \frac{x^2}{\sqrt{x^2+1}} = \sqrt2 y - \frac{y^2}{\sqrt{y^2+1}}= \sqrt2 x - \frac{z^2}{\sqrt{z^2-1}}$$
$\sqrt2 t - \dfrac{t^2}{\sqrt{t^2 + 1}}$ is strictly increasing $\implies x = y = z = 1$ for the only extremum, which is easily seen to be a minimum, from which the inequality follows.
| {
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I have a Hard time solving this system of nonlinear equations $x^2+y^2-z^2=20$
$x^4+y^4-z^4=560$
$x^3+y^3+z^3=3xyz$
I know the fact that if $x^3+y^3+z^3=3xyz$ then $x+y+z=0$ (coming from Euler's identity) and first equation can be written as
$(x+y-z)^2-2(xy-xz-yz)=20$ and since $x+y=-z$ then $(xy-xz-yz)=2z^2-10$.
And the second equation becomes
$(x^2+y^2-z^2)^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$.
Since we know $x^2+y^2-z^2=20$, therefore $20^2-2(x^2y^2-2x^2z^2-2y^2z^2)=560$ or
$(x^2y^2-2x^2z^2-2y^2z^2)=80$.
Now I am stuck! Help please!
| This is a well-knonw example for solving a system of polynomial equations via Groebner bases. The real solutions are given by (check this yourself):
$$
(x,y,z)=(5,-2,-3),\; (-2,5,-3),\;(2,-5,3),\;(-5,2,3).
$$
Over the complex numbers, there are another $8$ solutions for $z$, i.e., the roots of the polynomial
$$
9z^8 - 240z^6 + 1600z^4 + 57600z^2 + 230400=0,
$$
and $y=-\frac{1}{120(z^2 + 8)}(120xz^2 + 960x + 3z^5 - 280z^3 - 1440z)$, $x^2 =z^2- y^2 +20$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a^{-1}ba=b^2$ and $|a|=3$, find $|b|$.
Suppose $a$ and $b$ are group elements and $b\neq e$. If $a^{-1}ba=b^2$ and $|a|=3$, find $|b|$.
$b$ is also not of order $2$, otherwise $a^{-1}ba=e$, then $b=e$ contradiction.
$a^{-1}ba=b^2\Rightarrow (ab)^{3}=e$
also
$a^2ba=b^2\Rightarrow ab=a^2b^2a^2$
This leads to nowhere.
Do you know a useful manipulation ?
| Squaring the given relation, we get
$$
a^{-1}b^2a = b^4
$$
We know that $b^2 = a^{-1}ba$, so inserting that, we get
$$
a^{-2}ba^2 = b^4
$$
Squaring again, and applying the same trick, gives us
$$
a^{-3}ba^3 = b^8
$$
Since $|a| = 3$, this is the same as $b^8 = b$. Therefore $b^7 = e$, which means that since $b \neq e$, we must have $|b| = 7$.
| {
"language": "en",
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Find the limit of $\lim_{n\rightarrow\infty}(\frac{1}{2}+\frac{3}{2^2}+...+\frac{2n-1}{2^n})$ Can anybody help me with this limit?
$$\lim_{n\rightarrow\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right)$$
Abel's transform wouldn't work on this one I guess, because another serie diverges
As this contains a geometric series with ratio $1/2$, I had multiplied whole series by $1/2$ and then subtracted from original, but got original series back. Any ideas how can the limit be calculated?
| $S \ \ = \ \ \ \ \ \ \ \dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{5}{2^3}+\dfrac{7}{2^4}+\cdots$.
$2S = 1+\dfrac{3}{2}+\dfrac{5}{2^2}+\dfrac{7}{2^3} + \dfrac{9}{2^4} + \cdots$
Now, subtract the first equation from the second, but match up terms with the same denominator:
$S = 1+ \dfrac{2}{2} + \dfrac{2}{2^2} + \dfrac{2}{2^3} + \cdots$
This becomes $1$ plus a geometric series, which is easy to evaluate.
| {
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If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem :
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$
Solution :
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other
Thanks
| Let $f(x)=\sum_{k=0}^na_kx^k,f\neq0$, where $n$ is the degree of $f$, i.e. $a_n\neq0$. If $n=0$, then your equation is equivalent to $a_0^2=2a_0$, which implies $a_0=2$. So let's assume that $n\geq1$. Multiply both sides of your equation by $x^n$ and use the Cauchy product formula for finite sums. This yields after some simplifications
\begin{align*}
\sum_{j=0}^{n-k}a_ja_{j+k}&=a_k,\qquad\forall k\in\{1,\ldots,n\}, \\
\sum_{j=0}^na_j^2&=2a_0.
\end{align*}
For $k=n$ we get $a_0a_n=a_n$, and since $a_n\neq0$ we have $a_0=1$.
For $k=n-1$ we get $a_0a_{n-1}+a_1a_n=a_{n-1}$, which is equivalent to $a_1a_n=0$ and hence $a_1=0$. For $k=n-2$ we obtain $a_0a_{n-2}+a_1a_{n-1}+a_2a_n=a_{n-2}$, thus $a_2a_n=0$ and finally $a_2=0$. Continuing this process gives $a_k=0$ for each $k\in\{1,\ldots,n-1\}$. From the last equality we then have that $1+a_n^2=2$, which implies $a_n=\pm1$. Consequently, $f$ is of the form $f(x)=1\pm x^n$.
Using $f(4)=65$, we conclude $f(x)=x^3+1$ (see George V. Williams' answer).
| {
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Combinatorial proof for $ \sum _{r=1} ^n r^3 \binom nr = n^2(n+3) 2^{n-3}$
Find the combinatorial proof for $$ \sum _{r=1} ^n r^3 \binom nr = n^2(n+3) 2^{n-3}$$
After proving it using algebra, I'm unable to find a combinatorial argument for the above statement. Help would be appreciated!
| Identity $\boldsymbol{1}$: Consider that $r^3$ is the number of ways to pick $3$ items from a set of $r$, with replacement. Break that down into how many duplicates there are.
$0$ duplicates: there are $\binom{r}{3}$ choices of items and $3!=6$ choices of order. $\bbox[5px,border:2px solid #C00000]{6\binom{r}{3}\text{ ways}}$
$1$ duplicate: there are $\binom{r}{2}$ choices of items, $\binom{2}{1}=2$ choices of which is the duplicate and $\binom{3}{2}=3$ choices of order. $\bbox[5px,border:2px solid #C00000]{6\binom{r}{2}\text{ ways}}$
$2$ duplicates: there are $\binom{r}{1}$ choices of the item. $\bbox[5px,border:2px solid #C00000]{\binom{r}{1}\text{ ways}}$
Thus, we have shown
$$
r^3=6\binom{r}{3}+6\binom{r}{2}+\binom{r}{1}\tag{1}
$$
Identity $\boldsymbol{2}$: Consider the number of ways to choose a team of unspecified size with $3$ leaders from a group of $n$ people.
There are $\binom{n}{3}$ choices for the leaders and $2^{n-3}$ ways to choose the remainder of the team. $\bbox[5px,border:2px solid #C00000]{\binom{n}{3}2^{n-3}\text{ ways}}$
Alternatively, for each $r$, there are $\binom{n}{r}$ ways to choose the team and then $\binom{r}{3}$ ways to choose the leaders from the team. $\bbox[5px,border:2px solid #C00000]{\sum\limits_{r=0}^n\binom{n}{r}\binom{r}{3}\text{ ways}}$
Thus, we have shown
$$
\sum\limits_{r=0}^n\binom{n}{r}\binom{r}{3}=\binom{n}{3}2^{n-3}\tag{2}
$$
Conclusion: Putting all this together, we get
$$
\begin{align}
\sum_{r=0}^nr^3\binom{n}{r}
&=6\sum_{r=0}^n\binom{r}{3}\binom{n}{r}+6\sum_{r=0}^n\binom{r}{2}\binom{n}{r}+\sum_{r=0}^n\binom{r}{1}\binom{n}{r}\\
&=6\binom{n}{3}2^{n-3}+6\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\[8pt]
&=2^{n-3}\left(n^3+3n^2\right)\tag{3}
\end{align}
$$
| {
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How find this diophantine equation integer solution $a^3+b^3=(2ab+1)^2$
Find this following diophantine equation integer solution
$$a^3+b^3=(2ab+1)^2$$
I think this equation only have two following solution
$$(a,b)=(1,0),(0,1)$$
maybe this equation have no other solution? because can see wolframalpha.com
I think following idea is usefull
$$a^3+b^3=(a+b)(a^2-ab+b^2)=(2ab+1)^2$$
since
$(a+b,a^2-ab+b^2)=1$ so
$a+b,a^2-ab+b^2$ are square numbers?
| Let's rewrite equation in $s,t$ doetoe (edit: he deleted his answer, look at the end of mine) got as follows: $4t^2+(3s+4)t+(1-s^3)=0$, and try to solve it for $t$. Because $t$ is integer, we must get that discriminant is a perfect square: $\Delta=(3s+4)^2-4\cdot 4(1-s^3)=9s^2+24s+16-16+16s^3=s(16s^2+9s+24)$. It's easy to see that, assuming $a,b$ satisfy OP's equation, that $2\nmid s$, and we can check that $3\nmid s$: if $3\mid a,b$ then $3\mid a^3+b^3$ but $3\nmid (2ab+1)$, and if $a\equiv 1,b\equiv -1\mod 3$ then again $3\mid a^3+b^3$ but $3\nmid (2ab+1)$. So $s$ and $24$ are relatively prime, and thus $s$ and $16s^2+9s+24$ are, so, because these two multiply to a perfect square, both must be perfect squares. But $16s^2+9s+24=(4s+1)^2+s+23$. Now, for $s>\frac{20}{7}: (4s+1)^2<(4s+1)^2+s+23<(4s+2)^2$ and for $s<-23: (4s+1)^2>(4s+1)^2+s+23>(4s+2)^2$ (simple quadrtatic inequalities, can be checked with Wolfram Alpha). So, unless $-23\leq s\leq \frac{20}{7}$ $\Delta$ cannot be perfect square. One can check that in this interval $\Delta$ is square for $s=1$ or $s=-23$. For $s=1$ we get $t=0$ or $-\frac{7}{4}$. First leads to solutions you have found, second doesn't give integer solutions. $s=-23$ doesn't result in real solutions.
So only solutions are indeed $(0,1),(1,0)$.
EDIT: doetoe's answer has expressed OP's equality in terms of $s=a+b,t=ab$ and it wad form $4t^2+3ts+4t=s^3-1$, and equivalence can be easily verified by substitution and elementary transformations.
| {
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How prove this idenity this $mv-3nu=m-3u$ with unit circle
Assmue the $m,n,u,v$ be real numbers,and such
$$m^2+n^2=1,u^2+v^2=1,nv>0,m>0,u>0$$
and
$$5mu=3(1-nv)$$
show that
$$mv-3nu=m-3u$$
Following is My methods:
let
$$m=\cos{x},n=\sin{x},u=\cos{y},v=\sin{y}$$
and
$$5mu=3(1-nv)\Longrightarrow 5\cos{x}\cos{y}=3(1-\sin{x}\sin{y})$$
$$\Longrightarrow 5\cos{x}\cos{y}+3\sin{x}\sin{y}=3$$
$$4\cos{(x-y)}+\cos{(x+y)}=3\Longrightarrow -8\sin^2{\dfrac{x-y}{2}}+4+2\cos^2{\dfrac{x+y}{2}}-1=3$$
$$\Longrightarrow \cos{\dfrac{x+y}{2}}=2\sin{\dfrac{x-y}{2}}$$
$$\Longrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$
then we must show that
$$\Longleftrightarrow \cos{x}\sin{y}-3\sin{x}\cos{y}=\cos{x}-3\cos{y}$$
$$\Longleftrightarrow \cos{x}(1-\sin{y})=3\cos{y}(1-\sin{x})$$
$$\Longleftrightarrow\dfrac{\cos{x}}{1-\sin{x}}=3\dfrac{\cos{y}}{1-\sin{y}}\Longleftrightarrow \dfrac{1+\tan{\frac{x}{2}}}{1-\tan{\frac{x}{2}}}=3\dfrac{1+\tan{\frac{y}{2}}}{1-\tan{\frac{y}{2}}}\tag{2}$$
$$\Longleftrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$
so it is clear
My Question:have without this trigonometric methods?such this famous Lagrange's identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$
since $m^2+n^2=1,u^2+v^2=1$
$$(5mu+3nv)=3\Longrightarrow (mv+3u)=3nu+m$$
| It follows also directly by substituting the equations. Let $u= - 3(nv - 1)/(5m)$. Then the desired equation is given by
$$
9(nv - 1)(n - 1) + 5(v - 1)m^2=0.
$$
Now substitute $m^2:=1-n^2$. Then we need to show
$$
(4nv + 5n - 5v - 4)(n-1)=0.
$$
Since $n-1\neq 0$ because of $m>0$ we need to show $4nv + 5n - 5v - 4=0$.
This follows from $u^2+v^2=1$, which is equivalent to
$$
(4nv + 5n - 5v - 4)(4nv-5n+5v-4)=0.
$$
Here the second factor cannot be zero. Assume it is zero, i.e., $v=(5n+4)/(4n+5)$.
Then $u=-3(n-1)(n+1)/(4n+5)m)<0$, a contradiction. It follows that the first factor is zero, i.e.,
$$
4nv + 5n - 5v - 4=0.
$$
| {
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How to calculate the sum of digits of $2^n$? How do I find the sum of digits of $2^n$ in general?
Sum of digits of $2^1=2$ is $2$.
Sum of digits of $2^{10}=1024$ is $7$.
I have check there is no obvious pattern or any recurrence that i can find. Any ideas?
** i don't need repeated sums , just single time**
| $2^{1} = 2$
$2^{2} = 4$
$2^{3} = 8$
$2^{4} = 16 \rightarrow 7$
$2^{5} = 32 \rightarrow 5$
$2^{6} = 64 \rightarrow 10 \rightarrow 1$
Then..
$2^{7} = 128 \rightarrow 11 \rightarrow 2$
$2^{8} = 256 \rightarrow 13 \rightarrow 4$
$2^{9} = 512 \rightarrow 8$
$2^{10} = 1024 \rightarrow 7$
$2^{11} = 2048 \rightarrow 14 \rightarrow 5$
$2^{12} = 4096 \rightarrow 19 \rightarrow 10 \rightarrow 1$
If you keep this up, it seems you will find that
$2^{n} \equiv 2^{n+6}$
when summing the digits of the result of each.
| {
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$f=X^3+aX+b$ has pairwise different roots iff $-4a^3-27b^2\neq0$ I have to proof that $f=X^3+aX+b\in F$, where $f$ is a polynomial which is a product of linear terms in a field $F$, has pairwise different roots iff $d=-4a^3-27b^2\neq0$.
Now what I have done is $f=(x-\alpha)(x-\beta)(x-\gamma)$.
So I get $a=-\alpha\beta-\alpha^2-\beta^2$ and $b=\alpha^2\beta + \alpha\beta^2$ after using $-(\alpha + \beta+\gamma)=0$. Now for one direction I can plug in $\gamma=\alpha\Rightarrow\beta=-2\alpha$ or $\alpha=\beta$, so indeed $d=0$ if they arent pairwise different. Now I cant seem to find a quick argument. I did some calculations and I got (by brute force, plugging in $x\alpha$ for $\beta$): iff $g:=4x^6+12x^5-3x^4-26x^3-3x^2+12x+4=0$ then for $\beta=\alpha x$, $d$ will be zero (as $F$ is a field I will get all solutions with this substitution). So after I know $x=-2,1,-1/2$ I get through polynom division $g=(x-1)^2(x+2)^2(2x+1)^2$. So its proven now! Now this all seems way too much calculation for this exercise! Please tell me a more easy way!
| Note that a polynomial $f(X)$ has a multiple root if and only if it has a common root with its derivative $f'(X)$ (can be checked by writing $f$ as a product of linear factors).
In your case, $f(X)=X^3+aX+b$, so $f'(X)=3X^2+a$. If $a=0$, the polynomials have a common root if and only if $b=0$, so the result is true in this case and we can thus assume $a\not=0$.
If there is a root of $f(X)$ and $f'(X)$, it is a root of $f(X)-\frac{X}{3}f'(X)=\frac{2a}{3}X+b$, so is $X=-\frac{3b}{2a}$.
Computing
$$f'(-\frac{3b}{2a})=3\frac{9b^2}{4a^2}+a=\frac{27b^2+4a^3}{4a^2},$$
$$f(-\frac{3b}{2a})=-\frac{27b^3}{8a^3}-\frac{3b}{2}+b=\frac{-27b^3-12a^3b+8a^3b}{8a^3}=\frac{b(-27b^2-4a^3)}{8a^3},$$
you directly find that $f(X)$ has a multiple root if and only if $27b^2+4a^3=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{cos(x/n)-1}{x^4}dx$ Show that the following limit exists and compute it:
$$\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$$
Attempt: By using the integration by parts, I get the following result:
$$\begin{align*}\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx&=\lim_{k\to\infty}\int_{1}^{k}\frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\\\\
&=\lim_{k\to\infty}\left\{\left(\frac{k}{n}\right)^4\sin\left(\frac{k}{n}\right)+4\left(\frac{k}{n}\right)^3\cos\left(\frac{k}{n}\right)\right.\\
&\quad\quad\quad\quad-12\left(\frac{k}{n}\right)^2\sin\left(\frac{k}{n}\right)-24\left(\frac{k}{n}\right)\cos\left(\frac{k}{n}\right)\\
&\quad\quad\quad\quad+24\sin\left(\frac{k}{n}\right)+\frac{n^3}{3k^3}\\
&\quad\quad\quad\quad\left.-\left[\frac{1}{n^4}\sin\left(\frac{1}{n}\right)+\frac{4}{n^3}\cos\left(\frac{1}{n}\right)\right.\right.\\
&\quad\quad\quad\quad\left.\left.-12\left(\frac{1}{n}\right)^2\sin\left(\frac{1}{n}\right)-\frac{24}{n}\cos\left(\frac{1}{n}\right)\right.\right.\\
&\quad\quad\quad\quad\left.\left.+24\sin\left(\frac{1}{n}\right)+\frac{n^3}{3}\right]\right\}\end{align*}$$
But I could not compute this limit. And how can we show the existence of the limit $\displaystyle\lim_{n \to \infty}n^2\int_{1}^{\infty} \frac{\cos\left(\frac{x}{n}\right)-1}{x^4}\,dx$? Thanks!
| since $n^2(1-\cos(x/n)) = 2n^2\sin^2 \frac{x}{2n} \to \dfrac{x^2}{2}$ as $n \to \infty.$
so can i conclude that
$$\lim_{n \to \infty}n^2\int_1^\infty \dfrac{\cos( x/n) - 1}{x^4} \ dx= -\dfrac{1}{2} \int_1^\infty \dfrac{1}{x^2} \ dx = -\dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is the "i" disappearing? The task is:
Find the argument in its simplest form.
$$(\sin(x) +i(1-\cos(x)))^2$$
where $x$ is an acute angle.
I multiplied out the equation and let alpha be the required argument, then said that
$$\tan(\alpha) = \frac{2i\sin(x)(1-\cos(x))}{\sin(x)^2-(1-\cos(x))^2}.$$
However, the solutions says the same thing except there is no $i$ in $2i\sin(x)(1-\cos(x))$. So I was wondering where the $i$ went?
| $$(\sin(x) +i(1-\cos(x)))^2=$$
$$\left(\sin(x)+i\left(2\sin^2\left(\frac{x}{2}\right)\right)\right)^2=$$
$$\sin^2\left(\frac{x}{2}\right)(4\cos(x)+4i\sin(x))=$$
$$\sin^2\left(\frac{x}{2}\right)4\cos(x)+\sin^2\left(\frac{x}{2}\right)4i\sin(x)=$$
$$(2\cos(x)-\cos(2x)-1)+\left(4\sin^2\left(\frac{x}{2}\right)\sin(x)\right)i=$$
$$\left|(2\cos(x)-\cos(2x)-1)+\left(4\sin^2\left(\frac{x}{2}\right)\sin(x)\right)i\right|e^{\arg\left((2\cos(x)-\cos(2x)-1)+\left(4\sin^2\left(\frac{x}{2}\right)\sin(x)\right)i\right)i}=$$
$$\sqrt{\left(2\cos(x)-\cos(2x)-1\right)^2+\left(4\sin^2\left(\frac{x}{2}\right)\sin(x)\right)^2}e^{\arg\left((2\cos(x)-\cos(2x)-1)+\left(4\sin^2\left(\frac{x}{2}\right)\sin(x)\right)i\right)i}=$$
$$4\sqrt{\sin^4\left(\frac{x}{2}\right)}e^{\arg\left(e^{ix}\sin^2\left(\frac{x}{2}\right)\right)i}$$
$-------$
1) Argument if $\Re$ and $\Im$ are both positive:
$$\tan^{-1}\left(\frac{|\Im|}{|\Re|}\right)$$
2) Argument if $\Re$ and $\Im$ are both negative:
$$\pi+\tan^{-1}\left(\frac{|\Im|}{|\Re|}\right)$$
3) Argument if $\Re$ is negative and $\Im$ is positive:
$$\frac{1}{2}\pi +\tan^{-1}\left(\frac{|\Re|}{|\Im|}\right)$$
3) Argument if $\Re$ is positive and $\Im$ is negative:
$$-\tan^{-1}\left(\frac{|\Im|}{|\Re|}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit without L'Hopital's rule: $\lim_{x\to0} \frac{1-\cos^3 x}{x\sin2x}$ How can I solve the following problem without the use of the L'Hopitals's rule?
$$\lim_{x\to0} \frac{1-\cos^3(x)}{x\sin{(2x)}}$$
| Hint:
We have that $1-\cos^3x = (1-\cos x)(1+\cos x + \cos^2 x)$ and $x\sin 2x = 2x\sin x \cos x$, so by limit arithmetic
$$\lim_{x \to 0} \frac{1-\cos^3x}{x\sin 2x} = \lim_{x\to0} \frac{(1-\cos x)(1+\cos x + \cos^2 x)}{2x\sin x \cos x} = \frac{3}{2}\lim_{x \to 0}\frac{1-\cos x}{x\sin x}$$
From here, see if you can get a numerator in terms of $\sin x$.
| {
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"url": "https://math.stackexchange.com/questions/1111014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Reducing $\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$ for $n>2$
Simplify
$$\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$$ for $n>2$.
The answer is $(a+b+c)^n$, but I can't seen to get it.
Can someone help me?
Thanks
| Hint
$$a-c=(a-b)+(b-c)$$
so
\begin{align*}&(b-c)\cdot a^n-(a-b)\cdot b^n-(b-c)\cdot b^n+(a-b)\cdot c^n\\
&=(b-c)(a^n-b^n)+(a-b)(c^n-b^n)\\
&=(a-b)(b-c)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1}-c^{n-1}-c^{n-2}b-\cdots-b^{n-1})\\
&=(a-b)(b-c)[(a^{n-1}-c^{n-1})+(a^{n-2}b-c^{n-2}b)+\cdots+(ab^{n-2}-cb^{n-2})]\\
&=(a-b)(b-c)(a-c)[(a^{n-2}+\cdots+c^{n-2})+b(a^{n-3}+\cdots+c^{n-3})+\cdots+b^{n-2}]
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that {$r_n$} satisfies $r_n = 7r_{n-1} - 10r_{n-2}, n \geq 2$ Problem: For the sequence $r$ defined by
$$r_n = 3 \cdot 2^n - 4 \cdot 5^n, \ \ \ n \geq 0$$
Prove that {$r_n$} satisfies
$$r_n = 7r_{n-1} - 10r_{n-2}, \ \ \ n \geq 2$$
Can this problem be explained and broken down and show the process? I'd like to follow your steps on my own.
I've been studying this question but the opening multiplication is throwing me off.
Similar Problem
| Based on the input I was given by some members here's what I came up with following some of those steps and putting it in my own words in a way to make sure I understand what's happening:
$r_n = 3\cdot2^n - 4\cdot5^n$, $n \geq 0$
$r_n = 7r_{n-1} - 10r_{n-2}$, $n \geq 2$
$n = 1$
$1-1 = 1$
Working the left side first
$7\left(3\cdot2^1 - 4\cdot5^{1} \right)$
$7\left(\frac{3}{2}\cdot2 - \frac{4}{5}\cdot5 \right)$
$7\left(\left(1.5\cdot2\right) -\left(0.8\cdot5\right)\right)$
3, -4 <-- These are the first numbers in 3*2 and -4*5 (but ignoring -4 for now)
Moving the 3 the to the outside now. Repeat the process for the 7.
$3\left(7\cdot2^1 - 10\cdot2^{1} \right)$
$3\left(\frac{7}{2}\cdot2 - \frac{10}{4}\cdot2 \right)$ <-- Multiply the denominator by $2^n$ for the denominoator of the second part.
$3\left(\left(3.5\cdot2\right) -\left(2.5\cdot2\right)\right)$
$\left(7-5\right)$
$2$ <-- Bring down the 3
$3\cdot2$ <-- These are the first numbers in $r_n$
Now I have 3, 2 or $3 \cdot 2$ of $r_n$
Working the right side now
$-10\left(3\cdot2 - 4\cdot5 \right)$
$-10\left(\frac{3}{2}\cdot2 - \frac{4}{5}\cdot5 \right)$
$-10\left(\left(1.5\cdot2\right) -\left(0.8\cdot5\right)\right)$
3, -4 <-- We have the same values again this time we use the -4 and ignore the 3
Moving the -4 to the outside and repeating the same process above when we moved the 3 to the outside and moved the 7 inside.
$-4\left(7\cdot5 - 10\cdot5 \right)$
$-4\left(\frac{7}{5}\cdot5 - \frac{10}{25}\cdot5 \right)$ <-- Multiply the denominator by $5^n$ for the denominoator of the second part.
$-4\left(\left(1.4\cdot5\right) -\left(0.4\cdot5\right)\right)$
$\left(7-2\right)$
$5$ <-- Bring down the -4
$-4\cdot5$ <-- These are the first numbers in $r_n$
Now I have -4, 5 or $-4 \cdot 5$ of $r_n$
We then have: $r_n = 3\cdot2^n - 4\cdot5^n$, $n \geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Maclaurin Expansion of $f(x)=\ln (\cos x)$ Find up to the $x^4$ term in the Maclaurin expansion of $f(x)=\ln (\cos x)$
I've only just learnt about Maclaurin Series today so I'm not too familiar with them. Do I have to find up to the 4th derivative (I've tried it and it gets quite messy) or can I use one of the known expansions to do this?
| We have $\cos x = 1 + h(x)$, where $h(x) = -\frac{1}{2}x^2 + \frac{1}{24}x^4 + O(x^6)$ as $x \to 0$.
Since $h(x) \to 0$, we have
$$f(x) = \ln(\cos x) = \ln(1 + h(x)) = h(x) -\frac{1}{2}h(x)^2 + O(h(x)^3).$$
Since $h(x) \sim -\frac{1}{2}x^2$, we have $h(x)^3 \sim -\frac{1}{8}x^6$, so $O(h(x)^3) = O(x^6)$. Thus
$$
\begin{align*}
f(x) &= -\frac{1}{2}x^2 + \frac{1}{24}x^4 + O(x^6) -\frac{1}{2}[-\frac{1}{2}x^2 + O(x^4)]^2 + O(x^6) \\
&= - \frac{1}{2}x^2 + \frac{1}{24}x^4 + O(x^6) - \frac{1}{8} x^4 + O(x^6) + O(x^6) \\
&= -\frac{1}{2}x^2 - \frac{1}{12}x^4 + O(x^6).
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Matrix multiplication computation Any tips how to solve this?
$$
\left[ \begin{matrix}1 & 2 & 0 \\ -2 & -5 & 1 \\ 11&15&5 \end{matrix}\right] \times \mathbf{X} \times \left[ \begin{matrix} -4&5&1\\ -4&5&1 \\ -4&5&1\end{matrix}\right] =\left[ \begin{matrix}-12 & 15&3 0 \\ -24&-30&6 \\ -125&150&31 \end{matrix}\right]
$$
| First instinct would be to multiply on the left and right by inverses to get $A$ on its own on the LHS. However the rightmost of these matrices is not invertible. But we can get
$$
A \times \left[\begin{array}{ccc}-4 & 5 & 1\\-4 & 5 & 1\\-4 & 5 & 1 \end{array}\right] = \left[\begin{array}{ccc} 235 & 0 & -599 \\ -\frac{247}{2} & \frac{15}{2} & \frac{629}{2} \\ -\frac{343}{2} & \frac{15}{2} & \frac{761}{2}\end{array}\right]
$$
The system of equations for the components of $A$ is over-determined. This means that, letting
$$
A = \left[ \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]
$$
we see (for instance) that this requires
$$
-4 (a_1 + a_2 + a_3) = 235 \\
\implies a_1 + a_2 + a_3 = -\frac{235}{4}
$$
but
$$
5 (a_1 + a_2 + a_3) = 0 \\
\implies a_1 + a_2 + a_3 = 0
$$
which is clearly contradictory. So no solution exists.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5 $ with $a+b+c+d = 1$? For positive real numbers $a,b,c,d>0$ it seems to be true that:
if $$a+b+c+d = 1$$
then
$$ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d} \geq 0.5 $$
I can't think of a way to prove this statement.
Any help will be appreciated!
| Use Cauchy-Schwarz Inequality: $$(a+b+b+c+c+d+d+a)\left(\frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}\right) \ge (a+b+c+d)^2$$
Alternative approach: $$\sum\limits_{cyc} \frac{a^2}{a+b} = \sum\limits_{cyc} \left(\frac{a^2}{a+b} - (a-b)\right) = \sum\limits_{cyc} \frac{b^2}{a+b}$$
Therefore, $$\sum\limits_{cyc} \frac{a^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} \frac{a^2+b^2}{a+b} \ge \frac{1}{4}\sum\limits_{cyc} \frac{(a+b)^2}{a+b} = \frac{1}{2}\sum\limits_{cyc} a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve an inequality using Cauchy-Schwarz Inequality Le $a,b,c,d \in \mathbb{R^{+}}$. Using Cauchy-Schwarz Inequality prove that the following inequality holds:
$$\frac{1}{\frac{1}{a+c} + \frac{1}{b+d}} \ge \frac{1}{\frac 1a + \frac 1b} + \frac{1}{\frac 1c + \frac 1d}$$
I've been trying for a while, but I haven't been able to make any progress. Actually I have solved this ineqaulity using 2 methods, but both make no use of the Cauchy-Schwarz inequality. In the first solution I clear the denominators and it reduces to a rather simple inequality. The original inequality is equivalent to:
$$\frac{(a+c)(b+d)}{a+b+c+d} \ge \frac{ab}{a+b} + \frac{cd}{c+d} = \frac{ab(c+d) + cd(a+b)}{(a+b)(c+d)}$$
After clearing the denominators, multiplying and canceling everything out we are left with an inequality that's true from AM-GM:
$$b^2c^2 + a^2d^2 \ge 2abcd$$
The other solution makes use of the Minkowski Inequality. Substitute $r=1; s=-1; a_{11}=a; a_{12}=b; a_{21}=c; a_{22}=d; m=n=2$ in the following form of the Minkowski Inequality and we get the original inequality.
$$\left(\sum^{m}_{j=1}\left(\sum^{n}_{i=1} a_{ij}^r\right)^{\frac sr}\right)^{\frac 1s} \ge \left(\sum^{n}_{i=1}\left(\sum^{m}_{j=1} a_{ij}^s\right)^{\frac rs}\right)^{\frac 1r}$$
| Starting from the equivalent inequality
$$\frac{(a+c)(b+d)}{a+b+c+d} \ge \frac{ab}{a+b} + \frac{cd}{c+d} = \frac{ab(c+d) + cd(a+b)}{(a+b)(c+d)}$$
and cross-multiplying results in
$$\begin{align}(a+b)(a+c)(b+d)(c+d)&\geq\{(a+b)+(c+d)\}\{ab(c+d)+cd(a+b)\}\\&=cd(a+b)^2+ab(c+d)^2+(a+b)(c+d)(ab+cd)\end{align}$$
which (after subtracting $(a+b)(c+d)(ab+cd)$ from both sides) simplifies to
$$(a+b)(c+d)(bc+ad)\geq cd(a+b)^2+ab(c+d)^2$$
completing the square on the right hand side, and rearranging the cross term of the square, leads to the inequality being expressed as follows
$$(a+b)(c+d)(bc+ad+2\sqrt{abcd})\geq \{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)\}^2\\\Rightarrow(a+b)(c+d)(\sqrt{ad}+\sqrt{bc})^2\geq\{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)\}^2$$
Dividing both sides by positive value $(\sqrt{ad}+\sqrt{bc})^2$ leaves the direction of the inequality unchanged, resulting in
$$(a+b)(c+d)\geq\left[\frac{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)}{\sqrt{ad}+\sqrt{bc}}\right]^2\\=(\sqrt{ac}+\sqrt{bd})^2$$
Thus, we end up with the following inequality
$$(a+b)(c+d)\geq(\sqrt{ac}+\sqrt{bd})^2$$
which can be seen to be the Cauchy-Schwarz inequality,
$$\sum_{j=1}^N|x_j|^2\sum_{k=1}^N|y_k|^2\geq\left|\sum_{i=1}^Nx_i\bar{y_i}\right|^2$$ with $x_1=\sqrt{a},x_2=\sqrt{b}$ and $y_1=\sqrt{c},y_2=\sqrt{d}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve system of 3 equations $x+y+z=0$
$x^2+y^2+z^2=6ab$
$x^3+y^3+z^3=3(a^3+b^3)$
this is what i reasoned out so far;
$xyz=a^3+b^3$
$x^2+zx+z^2=3ab$
$y^2+zy+z^2=3ab$
$x^2+xy+y^2=3ab$
$y^2=3ab+zx$
$x^2=3ab+zy$
$z^2=3ab+xy$
I'd prefer a hint rather than a full answer - and how should I solve this kind of systems?
| A mechanical technique is offered by "Myself" in the comments. I'll sketch it.
Via Newton's identities we have
$$\begin{array}{ll} x^2+y^2+z^2 & =(x+y+z)^2-2(xy+yz+zx) \\ x^3+y^3+z^3 & =(x+y+z)^3-3(xy+yz+zx)(x+y+z)+3xyz \end{array}$$
(Read the background for information on how these can be obtained.)
Combining with the givens and Vieta's formulas, we have
$$(t-x)(t-y)(t-z)=t^3-3abt-(a^3+b^3).$$
Thus, $x,y,z$ are the roots of this polynomial. If you want to go further and obtain explicit formulas, you can use Cardano's method for the depressed cubic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using the distributive property to factor $(5^{-1}\cdot 5^x - 5^x - 5\cdot 5^x + 5^2\cdot 5^x)$ I can't seem to understand the distributive property.
Take this:
$$ 5^{-1}\cdot 5^x - 5^x - 5\cdot 5^x + 5^2\cdot 5^x$$
becoming this:
$$ 5^x\left(\frac 15 - 1 - 5 +25\right) $$
Help? :D
| Every term has a factor of $5^x$ which is factored out.
Also, note that:
$$5^{-1} = \frac 1{5^1} = \frac 15,\quad 5^2 = 25$$
$$\begin{align} 5^{-1}\cdot \color{blue}{5^x} - (1)\cdot \color{blue}{5^x} - 5\cdot \color{blue}{5^x} + 5^2\cdot \color{blue}{5^x} &= \frac 15\cdot \color{blue}{5^x} -(1)\cdot \color{blue}{5^x} - 5\cdot \color{blue}{5^x}+ 25\cdot \color{blue}{5^x}\\ \\ & =\color{blue}{5^x}\left(\frac 15 -1-5 + 25\right)\\ \\ & = \frac{96}5\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118082",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why does the following equality hold? $\sec^{-1}(2/\sqrt{2}) = \sec^{-1}(\sqrt{2})$? Why is $\sec^{-1}(2/\sqrt{2}) = \sec^{-1}(\sqrt{2})$ true?
| As Workaholic pointed out (here with more explanation):
\begin{align}
\frac{2}{\sqrt{2}} &= \frac{2}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\tag{$\frac{\sqrt{2}}{\sqrt{2}}=1$}\\[1em]
&= \frac{2\sqrt{2}}{2}\\[1em]
&= \sqrt{2}.
\end{align}
Since $\dfrac{2}{\sqrt{2}}=\sqrt{2}$, it must then be true that
$$
\sec^{-1}\left(\frac{2}{\sqrt{2}}\right)=\sec^{-1}(\sqrt{2}),
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can someone help check if I evaluated my integral right? The integral given to be evaluated is $\int 6x\cdot\mathrm{arctanh}(x)\,dx$.
I tried to evaluated and got the following answer:
$$3x^2\operatorname{arctanh}x + 3x -3\ln\,\bigl\lvert(1+x)/\sqrt{1-x^2}\bigr\rvert + c$$
I used some trig substitutions and integration by parts to find it. Did I get it right or no?
Thank you.
| Consider the function $y = \tanh^{-1} x$; that is to say, $$x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}.$$ Hence $$0 = x(e^y + e^{-y}) - e^y + e^{-y} = (x-1)e^y + (x+1)e^{-y},$$ or equivalently, $$e^{2y} = \frac{1+x}{1-x}.$$ Therefore, $$y = \tanh^{-1} x = \frac{1}{2} \log \frac{1+x}{1-x}, \quad |x| < 1.$$ Now observe that $$\begin{align*} \log \left| \frac{1+x}{\sqrt{1-x^2}} \right| &= \log \left| \frac{1+x}{\sqrt{1-x} \sqrt{1+x}} \right| = \log \left| \frac{\sqrt{1+x}}{\sqrt{1-x}} \right| = \log \left| \sqrt{\frac{1+x}{1-x}} \right| = \frac{1}{2} \log \left| \frac{1+x}{1-x} \right| \\ &= \tanh^{-1} x. \end{align*}$$ This immediately leads to the desired simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$? $$\int{\frac{1}{\sqrt{3-2x-x^2}}\,dx}$$
I tried to do it by substitution with no sucess. Anyone can solve it?
| Completing the square and using the substitution $u=x+1$ gives us
\begin{equation*}
\int\frac{1}{\sqrt{4-(x+1)^2}}dx=\int\frac{1}{\sqrt{4-u^2}}du=\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}.
\end{equation*}
Using the substitution $s=\frac{u}{2}$ gives
\begin{equation*}
\int\frac{1}{\sqrt{1-s^2}}=\sin^{-1}(s)+C.
\end{equation*}
Substituting back $s=\frac{u}{2}$ and $u=x+1$ gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| The answer is that you will need to have a constant curvature, which is the partial circle solution by robjohn. If you do want the curve within (0,1) then the rectangle + 1/2 circle solution by both rob and xan.
Why is that? it is actually a physics problem. The solution is a shape of a membrane under pressure.
Parametric function:
$$x=
\begin{cases}
1& 0\leq s\leq h\\
\frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) & h<s<h+\frac{2}{\pi}\\
0& h+\frac{2}{\pi}<s<2h+\frac{2}{\pi}\\
\end{cases}
$$
$$y=
\begin{cases}
s& 0\leq s\leq h\\
h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right)& h<s<h+\frac{2}{\pi}\\
\left( 2h+\frac{\pi}{2}-s \right)& h+\frac{\pi}{2}<s<2h+\frac{2}{\pi}\\
\end{cases}
$$
Area:
$$1=\int_{0}^{1}{y}dx$$
$$
1=\int_{h+\frac{\pi}{2}}^h{\left[ h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right) \right]}d\!\left[ \frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) \right]
$$
Take:
$$\theta=2(s-h)$$
$$4=\int_{\pi}^0{\left[ 2h+\sin \left( \theta \right) \right]}d\!\left( \cos \left( \theta \right) \right) =4h+\frac{\pi}{2}$$
$$h=1-{\pi \over 8}$$
$${\rm Length}=s_{\max}=2h+\frac{\pi}{2}=2+\frac{\pi}{4}=2.785398163...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 7
} |
Limit problems and quandaries: finding $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. Find $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. What I did is:
$\lim_\limits{n\to \infty }{({n^2-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty } {({n^2+1-1-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty }{(1-{1+n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty}[{(1-{1+n\over n^2+1})^{ n^2+1\over 1+n}}]^{(n+10)(1+n)\over (n^2+1)}$. Denoting: ${ n^2+1\over 1+n}=t$ ($t\to \infty$ as $n\to \infty$) we get: $\lim_\limits{n\to \infty}[{(1+{(-1)\over t})^{ t}}]^{(n+10)(1+n)\over (n^2+1)}=\lim_\limits{n\to \infty}e^{-{(n+10)(1+n)\over (n^2+1)}}=e^{\lim_\limits{n\to \infty}-{(n+10)(1+n)\over (n^2+1)}}=e^{-1}$.
My question is: how can I know it is defined and lawful? I know I can use the continuity of $e$, but I don't know if it is okay that I separated the limits that way(Because maybe the first limit is not even $e$ considering other things)
.I Would appreciate your help...
| Here is an alternative approach which only uses "elementary" calculations (i.e. no MacLaurin or any other kind of expansion, no continuity, no $\log$, no $\exp$):
Note that
$$\bigg(\frac{n^2-n}{n^2+1}\bigg)^{n+10}=\frac{\bigg(1-\frac{1}{n}\bigg)^n}{\bigg(1-\frac{1}{n^2}\bigg)^n}\cdot\frac{\bigg(1-\frac{1}{n}\bigg)^{10}}{\bigg(1-\frac{1}{n^2}\bigg)^{10}}$$
The last factor on the right hand side clearly converges to $1$. So, we only have to consider the term
$$\frac{\bigg(1-\frac{1}{n}\bigg)^n}{\bigg(1-\frac{1}{n^2}\bigg)^n}\tag{$*$}$$
Let us first compute the limit of the numerator of $(*)$:
$$\begin{align}
\bigg(1-\frac{1}{n}\bigg)^n & = \bigg(\frac{n}{n-1}\bigg)^{-n} \\
& = \bigg(1+\frac{1}{n-1}\bigg)^{-n} \\
& = \bigg(1+\frac{1}{n-1}\bigg)^{-1}\bigg(1+\frac{1}{n-1}\bigg)^{-(n-1)}
\end{align}$$
and this shows that the numerator tends to $e^{-1}$
Let us now compute the limit of the denominator of ($*$):
$$\begin{align}
\bigg(1-\frac{1}{n^2}\bigg)^n & = \bigg(\frac{n^2-1}{n^2}\bigg)^n \\
& = \bigg(\frac{n-1}{n}\cdot\frac{n+1}{n}\bigg)^n \\
& = \bigg(1-\frac{1}{n}\bigg)^n\bigg(1+\frac{1}{n}\bigg)^n \\
\end{align}$$
and therefore, the denominator tends to $1$.
This shows that the limit you are searching for is $e^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Volume of $y = 6\sqrt{\sin(x)}$ rotated around $y$-axis using triple integrals The problem is to find the volume of $y = 6\cdot \sqrt{\sin (x)}$ rotated around the $y$-axis when $0 \leq y \leq 6$.
I know this can be done by the sv-calc method of volumes of revolution but I wanted to see if a problem like this can be done by triple integrals. I tried it a few times myself but could not seem to get the limits of the integrals set up correctly.
| Maybe I miss the point of your question, but I think that the rotation of a function $y=f(x)$ around the $y$-axis using the formula
\begin{align*}
V=\pi\int_a^bx^2(y) dy\tag{1}
\end{align*}
and your Ansatz in the comment using a triple integral is essentially the same.
Since for
\begin{align*}
y=6\sqrt{\sin (x)}\qquad\longleftrightarrow\qquad x=\arcsin\left(\frac{y^2}{36}\right)
\end{align*}
We observe according to your comment
\begin{align*}
&\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}\int_0^{2\pi}r d\Theta dr dy\\
&\qquad=\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}\left. r\cdot\Theta\right|_0^{2\pi} dr dy\\
&\qquad=2\pi\int_0^6\int_0^{\arcsin(\frac{y^2}{36})}r dr dy\\
&\qquad=2\pi\int_0^6\left.\left(\frac{1}{2}r^2\right)\right|_0^{\arcsin(\frac{y^2}{36})}dy\\
&\qquad=\pi\int_0^6\arcsin^2\left(\frac{y^2}{36}\right)dy\tag{2}\\
\end{align*}
and (2) corresponds to the volume formula (1).
I think the difficulty lies in solving the integral (2) which seems to allow no simple closed representation. Wolfram alpha provides following solution:
\begin{align*}
&\pi\int\arcsin^2\left(\frac{y^2}{36}\right)dy\\
&\qquad=\frac{\pi^2y^5}{5184\sqrt{2}\Gamma\left(\frac{7}{4}\right)\Gamma\left(\frac{9}{4}\right)}
_{3}F_{2}\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};\left(\frac{y}{6}\right)^4\right)\\
&\qquad\qquad-\frac{2\pi y}{3} _2F_1\left(1,\frac{5}{4};\frac{7}{4};\left(\frac{y}{6}\right)^4\right)
\arcsin\left(\frac{y^2}{36}\right)\sin\left(2\arcsin\left(\frac{y^2}{36}\right)\right)\\
&\qquad\qquad+\pi y\arcsin^2\left(\frac{y^2}{36}\right)+C
\end{align*}
Note: You could perform a plausibility check, take a simpler integrand and you'll be able to calculate the volume in both ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Euler formula, trigonometry. Prove with Euler formula that
$$
\cos(x-y) = \cos(x)\cos(y) - \sin(x)\sin(y).
$$
I know how to find $\cos(x+y)$, but as for $\cos(x-y)$, I'm clueless.
Thanks.
| \begin{align*}
e^{ix} & = \cos x+ i \sin x\\
e^{i(x-y)} & = \frac{e^{ix}}{e^{iy}}\\
\frac{e^{ix}}{e^{iy}} & =\frac{\cos x + i \sin x}{\cos y+i\sin y}\\
\frac{\cos x + i \sin x}{\cos y+i\sin y} & =\frac{\cos x + i \sin x}{\cos y+i\sin y}\frac{\cos y - i \sin y}{\cos y -i\sin y}\\
& =\frac{\cos x \cos y+i\sin x\cos y-i\cos x\sin y+\sin x \sin y}{\cos^2 y+ \sin^2 y}\\
& ={\sin x\sin y + \cos x \cos y+i(\cos x \sin y -\sin x \cos y)}
\end{align*}
Equating real and imaginary parts:
$$\boxed{\cos [x-y]=\sin x\sin y + \cos x \cos y \text{ and } \sin[x-y]=\cos x \sin y -\sin x \cos y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\alpha \beta$ is the root of a polynomial Assuming that $\alpha, \beta$ are distinct roots of $P(x) = x^4+bx^3-1 = 0$, where $b \in \mathbb R$, show that $\alpha \beta$ is the root of $Q(x) = x^6+x^4+b^2x^3 -x^2 -1$.
I have already noticed that $P(0) < 0$ and $\deg P = 4$, so (from the intermediate value theorem) there are at least 2 real roots. It seems that the third and fourth root are complex, but I'm not sure, if this could help.
| We have $P(\alpha)=P(\beta)=0$, so we can express $b$ in terms of the distinct roots $\alpha$ and $\beta$
$$b=\frac{(1-\alpha^4)}{\alpha^3}=\frac{(1-\beta^4)}{\beta^3}$$
from which we can express $b^2$ as follows:-
$$b^2=\frac{(1-\alpha^4)(1-\beta^4)}{\alpha^3\beta^3} \text{ Eq.(1)}$$
In addition we obtain the following relationship, which can be further manipulated
$$\beta^3(1-\alpha^4)=\alpha^3(1-\beta^4)\\\Rightarrow \alpha^3\beta^3=\frac{\alpha^3-\beta^3}{\beta-\alpha}=-(\alpha^2+\alpha\beta+\beta^2)\\\Rightarrow(\alpha^3\beta^3+\alpha\beta)^2=(-(\alpha^2+\beta^2))^2\\\Rightarrow \alpha^6\beta^6+2\alpha^4\beta^4+\alpha^2\beta^2=\alpha^4+2\alpha^2\beta^2+\beta^4\\\Rightarrow \alpha^6\beta^6+2\alpha^4\beta^4-\alpha^2\beta^2-(\alpha^4+\beta^4)=0 \text{ Eq.(2)}$$
Using Equation (1) then (2), we can express $Q(\alpha\beta)$ as follows:-
$$Q(\alpha\beta)=\alpha^6\beta^6+\alpha^4\beta^4+\color{red}{\frac{(1-\alpha^4)(1-\beta^4)}{\alpha^3\beta^3}}\alpha^3\beta^3-\alpha^2\beta^2-1\\=\alpha^6\beta^6+2\alpha^4\beta^4-\alpha^2\beta^2-(\alpha^4+\beta^4)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{x\to\infty} ((x^5+x^4)^{1/6}-(x^5-x^4)^{1/6})$ I've been struggling with the following:
$$\lim_{x\to\infty} ((x^6+x^5)^{1/6}-(x^6-x^5)^{1/6})$$
Tried factoring out $x^{5/6}$ and then using L'hopital- which got me nowhere, tried multiplying by the conjugate, but it got messy- so either I'm scared of the algebra or there's a better way.
| When you have indeterminate type of form $\infty-\infty$ you need to rationalize your expression to get fraction:
$$\lim_{x\to\infty}\left(\sqrt[6]{x^5+x^4}-\sqrt[6]{x^5-x^4}\right)=\lim_{x\to\infty}\dfrac{\sqrt[3]{x^5+x^4}-\sqrt[3]{x^5-x^4}}{\sqrt[6]{x^5+x^4}+\sqrt[6]{x^5-x^4}}=\lim_{x\to0}\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x\sqrt[6]{1+x}+x\sqrt[6]{1-x}}$$
Now apply L'Hopital's rule
$$\lim_{x\to0}\dfrac{\dfrac13\cdot\dfrac1{(1+x)^{\frac23}}+\dfrac13\cdot\dfrac1{(1-x)^{\frac23}}}{\sqrt[6]{1+x}+x\cdot\dfrac16\cdot\dfrac{1}{(1+x)^{\frac56}}+\sqrt[6]{1-x}-x\cdot\dfrac16\cdot\dfrac{1}{(1-x)^{\frac56}}}=\dfrac{\frac13+\frac13}{1+1}=\dfrac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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1,3,7,13,21,31 what come next? It supposed to be some thing simple, but some how I don't get it.
I have a function with constant rate increase of 2. How can I calculate the n'th number?
It starts with 1, and it looks like this:
$$1,3,7,13,21,31,43...$$
| First observe that the second differences are constant.
Then we can assume it is $$f(x) = ax^2 + bx + c.$$
Subbing in $x = 0,1,2$ we have the system of equations
$$\pmatrix{1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 3 \\ 1 & 2 & 4 & 7} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 4 & 6} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 2 & 2} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1}.$$
Thus, we have $f(x) = x^2 + x + 1$.
For your interests.
If
$$a = 1 + 2 + \ldots + n, $$
then it is also true that
$$2a = (1 + 2 + \ldots + n) + (n + (n - 1) + \ldots + 1) = (n + 1) + \ldots (n + 1) = n(n +1).$$
Thus,
$$ 1 + 2 + \ldots + n = \frac{n(n+1)}{2}.$$
Finally, if $T(0) = 1$, then your sequence is just
$$T(n) = T(0) + 2 + 4 + \ldots + 2n = 1 + 2(1 + 2 + \ldots + n) = 1 + 2\frac{n(n+1)}{2} = n^2 + n + 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\
&=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\
&=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\
&=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc]
\end{align}$$
where $$s=\frac{a+b+c}{2}$$
I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
But I'm stuck there. Any help please?
| The polynomial $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}$$ vanishes for $c:=a+b$ and because it is cyclic it vanishes also at $a:=b+c$ and at $b:=a+c$. Notice that the same is true for $$(s-a)(s-b)(s-c).$$
In fact, for the second polynomial it is clear and for the first putting $c:=a+b$ in the first polynomial we get
$$\frac{2ba^2+2ab^2-2ab(a+b)}{8}\equiv 0.$$
Finally the two polynomials have the same leading coefficient $\frac{1}{8}$.
Therefore $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}=(s-a)(s-b)(s-c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $A^{100}+A^5$ Let $A= \begin{pmatrix} 1&-1 \\ 2&-2 \\ \end{pmatrix}$. We are to find $A^{100} +A^5$
My attempt : I found characteristic equation , $x^2+x=0$, so eigenvalues are $0$ and $-1$ hence its diagnol matrix is $$D= \begin{pmatrix} 0&0 \\ 0&-1 \\ \end{pmatrix}$$ So $D^{100}+D^5$ is null matrix.
But I am getting something else if I use method given here To Find $A^{50}$ by achille hui.
Please help in this matter. Thanks
| Observe the following computation:
$$
A^2 \;\; =\;\; \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ]\left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
-1 & 1 \\
-2 & 2\\
\end{array} \right ].
$$
Similarly if we compute $A^3$ we find
$$
A^3 \;\; =\;\; \left [ \begin{array}{cc}
-1 & 1 \\
-2 & 2\\
\end{array} \right ] \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ].
$$
Therefore we have that $A^{2n} = -A$ and $A^{2n+1} = A$ for all $n \in \mathbb{N}$. We conclude that
$$
A^{100} + A^5 \;\; =\;\; -A + A \;\; =\;\; \textbf{0}_{2\times 2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$ Let $a,b,c$ be the lenghts of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$.
My attempt:
I tried multiplying the whole thing but that didn't help at all. So, I tried to manipulate the triangle inequality and bring out the given form but that didn't help too. I am out of ideas now. Please help. Thank you.
| I could just show that it is less than 5 since I cannot comment I am just posting my partial answer.
$(a+1)(b+1)(c+1)=1+abc+a+b+c+ab+bc+ca$. So we have to just show that $abc+a+b+c<2$.
Now $a+b+c+abc \leq( a+b)(2+ab)$. Also not that $ab,bc,ca <1$. So atleast two sides must be less than $1$. wlog assume that they are $a$ and $b$. This implies $ab<a+b$. So $( a+b)(2+ab) < (a+b+1)^2-1 <3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How can I quickly find the determinant of this matrix $$
\begin{vmatrix}
14 & 2 & 1 & 3\\
31 & 4 & 5 & 6\\
26 & 3 & 7 & 4\\
10 & 1 & 3 & 2\\
\end{vmatrix}
=
\begin{vmatrix}
5\cdot2+1+3 & 2 & 1 & 3\\
5\cdot4+5+6 & 4 & 5 & 6\\
5\cdot3+7+4 & 3 & 7 & 4\\
5\cdot1+3+2 & 1 & 3 & 2\\
\end{vmatrix}$$$$
=
\begin{vmatrix}
5\cdot2 & 2 & 1 & 3\\
5\cdot4 & 4 & 5 & 6\\
5\cdot3 & 3 & 7 & 4\\
5\cdot1 & 1 & 3 & 2\\
\end{vmatrix}
+
\begin{vmatrix}
1 & 2 & 1 & 3\\
5 & 4 & 5 & 6\\
7 & 3 & 7 & 4\\
3 & 1 & 3 & 2\\
\end{vmatrix}
+
\begin{vmatrix}
3 & 2 & 1 & 3\\
6 & 4 & 5 & 6\\
4 & 3 & 7 & 4\\
2 & 1 & 3 & 2\\
\end{vmatrix}
$$
However I am not able to proceed beyond. The answer given is zero. Is there any simple determinant property that I am not able to guess?
| You have found—and expressed it in the first equality—that the first column is a linear combination of other columns: $$Col_1 = 5\cdot Col_2+Col_3+Col_4$$ And the simple determinant property you can't guess is: determinant with linearly dependent columns is $0$ (and vice versa, if it is zero, it has linearly dependent columns—and rows, too).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Natural solutions to $4^n + 2^{n + 1} = 2^{k}$ Is there such an $n$ and $k$ that
$$4^n + 2^{n + 1} = 2^{k}$$
with $n, k \in \mathbb N$.
I wrote a program and for $n, k < 5000$ have not found a solution.
Is this possible?
| It has already been mentioned in the previous answers, but I think it's worth highlighting: 4 is the square of 2, and all powers of 4 are also powers of 2. With just a little effort you should be able to verify that $4^n = 2^{2n}$.
Thus you want to solve $2^{2n} + 2^{n + 1} = 2^k$. But the only way two powers of 2 can add up to another power of 2 is if they're the same power of 2. This follows from the fact that $b + b = 2b$, and if $b$ is a power of 2, then $2b$ is the next higher power of 2. But if $b \neq c$, then $b + c \neq 2b$ nor $2c$ either, even if both $b$ and $c$ happen to be powers of 2.
So we have $2 + 2 = 4$, $4 + 4 = 8$, $8 + 8 = 16$, etc., but $2 + 4 = 2 \times 3$, $8 + 32 = 2^2 \times 5$, etc. Clearly $2^m + 2^m = 2(2^m) = 2^{m + 1}$.
This means you need to solve $2n = n + 1$, which leads to $k = 2n + 1$. There is only one possible solution in integers to $2n = n + 1$, and consequently only one solution to $2^{2n} + 2^{n + 1} = 2^{2n + 1}$, and that is $n = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of at least 3 sixes with 18 dice I am trying to logically think about this, not to just apply formulas that are readily available for similar problems. Can anyone help explain the math behind finding the probability of getting at least 3 sixes when rolling 18 (six sided) dice?
| The probability of not rolling any six is given by $\left(\frac{5}{6}\right)^{18}$, the probability of getting exactly one six is given by $18\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{17}$ and the probability of getting exactly two sixes is given by $\binom{18}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{16}$, so the probability of getting three or more sixes is given by:
$$ 1 - \left(\frac{5}{6}\right)^{18} - 18\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{17} - \binom{18}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{16} \approx \color{red}{59,73\%}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Mean value theorem for the second derivative, when the first derivative is zero at endpoints Suppose $f:[a,b]\to \mathbb R$ has derivative up to order $2$, and
$f'(a)=f'(b)=0$.
Prove there is a point $c$ at $(a,b)$ such that
$$
|f''(c)|\geq 4\frac{|f(b)-f(a)|}{(b-a)^2}.
$$
If it was factor 2, not 4, then I could use a Taylor expansion with Lagrange residue.
| Let $z = \frac{a + b}2$.
Taylor's expansion centered at $a$ is
$$
f(z) = f(a) + f'(a)(z - a) + \frac 12f''(\xi)(z - a)^2
$$
for some $\xi \in [a, z]$.
Since $f'(a) = 0$, we get
\begin{align}
f(z) - f(a) & = \frac 12 f''(\xi)(z - a)^2\\
\therefore f''(\xi) & = 2\frac{f(z) - f(a)}{(z - a)^2}. \tag{1}
\end{align}
Similarly, expansion centered at $b$ is
\begin{align}
f(z) & = f(b) + \frac 12f''(\eta)(z - b)^2\\
\therefore
f''(\eta) & = 2 \frac{f(z) - f(b)}{(z - b)^2} \tag{2}
\end{align}
for some $\eta \in [z, b]$.
Subtract (2) from (1), and divide by $2$:
\begin{align}
\frac{f''(\xi) - f''(\eta)}{2} = 4\frac{f(b) - f(a)}{(b - a)^2}.
\end{align}
Now, if $|f''(\xi)| \ge |f''(\eta)|$, it will follow that
$$
|f''(\xi)| \ge \frac{|f''(\xi)| + |f''(\eta)|}{2}
\ge \frac{|f''(\xi) - f''(\eta)|}{2} = 4\frac{|f(b) - f(a)|}{(b - a)^2}.
$$
Otherwise, $|f''(\xi)| < |f''(\eta)|$, and we get
$$
|f''(\eta)| \ge 4\frac{|f(b) - f(a)|}{(b - a)^2}
$$
by a similar reasoning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
polynomial remainder problem If a polynomial $f(x)$, of degree at least four, is such that
$$f(x) \equiv 3x+1 \mod (x^{2}-1)$$
and
$$f(x) \equiv 2x-3\mod (x^{2}+1),$$
find the remainder $g(x)$ such that
$$f(x) \equiv g(x) \mod (x^{2}-1)(x^{2}+1).$$
| The main steps are:
*
*We note that $\frac{1}{2} (x^2 + 1) - \frac{1}{2} (x^2 - 1) = 1$.
*Thus the inverse of $x^2 + 1 \mod x^2 -1$ is $1/2$ and the inverse of $x^2 - 1 \mod x^2 +1$ is $-1/2$.
*Thus a solution is $(3x+ 1) \frac{x^2 + 1}{2} + (2x-3) (-\frac{x^2 - 1}{2}) $.
The first step can be done in a more systematic way using the extended Euclidean algorithm. The general method is the Chinese Remainder Theorem for polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Prove that the limit of $a_n = \frac{2n^3+n^2}{(n+2)^3}$ is $2$. if $a_n = \dfrac{2n^3+n^2}{(n+2)^3}$, prove that the limit of $a_n$ (as $n$ tends to infinite) is $2$ using the definition of a limit.
My attempt was $\left |\dfrac{(2(n^3))+(n^2)}{(n+2)^3} - 2\right |=\left |\dfrac{(2(n^3))+(n^2)-2((n+2)^3)}{(n+2)^3}\right |$ and tried to simplify it as much as I could (though I couldn't get very far) then set it to less than Epsilon.
I'm not sure if that was the right approach though. Would anyone be able to help me out on this? Any help is greatly appreciated. Thank you!
| we will make a change of variable $u = n + 2, n = u - 2.$ then
$\begin{align}\frac{2n^3 + n^2}{(n+2)^3} &= \frac{2(u-2)^3 + (u-2)^2}{u^3}\\
& = \dfrac{2(u^3 - 6u^2+12u -8) +(u^2-4u+4)}{u^3}\\
&= 2 -\dfrac{11}{u}+\dfrac{20}{u^2}-\dfrac{12}{u^3}\end{align}$
pick an $\epsilon < 0.1.$ choose $N$ big enough so that for $u \ge N$ we have $$\dfrac{1}{u^3} \le \dfrac{1}{u^2}\le \dfrac{1}{u} \le\dfrac{\epsilon}{43} < 0.1$$
now $$|\frac{2(u-2)^3 + (u-2)^2}{u^3} -2|= |-\dfrac{11}{u}+\dfrac{20}{u^2}-\dfrac{12}{u^3}| \le \frac{43}{u} \le \epsilon$$
therefore $$ \lim_{n \to \infty}\frac{2n^3 + n^2}{(n+2)^3} = \lim_{u \to \infty} \frac{2(u-2)^3 + (u-2)^2}{u^3} = 2. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cannot solve an EL equation for a trivial textbook example Background. Retired engineer learning physics/need to learn calculus of variations as a tool.
Textbook: ML Boas Mathematical Methods for the Physical Sciences. Ed 3 Ch 9 Prob 2.6
Problem: Find $y(x)$ that MAKES
$$
\int (\frac{dy}{dx})^2 + \sqrt{y(x)} dx
$$
stationary.
I adopt the Euler_Lagrange route and thus need to solve what looks like a simple DE:
$$
\frac{d^2y}{dx^2}-\frac{1}{4\sqrt{y}}=0.
$$
The ELE functional is given by MLB as
$$
y'(x)^2+y^{\frac{1}{2}}.
$$
I can't do this by hand (always my first attempt). Mathematica 10 gives a massive, messy solution yet MLB claims that
$$
x+a = \frac{4}{3}(y^\frac{1}{2} -2b)(b+y^\frac{1}{2})^\frac{1}{2}
$$
'stationary-ides!!!) the ELE integral.
I apologise for this question to a community to which it is probably trivial but I know that mathematics is my toolkit for my study of physics and I must understand where I am going wrong. Thank you in advance for your help, David Mackay.
| Since your functional $F = y'^{2} + \sqrt{y}$ doesn't depend explicitly on $x$, there is a first integral of the form
$$\begin{align}
y' \frac{\partial F}{\partial y'} - F &= y' \cdot 2y' - (y'^{2} + \sqrt{y}) \\
&= y'^{2} - \sqrt{y} \\
&= C \\
\implies y'^{2} &= C + \sqrt{y} \\
\implies y' &= \pm \sqrt{C + \sqrt{y}}
\end{align}$$
Hence, separating and integrating gives us
$$\begin{align}
\int \frac{dy}{\sqrt{C + \sqrt{y}}} &= \pm \int dx \\
&= \pm (x + K) \\
\end{align}$$
To solve the LHS, we'll use a change of variable to simplify. Let
$$\begin{align}
y = u^{2} \implies dy &= 2u du\\
\end{align}$$
Hence
$$\begin{align}
\int \frac{dy}{\sqrt{C + \sqrt{y}}} &= \int \frac{2u du}{\sqrt{C + u}}
\end{align}$$
Using another change of variable, let
$$v = C + u \implies dv = du$$
Hence
$$\begin{align}
\int \frac{2u du}{\sqrt{C + u}} &= 2 \int \frac{(v - C) dv}{\sqrt{v}} \\
&= 2 \bigg[ \frac{2v^{\frac{3}{2}}}{3} - 2Cv^{\frac{1}{2}} \bigg ]
\end{align}$$
Unravelling, we find
$$\begin{align}
2 \bigg[ \frac{2v^{\frac{3}{2}}}{3} - 2Cv^{\frac{1}{2}} \bigg ] &= 2 \bigg[ \frac{2(C + u)^{\frac{3}{2}}}{3} - 2C(C + u)^{\frac{1}{2}} \bigg ] \\
&= 2 \bigg[ \frac{2(C + \sqrt{y})^{\frac{3}{2}}}{3} - 2C(C + \sqrt{y})^{\frac{1}{2}} \bigg ] \\
&= \pm (x + K)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Is there a Fibonacci identity? Let $F(n)$ be the n-th Fibonacci number. Is there any identity to simplify the following summation.
$$\sum_{i=1}^{n-1}F(i)^2F(n-i)^2 $$
| $$F(i)=\frac{a^i-b^i}{a-b},\ a=\frac{1+\sqrt{5}}{2},\ b=\frac{1-\sqrt{5}}{2},\\ F^2(i)=\frac{a^{2i}+b^{2i}-2(-1)^i}{5}\implies S(x)=\sum_{i\ge 1}F^{2i}x^i=\frac{\sum_{i\ge 1}a^{2i}x^i+\sum_{i\ge 1}b^{2i}x^i-2\sum_{i\ge 1}(-x)^i}{5}=\frac{1}{5}\left(\frac{1}{1-a^2x}+\frac{1}{1-b^2x}-\frac{2}{1+x}\right)=\frac{1}{5}\left(\frac{x^2+5x}{1-2x-2x^2+x^3}\right)$$ Then the given expression will be the coefficient of $x^{n}$ in the expansion of $S^2(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to expand $(x+y)^{-n}$? How to expand $(x+y)^{-n}$ by binomial theorem, where $n$ is a positive integer? Is there any limitation for $x$ and $y$? If it can be expanded, how to compute the coefficients? Many thanks in advance.
| This has a singularity along the line $y = -x$. The expansion in terms of powers of $y$ and $x$ will differ depending on whether you are above or below this line.
No binomial theorem, but note that
$$
\sum_{n = 0}^\infty z^n = \frac{1}{1-z}
$$
whenever $|z| < 1$.
If we relabel $y = -w$, and look at $(x - w)^{-1}$, we get
$$
\frac{1}{x-w} = \frac{1}{x}\frac{1}{1-\frac{w}{x}}
$$
and
$$
\frac{1}{x-w} = -\frac{1}{w}\frac{1}{1-\frac{x}{w}}
$$
which converge separately for $x > w$ and $w > x$ in each case (above and below the line x = w respectively).
But then $(x - w)^{-1}$ can be written as
$$
\sum_{n = 0}^{\infty} x^{-n -1} w^n
$$
and
$$
- \sum_{n = 0}^{\infty} x^n w^{-n-1}
$$
in each region.
Now,
$$
\partial_x^{n} (x - w)^{-1} = (-1)^{n} n! (x - w)^{-n-1}
$$
where I have written $\partial_x$ for the partial derivative with respect to $x$.
So
$$
(x - w)^{-n-1} = \frac{(-1)^n}{n!} \partial_x^n \left(\sum_{k > 0} x^{-k-1} w^k\right) \\
= \frac{(-1)^n}{n!} \sum_{k = 0}^\infty (-k-1)(-k-2)\cdots (-k - n) x^{-k-1-n}w^k \\
= (-1)^n \sum_{k = 0}^\infty (-1)^n \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1-n}w^k \\
= x^{-n} \sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1}w^k
$$
for the first case, and similarly,
$$
(x - w)^{-n-1} = \frac{(-1)^{n+1}}{n!} \partial_x^n \left(\sum_{k > 0} x^k w^{-k-1}\right) \\
= \frac{(-1)^{n+1}}{n!}\sum_{k \geq n} (k)(k-1)\cdots(k-n+1)x^{k-n} w^{-k-1} \\
= (-1)^{n+1} \sum_{k \geq n} \left(\begin{array}{c} k \\ n \end{array}\right)x^{k-n} w^{-k-1}\\
= (-1)^{n+1}w^{-n}\sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ n \end{array}\right)x^k w^{-k-1}
$$
for the second.
Note that this works for any values $x, y$ $(= -w)$ in $\mathbb{C}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $3*5*17*257-3^{16}$ using factorization formulas? look at this:
$2*4*10*82*6562-3^{16}$
It's easy to calculate it with elementary arithmetic.
but how to calculate it using factorization formulas?
| Hint $\ $ It is a radix $\,3\,$ analog of this radix $\,10\,$ (decimal) computation
$\qquad\qquad\qquad\qquad \underbrace{9\cdot 11}\cdot 101\cdot 10001 - 10^8 $
$\qquad\qquad\qquad\qquad\quad \underbrace{99 \cdot 101}$
$\qquad\qquad\qquad\qquad\qquad\ \ \underbrace{9999\cdot 10001}$
$\qquad\qquad\qquad\qquad\qquad\quad\ \ 99999999 - 10^8\, =\, -1$
Remark $\ $ At the heart of the matter is the following telescoping product (in your case $\,x=3).\,$ Notice that the same-colored terms all (diagonally) cancel out of the product.
$\qquad\qquad\, \displaystyle (x-1)(x+1)(x^2\!+1)(x^4\!+1)\quad\ \ \cdots\quad\ \ \ (x^{2^{\rm N}}\!+1)$
$\qquad\ \ \ = \ \displaystyle \frac{\color{#0a0}{x-1}}{\color{#90f}1} \frac{\color{brown}{x^2-1}}{\color{#0a0}{x-1}}\frac{\color{royalblue}{x^4-1}}{\color{brown}{x^2-1}}\frac{\phantom{f(3)}}{\color{royalblue}{x^4-1}}\, \cdots\, \frac{\color{#c00}{x^{2^{\rm N}}\!-1}}{\phantom{f(b)}}\frac{x^{2^{\rm N+1}}\!-1}{\color{#c00}{x^{\rm 2^N}\!-1}} \,=\, \frac{x^{2^{\rm N+1}}-1}{\color{#90f}1} $
For further telescopic intuition see my many posts on multiplicative telescopy.
| {
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"timestamp": "2023-03-29T00:00:00",
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