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Pascal triangle, binomial coefficients odd/even? I have managed to prove that $\dbinom{2^n}{k}$ is even for all $1\leq k \leq 2^n-n$ and that $\dbinom{2^n-1}{k}$ is odd for all $k$ using induction on $m$.
How can I prove that $\dbinom{2^n+r}{1+r},\dots,\dbinom{2^n+r}{2^m-1}$ are even for $0 \leq r \leq 2^m-2$ ?
How can I prove that $\dbinom{2^n+r}{r},\dbinom{2^n+r}{2^m}$ are odd for $0 \leq r \leq 2^m-1$?
| You've already proven all there is to prove about the case $r=0$, so I'll assume $1 \le r < 2^n$.
We can summarise the cases of interest as $$\binom{2^n+r}{m}, \quad r \le m \le 2^n$$ Since you're willing to assume Vandermonde's identity, we have $$\binom{2^n+r}{m} = \sum_{k=0}^m \binom{2^n}{k} \binom{r}{m-k}$$
Since $m \le 2^n$, all but the $k=0$ term and the $k=m$ term (in the case $m=2^n$) must be even, so $$\binom{2^n+r}{m} \equiv \binom{r}{m} + \binom{2^n}{m} \pmod 2$$
*
*When $m=r$ we get $\binom{2^n+r}{m} \equiv 1 + 0 \pmod 2$
*When $m=2^n>r$ we get $\binom{2^n+r}{m} \equiv 0 + 1 \pmod 2$
*When $r < m < 2^n$ we get $\binom{2^n+r}{m} \equiv 0 + 0 \pmod 2$.
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Surface tangent plane Determine a plane that is tangent to the surface $x ^{2}$ + $3y ^ 2$ + $2z ^ 2$ = $\frac{11}{6}$ and parallel to the plane x + y + z = 10
$$\nabla f (x, y, z) = (2x, 6y, 4z)$$
Quotation vector $X + Y + Z = 10 \implies (1,1,1)$
$$(2x, 6y, 4z) = K (1,1,1)$$
for $K = 1$
*
*$2x = 1 x = \frac{1}{2}$
*$6y = 1 y = \frac{1}{6}$
*$4z = 1 z = \frac{1}{4}$
Putting the plane equation together: $(1,1,1) [(x - 1/2) + (y-1/6) + (z-1/4)]$
My plan equation gave: $X + Y + Z = \frac{-11}{12}$
But the answer of the book is $X + Y + Z = \frac{-11}{6}$or X + Y + Z = $\frac{11}{6}$
| You cannot impose that $K=1$. You're after a point $(x,y,z)$ of the surface $x^2+3y^2+2z^2=\frac{11}6$ which is of the form $\left(\frac K2,\frac K6,\frac K4\right)$. So, you solve the equation$$\left(\frac K2\right)^2+3\left(\frac K6\right)^2+2\left(\frac K4\right)^2=\frac{11}6.$$You will get $K=\pm2$. Can you take it from here?
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Perfect numbers (prime number theory) Calculate x and y, so that $2 ^ x. 3 ^ y$ be a perfect number.
I tried to produce using this formula of the sum of the divisors of a number: Let $a ^ x, b ^ y,$ and $c ^ z$ prime factors of a number: $\frac{ a^{x+1} -1}{ a-1}. \frac{b^{y-1} -1}{b-1}. \frac{c^{c +1} -1}{c-1}$
It replaces $ 2 ^ x = a$ and $ 3^y = b $, but it can't produce much else: $ a.b $ = $ \frac {(2a-1) (3b-1)} {2} $
| The sum of all divisors of $2^x 3^y$ is $(2^{x+1}-1)(3^{y+1}-1)/2$. If $2^x 3^y$ is perfect, it means that:
$$(2^{x+1}-1)(3^{y+1}-1)/2=2\cdot 2^x3^y=2^{x+1}3^y$$
or
$$(2^{x+1}-1)(3^{y+1}-1)=2^{x+2}3^y$$
Now, this means that both $2^{x+1}-1$ and $3^{y+1}-1$ are products of only twos and threes; however, $2^{x+1}-1$ is not divisible by $2$ and $3^{y+1}-1$ is not divisible by $3$, so it follows that all the twos in the right-hand side ($2^{x+2}3^y$) "go" into $3^{y+1}-1$ and all the threes "go" into $2^{x+1}-1$, in other words:
$$\begin{array}{c}2^{x+1}-1=3^y\\3^{y+1}-1=2^{x+2}\end{array}$$
Now if you set $a=2^x, b=3^y$, this leads to the system of equations:
$$\begin{array}{c}2a-1=b\\3b-1=4a\end{array}$$
which has a unique solution $a=2, b=3$, so the only solution is $2^x 3^y=ab=6$.
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Prove that there are integers $n$ such that $\sqrt 2[f(1) + \cdots + f(n)] \in \mathbb Z$, $f(x) = \sqrt{1 - \frac{1}{2x^2 + \sqrt{4x^4 + 1}}}$.
Define function $f(x)$ as followed $$\large f(x) = \sqrt{1 - \dfrac{1}{2x^2 + \sqrt{4x^4 + 1}}}, \forall x \in \mathbb Z^+$$. Prove that there are infinitely many positive integers $n$ such that $$\large \sqrt 2[f(1) + f(2) + \cdots + f(n - 1) + f(n)] \in \mathbb Z^+$$
Let $g(x) = f(1) + f(2) + \cdots + f(n - 1) + f(n)$
We have that $$\sqrt{1 - \dfrac{1}{2x^2 + \sqrt{4x^4 + 1}}} = \sqrt{1 + 2x^2 - \sqrt{4x^4 + 1}} = \frac{\sqrt{2x^2 + 2x + 1} - \sqrt{2x^2 - 2x + 1}}{\sqrt 2}$$
Furthermore, we have that $2(n + 1)^2 - 2(n + 1) + 1 = 2n^2 + 2n + 1, \forall n \in \mathbb Z^+$
$$ \implies g(n) = f(1) + f(2) + \cdots + f(n - 1) + f(n) = \frac{\sqrt{2n^2 + 2n + 1} - \sqrt{2 \cdot 1^2 - 2 \cdot 1 + 1}}{\sqrt 2}$$
$$ \implies \sqrt{2} \cdot g(n) = \sqrt{2n^2 + 2n + 1} - 1$$
I'm not sure if my solution is incorrect or there is anything wrong with the problem.
| You have done great so far. Now note that $$2n^2+2n+1=n^2+(n+1)^2,$$ so you just need an infinite family of Pythagorean triplets of the form $(n,n+1,m)$. But if $(n,n+1,m)$ works, then $$(3n+2m+1,3n+2m+2, 4n+3m+2)\ \ \ \ \ (1)$$ also works.
Here is how to get the triplet (1). Let $2n^2+2n+1=m^2$. Then
$$(2n+1)^2-2m^2=2(2n^2+2n+1-m^2)-1=-1.$$
Thus $(2n+1,m)$ satisfies the following Pell-type equation: $$x^2-2y^2=-1.\ \ \ \ \ (2)$$ All integer solutions $(x,y)$ to (2) are known and they are given by
$$|x|+\sqrt{2}|y|=(1+\sqrt{2})^{2k+1}$$
for some non-negative integer $k$. But let's not bother with that. Observe that if $(x,y)$ is a solution to (2), then $(x',y')$ is again a solution where
$$x'+\sqrt{2}y'=(1+\sqrt{2})^2(x+\sqrt{2}y)=(3+2\sqrt2)(x+\sqrt2y).$$
So $(x,y)=(2n+1,m)=\big(n+(n+1),m\big)$ gives
$$(x',y')=(6n+4m+3,4n+3m+2)=\big((3n+2m+1)+(3n+2m+2),4n+3m+2\big).$$
In fact for a non-negative integer $n$, $\sqrt2 g(n)$ is an integer if and only if $n=n_k$ where
$$n_k=\frac{(1+\sqrt2)^{2k+1}-2+(1-\sqrt{2})^{2k+1}}{4},$$
for some non-negative integer $k$. We have $n_0=0$, $n_1=3$, and
$$n_{k+2}=6n_{k+1}-n_k+2.$$
So $n_2=20$ and $n_3=119$ for example. The corresponding $m_k$ to $n=n_k$ (i.e., $g(n_k)=m_k-1$) is
$$m_k=\frac{(1+\sqrt2)^{2k+1}-(1-\sqrt2)^{2k+1})}{2\sqrt{2}},$$
which can be obtained from the following recursion: $m_0=1$, $m_1=5$, and
$$m_{k+2}=6m_{k+1}-m_k.$$
You can also compute $g_k=g(n_k)$ directly from the recursion $g_0=0$, $g_1=4$, and
$$g_{k+2}=6g_{k+1}-g_k+4.$$
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Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| I don't think an expansion is needed at all. Here is an even more trivial proof.
Consider that if $a>b$, $a^3 > b^3$ and $a^5> b^5$ since $a$ and $b$ are positive.
So, we have that $$(a^3-b^3)(a^5-b^5) > 0$$ when $a>b\geq 0 \tag{1}$.
Now, if $a\leq b$,
$b^5\geq a^5$ and $b^3 \geq a^3$
so we have that $$(b^3-a^3)(b^5-a^5) = (a^3-b^3)(a^5-b^5) \geq 0 \tag{2}$$ when $b\geq a\geq 0$.
$(1)$ and $(2)$ are together necessary and sufficient condition to prove the required inequality, i.e. that $$\boxed{a^8+b^8 \geq a^3b^5 + a^5b^3}
~~~\blacksquare$$
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Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ Find a simple closed form of :
$\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
My try :
Let :
$A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
And
$B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
Now :
$A^{3}+B^{3}=56$
But how I can now find $A$ and $B$ ?
| As you did, let
$$ A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}, B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}. $$
Then
$$ A^3+B^3=56, AB=-\frac23 (18504483479)^{1/3}. $$
Let $z=A+B$ and then
$$ A^3+B^3=(A+B)[(A+B)^2-3AB]=z^3+2(18504483479)^{1/3}z=56. $$
So $z=A+B$ satisfies
$$ z^3+2(18504483479)^{1/3}z-56 = 0. $$
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Exercise from Knuth's TAoCP - Find a formula First of all, I think someone will probably flag this as duplicate due to the past post with same question :An exercise from Knuth's book - Proving a formula by induction
$\frac{1^3}{1^4+4}-\frac{3^3}{3^4+4}+\frac{5^3}{5^4+4}-...+\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}$
Which the proving by induction part has already been answered in the above post but so far no answer has been made on how to derive from the sum above to this (at least in the above post if i am missing out anything please flag this post)
$\frac{(-1)^n(n+1)}{4(n+1)^2+1}$
Any help on how to derive this would be extremely useful.Thanks alot and sorry for my English.
| The summand can be decomposed via partial fractions if one recalls the Sophie Germain factorization identity $$x^4 + 4 = (x^2 - 2x + 2)(x^2 + 2x + 2),$$ and noting that $$x^2 + 2x + 2 = (x+2)^2 - 2(x+2) + 2.$$ Hence $$(2n+1)^4 + 4 = \left((2n+1)^2 - 2(2n+1) + 2\right)\left((2n+3)^2 - 2(2n+3) + 2\right),$$ and we seek a partial fraction decomposition of the form $$\frac{(2n+1)^3}{(2n+1)^4 + 4} = \frac{A(2n+1)+B}{(2n+1)^2 - 2(2n+1) + 2} + \frac{A(2n+3)+B}{(2n+3)^2 - 2(2n+3) + 2}$$ for some constants $A$, $B$ with respect to $n$. Cross multiplication gives $$(2n+1)^3 = 16An^3 + (32A+8B)n^2 + (20A+8B)n + (8A+6B),$$ and equating coefficients yields $$A = 1/2, \quad B = -1/2;$$ hence $$\frac{(2n+1)^3}{(2n+1)^4 + 4} = \frac{f(n) + f(n+1)}{2},$$ where $$f(n) = \frac{(2n+1) - 1}{(2n+1)^2 - 2(2n+1) + 2}.$$ It follows that $$\begin{align*}
\sum_{k=0}^n \frac{(-1)^k(2k+1)^3}{(2k+1)^4 + 4}
&= \frac{1}{2} \sum_{k=0}^n (-1)^k \left( f(k) + f(k+1) \right) \\
&= \frac{1}{2}\left( f(0) + (-1)^n f(n+1)\right) \\
&= \frac{(-1)^n (n+1)}{4(n+1)^2 + 1}. \end{align*}$$
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A unit circle inscribed in parabola $y=x^2$ has center $(0,5/4)$. Generalize this result to $y=ax^2+bx+c$.
A circle with radius $1$ inscribed in the parabola $y = x^2$. If the coordinates of the center of the circle in this case is found to be $(0, 5/4)$. Can you generalize this problem for an arbitrary quadratic $y = ax^2 + bx + c$? If so, propose a mathematically precise generalization and prove it.
(Recall that the equation of a circle or radius $r$ centered at the point $(h,k)$ is given by $(x −h)^2 +(y −k)^2 = r^2$.)
| Rewrite the equation of the general parabola equation as,
$$y = a\left(x+\frac{b}{2a}\right)^2- \frac{b^2}{4a}+c$$
and let the corresponding equation of the circle
$$\left(x+\frac{b}{2a}\right)^2+(y-m)^2=1$$
The center of the circle is $(-\frac{b}{2a}, m)$ with $m$ to be solved next. Combine the two equations above to obtain the quadratic equation for $y$,
$$y^2+\left(\frac 1a -2m\right)y + \frac 1a \left(\frac {b^2}{4a} -c\right)+ m^2 -1=0$$
Since the parabola and the circle are tangential to each other, the discriminant of above quadratic equation is zero, which leads to the solution for the general case,
$$m= \frac{1}{4a}(1+4a^2-b^2+4ac)$$
In the special case of $y=x^2$, the center becomes $(-\frac{b}{2a}, m)=(0,\frac 54)$.
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Given points $A$, $B$, $C$, $D$ in a straight line, find $O$ in the line such that $OA:OB=OC:OD$.
Given four points $A$, $B$, $C$, $D$ in a straight line, find a point $O$ in the same straight line such that $OA:OB=OC:OD$.
I tried doing this by drawing a ray through $A$ and drawing lines through $B$, $C$, and $D$ parallel to each other. That way we get three triangles. I do not know what to do from here.
Am I going in the right direction? And if no, provide a hint.
Thanks
| I will play with the case where Larry's great solution fails to find a point (or you can say that Larry's solution yields the point $O$ at infinity, but there are two finite points that also satisfy the condition). That is, when the midpoint of the line segment $AD$ coincides with the midpoint of the line segment $BC$. Let $M$ denote this common midpoint.
WLOG, suppose that the points $A,B,M,C,D$ are arranged from left to right on the given line $\ell$. Let $P$ be the midpoint of $AB$. Construct a right triangle $XYZ$ with $\angle XYZ=\pi/2$, $XY=AP$, and $YZ=MP$. Let $O_1,O_2$ be two distinct points on the line $\ell$ that are at distance $XZ$ away from $M$. Then $O=O_1$ or $O=O_2$ will give you a desired point $O$.
To show these indeed are solutions, we suppose that $AM=a$ and $BM=b$. Then $XY=MP=(a+b)/2$ and $YZ=AP=(a-b)/2$. Hence, $$MO_1=MO_2=XZ=\sqrt{XY^2+YZ^2}=\sqrt{\frac{a^2+b^2}2}.$$
WLOG, suppose that $O_1$ is on the left of $M$ and $O_2$ is on the right. Then
$$\frac{O_1A}{O_1B}=\frac{a-\sqrt{(a^2+b^2)/2}}{\sqrt{(a^2+b^2)/2}-b}=\frac{\frac{(a^2-b^2)/2}{a+\sqrt{(a^2+b^2)/2}}}{\frac{(a^2-b^2)/2}{\sqrt{(a^2+b^2)/2}}+b}=\frac{\sqrt{(a^2+b^2)/2}+b}{a+\sqrt{(a^2+b^2)/2}}.$$
However
$$\frac{O_1C}{O_1D}=\frac{\sqrt{(a^2+b^2)/2}+b}{a+\sqrt{(a^2+b^2)/2}}.$$
We have a similar proof for $O_2$. So both $O_1$ and $O_2$ satisfy the condition.
In general, let $R_1$ be the midpoint of $AD$, $R_2$ the midpoint of $BC$, and $M$ the midpoint of $R_1R_2$. Suppose now that $A,B,C,D$ are arranged from left to right on the given line $\ell$. Now we measure distance with signs that is $UV=-VU$ and $UV$ is positive if $U$ is on the left of $V$. Let $AM=a$, $BM=b$, $CM=c$, and $DM=d$.
Then apart from point $O_0$ obtained in Larry's solution, there are two other points $O_1,O_2$ if and only if $ad+bc\le 0$. In such cases, the other two points satisfies
$$O_1M=-\sqrt{-\frac{ad+bc}{2}}$$
and $$O_2M=\sqrt{-\frac{ad+bc}{2}}.$$
You can construct $\delta=\sqrt{-\frac{ad+bc}{2}}$ via
$$\delta=\sqrt{\left(\frac{a+b-c-d}{4}\right)^2+\left(\frac{a-b+c-d}{4}\right)^2-\left(\frac{a+b+c+d}{4}\right)^2-\left(\frac{a-b-c+d}{4}\right)^2}.$$
Notice that $a+b+c+d=0$. Let $P_1$ be the midpoint of $AB$, $P_2$ the midpoint of $CD$, $Q_1$ be the midpoint of $AC$, and $Q_2$ the midpoint of $BD$. Then,
$$\delta=\sqrt{MP_1^2+MQ_1^2-MR_1^2}.$$
We then construct $\delta$ as follows. First draw a right triangle $XYZ$ with $\angle XYZ=\pi/2$, $XY=|MP_1|$, and $YZ=|MQ_1|$. Then on the circumcircle of the triangle $XYZ$, locate a point $W$ such that $XW=|MR_1|$. Then, $\delta=ZW$.
The points $O_1$ and $O_2$ work simply because the roots of
$$t^2-\left(-\frac{ad+bc}{2}\right)=\frac{(t-a)(t-d)+(t-b)(t-c)}{2}=0$$
are precisely $\pm\delta$. This means
$$\frac{|t-a|}{|t-b|}=\frac{|t-c|}{|t-d|}$$
for $t=\pm\delta$.
In some degenerate cases such as when three of the points $A,B,C,D$ coincide, say $B=C=D$, one of the points $O_1$ and $O_2$, say $O_1$, is the midpoint of $AD$. So we do not have division by $0$ in $\frac{O_1A}{O_1B}=\frac{O_1C}{O_1D}$ but Larry's solution yields $O_0=B=D$. Hence, it is useful to know how to construct $O_1$ and $O_2$ just in case things fail (and in contest math, sometimes you have to be careful to cover all bases). In a real contest, since the problem doesn't state that $A,B,C,D$ are necessarily distinct points, we should perhaps include the case $A=B$ and $C=D$ where the point $O$ is any point on the straight line not equal to $A=B$ or $C=D$, just to be careful.
There are exactly three degenerate cases where $O$ does not exist (as a finite point). In the case when $A=B$ is the midpoint of the segment $CD$, it can be shown that the equality $OA\cdot OD=OB\cdot OC$ holds if and only if $O=A=B$, but then we have division by $0$ in $\frac{OA}{OB}=\frac{OC}{OD}$. Similarly when $B=D$ is the midpoint of $AC$ or when $C=D$ is the midpoint of $AB$, we have a similar problem.
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A box contains m white balls and n black balls This is an example from Probability, Random Variables and Stochastic Process by Papoulis
A box contains m white balls and n black balls. Balls are drawn at random one at a time without replacement. Find the probability of encountering a white ball by the kth draw.
$W_k=$ {a white ball is drawn by the kth draw}
$X_i=$ {i black balls followed by a white ball are drawn}
$W_k=X_0\cup X_1\cup... \cup X_{k-1}$
The probability for the first white ball to be drawn is:
$P(X_0)=\dfrac{m}{m+n}$
$P(X_1)=\dfrac{n}{m+n}.\dfrac{m}{m+n-1}$
This is what offered in the book, I try to write down a few more examples to get a grip of it:
$P(X_2)=\dfrac{n}{m+n}.\dfrac{n-1}{m+n-1}.\dfrac{m}{m+n-2}$
$P(X_3)=\dfrac{n}{m+n}.\dfrac{n-1}{m+n-1}.\dfrac{n-2}{m+n-2}.\dfrac{m}{m+n-3}$
The book writes:
$P(X_{k-1})=\dfrac{n(n-1)...(n-k+1)m}{(m+n)(m+n-1)...(m+n-k+1)}.$
I don't understand this part, which is a general form of the above result.
Then the author writes that:
$P(W_k)=\dfrac{m}{m+n}(1+\dfrac{n}{m+n-1}+\dfrac{n(n-1)}{(m+n-1)(m+n-2)}+...\dfrac{n(n-1)...(n-k+1)}{(m+n-1)(m+n-2)...(m+n-k+1)}$
I don't understand this part as well.
By the $(n+1)$st draw, we must have a white ball, hence $P(W_{n+1})=1$
Why is it equal 1, I don't get this part.
The last part of this example is also beyond my understanding:
$1+\dfrac{n}{m+n-1}+\dfrac{n(n-1)}{(m+n-1)(m+n-2)}+...+\dfrac{n(n-1)...2.1}{(m+n-1)(m+n-2)...(m+1)m}=\dfrac{m+n}{m}$
I don't understand this equality, how can the left side is equal to right side?
| Your calculations for $\Pr(X_2)$ and $\Pr(X_3)$ are correct.
By definition, $\Pr(X_{k - 1})$ is the probability that $k - 1$ black balls are drawn before a white ball is drawn on the $k$th draw, which is
\begin{align*}
\Pr(X_{k - 1}) & = \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{n - 2}{m + n - 2} \ldots \frac{n - (k - 2)}{m + n - (k - 2)} \cdot \frac{m}{m + n - (k - 1)}\\
& = \frac{n(n - 1)(n - 2) \ldots (n - k + 2)m}{(m + n)(m + n - 1)(m + n - 2) \ldots (m + n - k + 2)(m + n - k + 1)}
\end{align*}
since there are $n - (j - 1)$ black balls left from which to choose on the $j$th draw, where $1 \leq j \leq k - 1$, out of $m + n - (j - 1)$ balls in total, and there are $m$ white balls from which to draw out of $m + n - (k - 1)$ balls in total on the $k$th draw.
Therefore, the book appears to have an indexing error.
Since $\Pr(W_k)$ represents the probability that a white ball is drawn by the $k$th draw,
\begin{align*}
\Pr(W_k) & = \Pr(X_0) + \Pr(X_1) + \Pr(X_2) + \cdots + \Pr(X_{k - 1})\\
& = \frac{m}{m + n} + \frac{n}{m + n} \cdot \frac{m}{m + n - 1} + \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{m}{m + n - 2}\\
& \qquad + \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{n - 2}{m + n - 2} \ldots \frac{n - (k - 2)}{m + n - (k - 2)} \cdot \frac{m}{m + n - (k - 1)}\\
& = \frac{m}{m + n}\left[1 + \frac{n}{m + n - 1} + \frac{n}{m + n - 1} \cdot \frac{n - 1}{m + n - 2} + \ldots \right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{n - (k - 2)}{m + n - (k - 1)}\right]
\end{align*}
where we have factored out an $m$ from the last term in the numerator and an $m + n$ from the first term in the denominator.
Since there are only $n$ black balls, we are guaranteed to obtain a white ball by the $(n + 1)$st draw. Since $W_{n + 1}$ is the event that we draw a white ball by the $(n + 1)$st draw, $\Pr(W_{n + 1}) = 1$.
Notice that
\begin{align*}
\Pr(W_{n + 1}) & = \Pr(X_0) + \Pr(X_1) + \Pr(X_2) + \cdots + \Pr(X_{k - 1}) + \Pr(X_n)\\
& = \frac{m}{m + n} + \frac{n}{m + n} \cdot \frac{m}{m + n - 1} + \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{m}{m + n - 2}\\
& \quad + \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{n - 2}{m + n - 2} \ldots \frac{n - (n - 2)}{m + n - (n - 2)} \cdot \frac{m}{m + n - (n - 1)}\\
& \qquad + \frac{n}{m + n} \cdot \frac{n - 1}{m + n - 1} \cdot \frac{n - 2}{m + n - 2} \ldots \frac{n - (n - 2)}{m + n - (n - 2)} \cdot \frac{n - (n - 1)}{m + n - (n - 1)} \cdot \frac{m}{m + n - n}\\
& = \frac{m}{m + n}\left[1 + \frac{n}{m + n - 1} + \frac{n}{m + n - 1} \cdot \frac{n - 1}{m + n - 2} + \ldots \right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{n - (n - 2)}{m + n - (n - 1)}\right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{2}{m + 1} \cdot \frac{1}{m}\right]
\end{align*}
Substituting $1$ for $W_{n + 1}$ yields
\begin{align*}
1 & = \frac{m}{m + n}\left[1 + \frac{n}{m + n - 1} + \frac{n}{m + n - 1} \cdot \frac{n - 1}{m + n - 2} + \ldots \right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{n - (n - 2)}{m + n - (n - 1)}\right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{2}{m + 1} \cdot \frac{1}{m}\right]
\end{align*}
from which we conclude that
\begin{align*}
\frac{m + n}{m} & = \left[1 + \frac{n}{m + n - 1} + \frac{n}{m + n - 1} \cdot \frac{n - 1}{m + n - 2} + \ldots \right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{n - (n - 2)}{m + n - (n - 1)}\right.\\
& \qquad \left. + \frac{n}{m + n - 1} \cdot \frac{n - 2}{m + n - 1} \cdot \frac{n - 3}{m + n - 2} \ldots \frac{2}{m + 1} \cdot \frac{1}{m}\right]
\end{align*}
Having said all that, I agree with the comment posted by JMoravitz. It is also how I approach such problems.
| {
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"url": "https://math.stackexchange.com/questions/3383379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Irreducible representation of the dihedral group of order 6 I know that up to isomorphism there is only one irreducible representation of $D_3$ of dimension 2. Write $D_3=\{1, x, x^2, xy, x^2y, y\}$, where $x^3 = 1$, $y^2 = 1$, $xy = yx^2$. Given any $\theta\in[0, 2\pi)$, I think the following is one of such irreducible representation:
$R_y = \left[ \begin{array}{cc}
1 & 0 \\ 0 & -1 \end{array} \right],$
$R_x = \left[ \begin{array}{cc}
-\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}e^{-i\theta} \\ \dfrac{\sqrt{3}}{2}e^{i\theta} & -\dfrac{1}{2} \end{array} \right].$
Am I correct? If the answer is yes, then what is the isomorphism $T:\mathbb{C^2}\rightarrow\mathbb{C^2}$ between this representation and the standard representation
$R'_y = \left[ \begin{array}{cc}
1 & 0 \\ 0 & -1 \end{array} \right],$
$R'_x = \left[ \begin{array}{cc}
-\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{array} \right].$
On the other hand, if the answer is no, then what forces $\theta$ to be $0$?
Thanks!
| You have $R_xA=AR'_x$, where
$$
A=\begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}
$$
So the two representations are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Homework help: Exponential and Logarithmic Equations I am kinda stuck with the last two equations. Just need a kick/hint in the right direction to get it done.
My question is: How to simplify them to some base form from which I can get the x value. I want to know the method.
$8^{x+3}=6\cdot 2^x+4\cdot 4^{x+2}$
$2\cdot\ln\left(x\right)-\ln\left(x-5\right)=\ln\left(x+1\right)$
| For the equation
$$8^{x + 3} = 6 \cdot 2^x + 4 \cdot 4^{x + 2}$$
use $2$ as the common base.
\begin{align*}
8^{x + 3} & = 6 \cdot 2^x + 4 \cdot 4^{x + 2}\\
(2^3)^{x + 3} & = 6 \cdot 2^x + 4 \cdot (2^2)^{x + 2}\\
2^{3x + 9} & = 6 \cdot 2^x + 4 \cdot 2^{2x + 4}\\
2^9 \cdot 2^{3x} & = 6 \cdot 2^x + 4 \cdot 2^4 \cdot 2^{2x}\\
512 \cdot 2^{3x} & = 6 \cdot 2^x + 64 \cdot 2^{2x}\\
256 \cdot 2^{3x} & = 3 \cdot 2^x + 32 \cdot 2^{2x}
\end{align*}
Let $u = 2^x$ to obtain
$$256u^3 = 3u + 32u^2$$
and note that $u = 2^x > 0$ for every real number $x$, so $u = 0$ is not a solution of the cubic.
For the equation
$$2\ln x - \ln (x - 5) = \ln (x + 1)$$
we obtain
\begin{align*}
\ln x^2 - \ln (x - 5) & = \ln (x + 1)\\
\ln\left(\frac{x^2}{x - 5}\right) & = \ln (x + 1)\\
e^{\ln\left(\frac{x^2}{x - 5}\right)} & = e^{\ln (x + 1)}\\
\frac{x^2}{x - 5} & = x + 1
\end{align*}
Keep in mind that $\ln x$ is only defined when $x > 0$, $\ln (x - 5)$ is only defined when $x - 5 > 0$, and $\ln (x + 1)$ is only defined when $x + 1 > 0$. Your final answer must satisfy all three of those restrictions.
| {
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"url": "https://math.stackexchange.com/questions/3384480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| Correct answer:
Let $y=2x-3$. Then $f(y)=4x-2=2(2x-3)+4=2y+4.$
Where you went wrong:
If $f(2x-3)=4x-2=2(2x)-2$, then $f(2x)=2(2x+3)-2=4x+4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreciated.
| Fix an $n$. By analytic continuation, it suffices to assume $y$ be a small positive number, so small such that $0<\frac{\pi(2k+1)}{2n}+y<\pi$ for $k=0,\cdots,n-1$.
Integrate $$f_m(z) = \tan (n(z-y)) \csc^{2m} z$$ around rectangle with vertices $\pm \infty i, \pi \pm \infty i$ (Indention has to be made at $0$ and $\pi$), integrals along two horizontal lines tend to $0$, integrals along two vertical lines cancel by periodicity $f_m(z)=f_m(z+1)$.
$f_m(z)$ has poles at $$z=0 \qquad z = \frac{\pi(2k+1)}{2n}+y\quad \text{ for }k=0,\cdots,n-1$$
therefore
$$\text{Res}[f_m(z),z=0] -\frac{1}{n}\sum _{k=0}^{n-1} \csc^{2m}\left(\frac{\pi (2 k+1)}{2 n}+y\right) = 0$$
Calculating residues proves your result, along with instances of higher powers:
$$\begin{aligned}&\sum _{k=0}^{n-1} \csc^{6}\left(\frac{\pi (2 k+1)}{2
n}+y\right)=\\ &\frac{1}{15} n^2 \sec ^2(n y) \left(15 n^4 \sec ^4(n y)-15 n^4 \sec ^2(n y)+2 n^4+15 n^2 \sec ^2(n y)-10 n^2+8\right)\\ &\sum _{k=0}^{n-1} \csc^{8}\left(\frac{\pi (2 k+1)}{2
n}+y\right)=\\ &\frac{1}{315} n^2 \sec ^2(n y) \left(315 n^6 \sec ^6(n
y)-420 n^6 \sec ^4(n y)+126 n^6 \sec ^2(n y)-4 n^6+420 n^4 \sec ^4(n
y)-420 n^4 \sec ^2(n y)+56 n^4+294 n^2 \sec ^2(n y)-196
n^2+144\right) \end{aligned}$$
Taking $y=0$, one can also easily deduce $\zeta(2n)$ from such finite trigonometric sums.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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piecewise ceiling function and what $g(x) = f(1/x)$ means I have some limits I am supposed to find and I'm having trouble understanding the question.
Given:
$$f(x) = \begin{cases} 2 & \text{ if } \left \lceil x \right \rceil
\text{is even}\\-1 & \text{ if } \left \lceil x \right \rceil
\text{is odd} \end{cases}$$
$g(x) = f\left(\frac{1}{x}\right)$
what does $g(x) = f\left(\frac{1}{x}\right)$ mean? (maybe I'm making it more complicated than it should be in my head, it's supposed to be a 'challenge' problem.)
example: $g\left(\frac{1}{3}\right)$, does that mean $g\left(\frac{1}{\frac{1}{3}}\right)$ so then $g(3) = -1$ ?
or like a composite function where $g\left(\frac{1}{3}\right)$ is odd so $f\left(\frac{1}{3}\right) = -1$ then $g(x) = \frac{1}{-1} = -1$?
I realize both come out to $-1$, but I think it makes a difference for when I have to find the limits. Or maybe not?
| $g(x) = f(\frac 1x)$ means just what it looks like:
$g:x\to f(\frac 1x)$.
So for example: If $x = \frac {\sqrt 3}4$ then
$g(x)=g(\frac {\sqrt 3}4) = f(\frac 4{\sqrt 3}) = \begin{cases}2&\text{if }\lceil \frac 4{\sqrt 3} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 4{\sqrt 3} \rceil \text{ is odd}\end{cases}$
$=2$
Or $g(52) = f(\frac 1{52}) = \begin{cases}2&\text{if }\lceil \frac 1{52} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 1{52} \rceil \text{ is odd}\end{cases}$
$=-1$
......
Or we could simple say $g(x) =\begin{cases}2&\text{if }\lceil \frac 1{x} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 1{x} \rceil \text{ is odd}\end{cases}$
.....
Might, or might not, be worth noting:
If $0< x \le 1$ then $f(x) = -1$ and if $-1 < x \le 0$ then $f(x) = 2$.
So if $x \ge 1$ then $0< \frac 1x \le 1$ so $g(x) = -1$. And if $x \le -1$ then $-1 \le \frac 1x < 0$ so $g(x)=2$.
And $g(0)$ is not defined.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conditional probability with two inequalities Let $(X,Y)$ a random vector with density $f(x,y)=\frac{6}{7}(x^2+\frac{xy}{2});0<x<1,0<y<2$.
After checking that is a density, finding the marginal $f_X(x)=\frac{12}{7}x^2+\frac{6}{7}x$ and the probability $\mathbb{P}(X>Y)=\frac{15}{56}$, i have to find $\mathbb{P}(Y>\frac{1}{2}|X<\frac{1}{2})$.
I know that the domain is $0<x<\frac{1}{2}<y<2$, so i applied the conditional probability $\mathbb{P}(Y>\frac{1}{2}|X<\frac{1}{2})=\frac{\mathbb{P}((Y>\frac{1}{2})\cap (X<\frac{1}{2}))}{\mathbb{P}(X<\frac{1}{2})}$ and i tried to integrate in the most different ways…
1) $\frac{\int_{0}^{3/2}(\int_{0}^{5/2}f(x,y)dy)dx}{\int_{0}^{1/2}f_X(x)dx}$
2) $\frac{\int_{0}^{1/2}(\int_{1/2}^{2}f(x,y)dy)dx}{\int_{0}^{1/2}f_X(x)dx}$
3) ${\int_{0}^{1/2}(\int_{1/2}^{2}\frac{f(x,y)}{f_X(x)}dy)}dx$
…but to no avail.
How to act when the conditional probability provides for two disequalities?
EDIT: Trivially, you might say "Making the graph". Well, I'm having a hard time to understand how to draw it correctly.
In the first probability that i said, the graph was fairly simple because I could use the 1°-3° bisector to separate the region to integrate.
| For $a\in\left[0,1\right]$ and $b\in\left[0,2\right]$ define:
$$\begin{aligned}f\left(a,b\right) & :=P\left(X<a,Y>b,\right)\\
& =\frac{6}{7}\int_{0}^{a}\int_{b}^{2}x^{2}+\frac{1}{2}xy\;dydx\\
& =\frac{6}{7}\int_{0}^{a}\left[x^{2}y+\frac{1}{4}xy^{2}\right]_{b}^{2}\;dx\\
& =\frac{6}{7}\int_{0}^{a}\left(2-b\right)x^{2}+\frac{1}{4}\left(4-b^{2}\right)x\;dx\\
& =\frac{6}{7}\left[\frac{1}{3}\left(2-b\right)x^{3}+\frac{1}{8}\left(4-b^{2}\right)x^{2}\right]_{0}^{a}\\
& =\frac{6}{7}\left(\frac{1}{3}\left(2-b\right)a^{3}+\frac{1}{8}\left(4-b^{2}\right)a^{2}\right)\\
& =\frac{1}{28}a^{2}\left(2-b\right)\left(8a+6+3b\right)
\end{aligned}
$$
Then $P\left(X<\frac{1}{2}\right)=f\left(\frac{1}{2},0\right)$ and
$P\left(Y>\frac{1}{2},X<\frac{1}{2}\right)=f\left(\frac{1}{2},\frac{1}{2}\right)$
so that: $$P\left(Y>\frac{1}{2}\mid X<\frac{1}{2}\right)=\frac{f\left(\frac{1}{2},\frac{1}{2}\right)}{f\left(\frac{1}{2},0\right)}$$
Just substitute.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $x \cdot \sin \left( \frac{1}{x} \right) = 1$ has solutions? A look to the plot show to you that the function $f(x) = x \cdot \sin \left( \frac{1}{x} \right) - 1 $ has no zero near the origin. Wolframalpha software says that $ x \cdot \sin \left( \frac{1}{x} \right) = 1$ if $x \approx 5.16\cdot 10^{15}$ but I suspect this huge number is a wrong solution due to numerical problems. Is it true that for some $x$, far away from the origin, the equation $$ x \cdot \sin \left( \frac{1}{x} \right) = 1 $$ is satisfied? Can, far away, $f(x)= x \cdot \sin \left( \frac{1}{x} \right)$ exceeds 1?
| $$x \sin (1/x) =1 \implies \sin (\frac{1}{x})=\frac{1}{x} \implies \sin y =y.$$
But $$y > \sin y,~ if ~ y>0 ;~ y < \sin y ~ ~ if ~ ~ y<0.$$~
The only solution the last equation may have is $y=0$, but it is not allowed here as it would mean $x=\infty$. So the given equation cannot have a real root.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{0}^{x^2 (1+x)} f(t) dt = x$ According to the second fundamental theorem of Calculus, for this problem $F(x^2(1+x))-F(0)=x$ can be used.
But I can't find a function F that works for this problem. Can someone help me with this?
| Hint. You already have $$F(x^2(1+x))-F(0)=x.$$ Now, differentiate both sides to get $$x(3x+2)f(x^2(1+x))=1.$$
You can now solve for $f(x^2(1+x))$ explicitly, from which you can solve for $f(x)$, as long as you choose your domain wisely (since $x\mapsto x^2(1+x)$ is not a bijection from $\mathbb R$ to itself).
To do the last part, let's substitute $y=x^2(1+x)$, so that $x^3+x^2-y=0$. We want to solve this for $x$. This is quite troublesome to solve analytically (since it's a cubic), but can be done: let $x=1/a$ and multiply throughout by $a^3$ to get
$$1+a-ya^3=0.$$
The idea here is to use the triple angle identity for cosine, $\cos(3\theta)=4\cos^3\theta-3\cos\theta$. We substitute $a=k\cos\theta$ for some constant $k$ to get
$$yk^3\cos^3\theta-k\cos\theta+1=0.$$
We need the ratio of coefficients of $\cos^3\theta$ and $\cos\theta$ to be $4:-3$, so that $yk^2=4/3$, or $k=\sqrt{4/(3y)}$. So
$$\frac43\sqrt{\frac{4}{3y}}\cos^3\theta-\sqrt{\frac{4}{3y}}\cos\theta+1=0\iff\frac{2}{3\sqrt{3y}}\cos(3\theta)=-1.$$
Finally we can solve for $\cos(3\theta)$ in terms of $y$. Then, once we appropriately restrict the domain, we can find $\theta$ and hence $a$ and hence $x$ analytically. Once we have $x$ in terms of $y$, substituting back into
$$f(x^2(1+x))=\frac1{x(3x+2)}$$
will give us the answer, since the LHS will become $f(y)$ and the RHS will become something in terms of $y$. This provides an explicit formula for $f$.
| {
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For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$ Find all polynomials $P(x)$ with real coefficients such that:- $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
This is $a$ creating some big problems for me. I don't know what to do.
I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.
If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.
Any help would be appreciated
| the function is not a polynomial
$$ (x^2-ax+18)P(x) = (x^2+3x)P(x-3) $$
Put $x=-3$,
$$ (27+3a)P(-3) = (9-9)P(-6) $$ $ (27+3a) \cdot P(-3)=0$, $27+3a=0$, then $a= -9$
$$ (x²-(-9)x+18)P(x) = (x²+3x)P(x-3) $$
$$(x^2+9x+18)P(x) = (x^2+3x)P(x-3) $$
$$(x+3)(x+6)P(x) = x(x+3)P(x-3) $$
$$(x+6)P(x) = xP(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+P(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+ \frac{(x+6)}{x} P(x)$$
$$6P'(x) = \frac{(x+6)}{x} P(x)-xP(x) $$
$$6P'(x) = ( 1+ \frac{6}{x}-x)P(x) $$
$$ \frac{P'(x)}{P(x)} = \frac{1}{6}+\frac{1}{x}-\frac{x}{6} $$
$$ \int \frac{P'(x)}{P(x)} = \int ( \frac{1}{6}+\frac{1}{x}-\frac{x}{6} ) $$
$$\log(P(x))+c = \frac{x}{6}-\frac{x^2}{12}+\log(x) $$
$$\log(P(x)) = \frac{x}{6}-\frac{x^2}{12}+\log(x)-c $$
$$ P(x) = \exp ( \frac{x}{6}-\frac{x^2}{12}+\log(x)-c ) $$
$$P(x) = x \cdot \exp( \frac{x}{6}-\frac{x^2}{12}-c) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
| Yes we can eliminate it indeed
*
*$x\equiv 1 \pmod 2 \implies x=2k+1$
*$x\equiv 2 \pmod 3 \implies 2k+1\equiv 2 \pmod 3\implies k\equiv 2 \pmod 3\implies k=3h+2$
that is
$$x=6k+5 \implies x\equiv 5\pmod 6$$
Note that we can also eliminate $x\equiv 1 \pmod 2$ since we have that $x\equiv 3 \pmod 4$.
Then we can use CRT to solve the system:
*
*$x\equiv 2 \pmod 3$
*$x\equiv 3 \pmod 4$
*$x\equiv 4 \pmod 5$
*$x\equiv 0 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int x \sqrt{x+7}\,dx$ Consider the integral
$$\int x \sqrt{x+7} \,dx$$
Let $u=x+7$, so that $x=u-7$, $du=dx$. Then the integral becomes
$$\int(u-7)u^{1/2}du $$
$$=\int\left(u^{3/2}-7u^{1/2}\right) du, $$
which evaluates to
$$\frac{2}{5}(x+7)^{5/2}-\frac{14}{3}(x+7)^{3/2}+C.$$
Now even though this is correct, the answer that my professor is looking for is
$$\frac{2}{15}(x+7)^{3/2}(3x-14)+C $$
So my question is how do I manipulate my answer to get the one above?
| Simply observe that $(x+7)^{3/2}$ is "common throughout":
\begin{align*}
\frac{2}{5}(x+7)^{5/2}-\frac{14}{3}(x+7)^{3/2} &= (x+7)^{3/2}\left[\frac25(x+7)^1-\frac{14}3\cdot 1\right]\\[3pt]
&= \frac2{15}(x+7)^{3/2}(3x-14).
\end{align*}
Another easy illustration of this is, for example, $x^{1/2} + x^{5/2} = x^{1/2}(1+x^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$
How to evaluate the following challenging sum:
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+i)?$$
where $H_{n}$ is the harmonic number, $\operatorname{Li}_n$ is the polylogarithm function and $G$ is Catalan constant.
This problem was proposed by a friend and here is my approach:
Using the identity
$$\int_0^1\frac{x^n}{1+x}\ dx=H_{n/2}-H_n+\ln2, \quad x\mapsto x^2$$
$$2\int_0^1\frac{x^{2n+1}}{1+x^2}\ dx=H_{n/2}-H_n+\ln2$$
Multiply both sides by $\frac{(-1)^n}{(2n+1)^2}$ to get
$$\sum_{n=0}^\infty (-1)^n\frac{H_{n/2}-H_n+\ln2}{(2n+1)^2}=2\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}\ dx\\=2\int_0^1\frac{1}{1+x^2}\Im\sum_{n=0}^\infty\frac{(i)^nx^n}{n^2}\ dx=2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
Rearranging the terms to get
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\ln2G+2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
where
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}=-\int_0^1\ln x\sum_{n=0}^\infty(-1)^n x^{2n}H_n\ dx=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx$$
which is calculated here:
$$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$
And the question here is how to evaluate the dilogarithm integral or a different way to compute $S$? Thank you.
| We have that:$$\Im\operatorname{Li}_2 (ix)=\operatorname{Ti}_2 (x)=\int_0^x \frac{\arctan t}{t}dt=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^2}$$
So let's rewrite the integral in terms of the inverse tangent function:$$\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx$$
$$\overset{IBP}=\arctan x\operatorname{Ti}_2(x)\bigg|_0^1-\int_0^1 \frac{\arctan^2 x}{x}dx=\frac{\pi}{4}G-\mathcal J$$
$$\mathcal J=\int_0^1\frac{\arctan^2 x}{x}dx\overset{x=\tan t}=2\int_0^\frac{\pi}{4} \frac{t^2}{\sin(2t)}dt\overset{2t=x}=\frac14\int_0^\frac{\pi}{2}\frac{x^2}{\sin x}dx=\frac{\pi}{2}G-\frac{7}{8}\zeta(3)$$
The last integral is nicely evaluated here.
$$\Rightarrow \boxed{\Im\int_0^1 \frac{\operatorname{Li}_2(ix)}{1+x^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx=\frac{7}{8}\zeta(3)-\frac{\pi}{4}G}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the summation of a series Determine the following sum
$$\frac{5}{3 \cdot 6 }\cdot \frac{1}{4^2} + \frac{5\cdot 8}{3 \cdot 6 \cdot 9 }\cdot \frac{1}{4^3} + \frac{5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot \frac{1}{4^4} + \frac{5\cdot8\cdot 11\cdot 14}{3 \cdot 6\cdot 9\cdot 12 \cdot 15}\cdot \frac{1}{4^5} + \dots \dots$$
I tried to use Generalised Binomial Theorem, but I am unable to find $x,y,r$
The Generalized Binomial Theorem
$$(x+y)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}$$ where the ${r \choose k }$ denotes the falling factorial.
I notice that the factors in the numerator is increasing, hence I am unable to use it.
| As the denominator has one extra multiplicand
and the number of terms in denominator $=$ the exponent of $\dfrac14$
$$2S=\dfrac{2\cdot5}{3\cdot6}\left(\dfrac14\right)^2+\dfrac{2\cdot5\cdot8}{3\cdot6\cdot9}\left(\dfrac14\right)^3+\cdots$$
Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
$$2S+1=1+\dfrac{\dfrac23\cdot\dfrac53}{2!}\left(\dfrac14\right)^2+\cdots$$
$$=1+\dfrac{\left(-\dfrac23\right)\cdot\left(-\dfrac23-1\right)}
{2!}\left(-\dfrac14\right)^2+\cdots$$
$$=\left(1-\dfrac14\right)^{-2/3}$$ using Binomial series
We can use Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
to find $x,n$ to be $-\dfrac14-\dfrac23$ respectively in $$2S+1=(1+x)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25} \le A(a) < 2$$
Note that the bounds are rather tight. I encountered this issue in determining a narrow range of an angle in a geometry problem. I was only able to examine certain limits and identified the correct answer. But, I did not manage to fully prove the above inequalities.
Since I am not all that versed in dealing with such type of problems and would appreciate if anyone could suggest solutions for the proof.
| The left inequality.
We need to prove that
$$25(a^2+a+1)^2\geq9(2a^2+2a+1)(a^2+2a+2)$$ and since by AM-GM
$$(2a^2+2a+1)(a^2+2a+2)\leq\left(\frac{2a^2+2a+1+a^2+2a+2}{2}\right)^2,$$ it's enough to prove that
$$100(a^2+a+1)^2\geq9(3a^2+4a+3)^2$$ or
$$10(a^2+a+1)\geq3(3a^2+4a+3)$$ or
$$(a-1)^2\geq0$$ and we are done.
The right inequality.
We need to prove that
$$\left(a^2+a+\frac{1}{2}\right)(a^2+2a+2)>(a^2+a+1)^2,$$ which is true by C-S:
$$\left(a^2+a+\frac{1}{2}\right)(a^2+2a+2)\geq\left(a^2+\sqrt2a+1\right)^2>(a^2+a+1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to prove $-\sqrt a < x<\sqrt a$ from $x^2I want to prove the following:
\begin{align}
x^2<a \iff \tag 1 \\
-\sqrt a < x<\sqrt a \tag 2
\end{align}
Update:
Do I need some restriction on $a$ or is $a\in\mathbb R$?
Between $(1)$ and $(2)$, is it correct to use "equivalence" or is it just "implies"?
My attempt:
Case 1: $x>0$:
We have
\begin{align}
x^2&<a\iff \tag 3\\
\sqrt{ x^2} &<\pm \sqrt a \tag 4
\end{align}
But $\sqrt{ x^2}=\lvert x \rvert$? This looks weird?
Case 2: $x<0$.
If $x<0$ then $-x>0$ is positive, so
\begin{align}
(-x)^2=x^2&<a \iff \tag 5\\
x&<\pm \sqrt a \tag 6
\end{align}
| First of all, do you know that $0 < m < n$ means $\sqrt{m} < \sqrt{n}$? Can you take that as a given. If so, can you take for $m>0, n> 0$ then $m< n \iff \sqrt{m} < \sqrt{n}$ is a given?
In general: You have an axiom that if $a < b$ and $c > 0$ then $ac < bc$. (as well as if $a < b$ then $a+d < b+d$
From that you can prove many propositions including i) if $a > 0$ then $-a < 0$; ii) if $a<b$ and $c < 0$ then $ac > bc$ and that iii) $x^2 = 0$ if $x=0$ and $x^2 > 0$ if $x \ne 0$, iv) if $a > 0$ there are two $x_0, x_1$ so that $x_i^2 = a$. $x_0 > 0$ and $x_1 = -x_0 <0$; we call $x_0 := \sqrt{x_0}$ and $x_1=-\sqrt a$. v) $\sqrt{x^2} = |x|$. $|x| =x$ if $x\ge 0$ and $|x|=-x$ if $x \le 0$. etc. These are basic I won't go over them but I think it worth pointing out that
For $a > 0; b> 0$ then 1) $a < b \iff a^n < b^n\iff \sqrt[k]a < \sqrt[k]b$ for all $n,k \in \mathbb N$, is worth noting why it is true.
$a < b \implies a^2 = a*a < a*b < b*b < b^2$ and so by induction if $a^{n-1} < b^{n-1}$ then $a^n = a^{n-1}*a < b^{n-1}*a < b^{n-1}*b = b^n$. And if $a \ge b$ then the same argument shows $a^n \ge b^n$ and so $a<b \iff a^n < b^n$. And therefore $\sqrt[k]a < \sqrt[k]b \iff a=(\sqrt[k]a)^k< (\sqrt[k]b)^k = b \iff a^n < b^n$.
...
Okay, now we can begin:
To prove $x^2 < a \implies -\sqrt a < x <\sqrt a$.
$x^2 \ge 0$ so $a > 0$. So $\sqrt a$ exists and $\sqrt a > 0$.
$x^2 \ge a$ so $\sqrt{x^2}=|x| > \sqrt a \ge 0$
If $x \ge 0$ we have $0 \le x=|x| = \sqrt{x^2} < \sqrt a$.
... and so $-\sqrt a < 0 \le x < \sqrt a$.
If $x < 0$ we have $0 \le -x=|x| =\sqrt{x^2} < \sqrt a$ and therefore:
$-\sqrt a < -\sqrt{x^2} = -|x| = x < 0$.
.... and so $-\sqrt a < x < 0 < \sqrt a$.
In either case we have $-\sqrt a < x < \sqrt a$.
...
to prove $-\sqrt a < x < \sqrt a\implies x^2 < a$.
(Note: if $\sqrt a$ exists at all, that means $a \ge 0$.)
If $x \ge 0$ then $0 \le x < \sqrt a$ and so $x^2 < (\sqrt a)^2 = a$
If $x < 0$ then $-\sqrt a < x < 0$ means $0 < -x < \sqrt a$. And so $x^2 = (-x)^2 < (\sqrt a)^2 = a$.
Either way, $x^2 < a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Using similitudes to prove Pythagoras' Theorem Right triangle collage
(See above link for the image I refer to in this problem.)
I'm teaching myself about similitudes, and it's been bothering me that I've been struggling to come up with the similitude that represents the map from the bigger right triangle to the smaller blue right triangle. The original problem asked to use similitudes to prove Pythagoras' Theorem, but I want to make sure I first understand similitudes.
I'm trying this out with a specific example, by letting the big triangle be a 3-4-5 right triangle $(b=3, a=4, c=5)$ where I set the right angle of the triangle at the origin where the $x$- and $y$-axis intersect. I worked with the general similitude form:
$
s\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right)=
\left( {\begin{array}{cc}
r\cos(\theta) & -r\sin(\theta) \\
r\sin(\theta) & r\cos(\theta) \\
\end{array} } \right)\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right) + \left( {\begin{array}{c}
e \\
f \\
\end{array} } \right)
$
Since I want to map to the small blue triangle, I let $r = 3/5$. So, mapping from the big triangle to the small triangle, we go from the point $(0,0)$ to $(\frac{12}{5}\cos(53.13^\circ),\frac{12}{5}\sin(53.13^\circ))$ - I did this through some triangle similarity & right-triangle trig. The point $(0,3)$ stays at $(0,3)$, and $(4,0)$ goes to $(0,0)$.
I said $e=\frac{12}{5}\cos(53.13^\circ)$ and $f=\frac{12}{5}\sin(53.13^\circ)$. Intuitively I thought we would need to rotate the larger triangle $223.18^\circ$, but solving for $\theta$ leaves me with $53.18^\circ$. Either way, I cannot get the transformation from $(4,0)$ to $(0,0)$ to work out with my value of $\theta, e,$ and $f$.
Is there something glaringly wrong about how I'm thinking about similitudes? (Teaching myself)
| It seems like the approach you're taking is by composing basic affine transformations with each other.
$
s\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right)=
\color{red}{\left( {\begin{array}{cc}
r\cos \theta & -r\sin\theta \\
r\sin\theta & r\cos\theta \\
\end{array} } \right)}\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right) + \color{green}{\left( {\begin{array}{c}
e \\
f \\
\end{array} } \right)}
$
Multiplying by the red is a scaling and a rotation, and adding the green is a translation.
You overlooked one detail: You cannot transform the big right triangle to either of the smaller blue and red triangles with only scaling, rotation and translation. You need a reflection as well.
Since any reflection works, let's use a basic one where we flip over the $y$-axis first.
$
s\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right)= \color{blue}{\left( {\begin{array}{cc}
-1 & 0 \\
0 & 1 \\
\end{array} } \right)}
\color{red}{\left( {\begin{array}{cc}
r\cos \theta & -r\sin\theta \\
r\sin\theta & r\cos\theta \\
\end{array} } \right)}\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right) + \color{green}{\left( {\begin{array}{c}
e \\
f \\
\end{array} } \right)}
$
$
s\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right)=
\color{purple}{\left( {\begin{array}{cc}
-r\cos \theta & r\sin\theta \\
r\sin\theta & r\cos\theta \\
\end{array} } \right)}\left( {\begin{array}{c}
x \\
y \\
\end{array} } \right) + \color{green}{\left( {\begin{array}{c}
e \\
f \\
\end{array} } \right)}
$
This may also be why your intuition of what $\theta$ should be doesn't match the value you solved for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int \frac{dx}{1-\sin x+\cos x}$ Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$
I tried:
$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos x+\sin x}{2 \cos x(\cos x+1)}$$ $$=\int (\frac{1}{2 \cos x}+ \frac{\sin x}{2 \cos x(\cos x+1)})dx=\frac{1}{2} \ln(\sec x+\tan x)+ \int \frac{\sin x}{2\cos x(\cos x+1)}dx$$
$ \cos x=u$ , $du=-\sin xdx$
$$\int \frac{-du}{2u(u+1)}= \frac{-1}{2} \int (\frac{1}{u}-\frac{1}{u+1})= \frac{-1}{2}(\ln(\cos x)-\ln(\cos x+1))$$
final answer : $$\frac{1}{2} \ln(\sec x+\tan x)-\frac{1}{2} \ln(\frac{\cos x}{1+\cos x})+ c$$
First: Is my answer right?
Second: Is there another approach or easier approach to solve this integral?
| Your answer is correct
You may try the following as well.
Substitute $$\sin x =\frac {2\tan (x/2)}{1+\tan^2(x/2)}$$ and $$\cos x =\frac {1-\tan^2 (x/2)}{1+\tan^2(x/2)}$$
Then let $u=\tan (x/2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}$
Does anyone know how to solve this?
| $x^3-y^3=(x-y)(x^2+xy+y^2)=0$ $\implies$ $x^3=y^3$
$\left(\dfrac x{x+y}\right)^{2013}=\left(\dfrac{x^2}{x^2+xy}\right)^{2013}=\left(\dfrac{x^2}{-y^2}\right)^{2013}=-\left(\dfrac{x^3}{y^3}\right)^{1342}=-1$
Similarly, $\left(\dfrac y{x+y}\right)^{2013}=-1$.
The sum is $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
how to simplify $\frac{2+\sqrt{3}}{6+4\sqrt{3}}$ among solving a problem I need to simplify: $$\frac{2+\sqrt{3}}{6+4\sqrt{3}}$$
I know it is $\frac{1}{2\sqrt{3}}$ but I dont know how to show that in mathematic way for example I can write $$\frac{2+\sqrt{3}}{2\sqrt{3}(\sqrt{3}+2)}$$ but still not obvious.Is there better way to prove/show that?
| If in doubt, use conjugates:
$$ \frac{2+\sqrt 3}{6+4\sqrt 3}=\frac{2+\sqrt 3}{2\cdot (3+2\sqrt 3)}=\frac{(2+\sqrt 3)(3-2\sqrt 3)}{2\cdot (3+2\sqrt 3)(3-2\sqrt 3)}=\frac{-\sqrt 3}{-6}=\frac1{2\sqrt 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine differentiability for a piecewise function
Def.(Differentiable and Gradient)
Let $f:S\mapsto\mathbb{R}$ where $S=S^\circ\subseteq \mathbb{R},x\in S$
$$\exists m\in\mathbb{R},s.t.\lim_{h\to0}\frac{f(x+h)-f(x)-m\cdot h}{\Vert h\Vert}=0$$
$$\Leftrightarrow\nabla f(x)=m$$
Then $f$ is differentiable at $x$ with gradient $m$
Let $f:\mathbb{R}^2\mapsto\mathbb{R}$
$$f(x,y)=
\begin{cases}
\frac{y^3-x^8y}{x^6+y^2}& (x,y)\neq(0,0)\\\\
0& (x,y)=(0,0)
\end{cases} $$
Is $f$ differentiable at $(0,0)?$
Since for all $x,y\text{ in }\mathbb{R},f(x,0)=0\text{ and }f(0,y)=y$
Then if $\nabla f(0)\text{ exists}$ implies $\nabla f(0)=(0,1)$
That if the following limit exists and equals to $0$, $f$ must be differentiable at $(0,0)$
\begin{align}
& \lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-(0,1)\cdot(x,y)}{\sqrt{x^2+y^2}}\\
& =\lim_{(x,y)\to(0,0)}\frac{f(x,y)-y}{\sqrt{x^2+y^2}} \\
& =\lim_{(x,y)\to(0,0)}\frac{\frac{y^3-x^8y}{x^6+y^2}-y}{\sqrt{x^2+y^2}} \\
& =\lim_{(x,y)\to(0,0)}-\frac{\left(yx^8+yx^6\right)\sqrt{y^2+x^2}}{\left(y^2+x^6\right)\left(y^2+x^2\right)}\\
& =\lim _{\left(x,\:y\right)\to \left(0,\:0\right)}-\frac{y}{\sqrt{y^2+x^2}}\cdot \frac{x^8+x^6}{y^2+x^6}\\
\end{align}
$\text{Update:}$
$\text{Let $x=r\cos(\theta),y=r\sin(\theta)$ we have}$
\begin{align}
& =\lim _{r\to \:0}-\frac{r\sin \left(θ\right)}{\sqrt{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot \frac{\left(r\cos \left(θ\right)\right)^8+\left(r\cos \left(θ\right)\right)^6}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^6}\\
& =-\sin \left(θ\right)\cdot \lim _{r\to \:0}\frac{r}{\sqrt{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot \lim _{r\to \:0}\frac{\left(r\cos \left(θ\right)\right)^8+\left(r\cos \left(θ\right)\right)^6}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^6}\\
& \neq-\sin \left(θ\right)\cdot\sqrt{\lim _{r\to \:0}\frac{r^2}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot\lim _{r\to \:0}\frac{r^4\cos ^6\left(θ\right)\left(r^2\cos ^2\left(θ\right)+1\right)}{\sin ^2\left(θ\right)+r^4\cos ^6\left(θ\right)}\\
& =-\sin(\theta)\cdot \text{d.n.e}\cdot0=\text{d.n.e}
\end{align}
Therefore $f$ isn't differentiable at $(0,0).$$\tag*{$\square$}$
Is this calculation correct $?$
Any help would be appreciated.
| We have that limit doesn't exist
$$-\frac{\left(yx^8+yx^6\right)\sqrt{y^2+x^2}}{\left(y^2+x^6\right)\left(y^2+x^2\right)}
=-\frac{y}{\sqrt{y^2+x^2}}\cdot \frac{x^8+x^6}{y^2+x^6}
$$
indeed
$$-\frac{y}{\sqrt{y^2+x^2}}=-\sin \theta$$
and as $y=t$ and $x=t\to 0$
$$\frac{x^8+x^6}{y^2+x^6}=\frac{t^8+t^6}{t^2+t^6}\to 0$$
but as $y=t^3$ and $x=t\to 0$
$$\frac{x^8+x^6}{y^2+x^6}=\frac{t^8+t^6}{2t^6}\to \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0$. Show that
\begin{equation*}
\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0.
\end{equation*}
Let's show this formally using an $\epsilon-\delta$ proof. For $(x,y)\neq (0,0)$, let $\epsilon > 0$. Then if $(x,y)\in \mathbb{R}^2$ and $|(x,y)| < \frac{\epsilon}{3}$, then
\begin{equation*}
|y|^3\leq x^2+y^4 < \epsilon^3,
\end{equation*}
so $|y| < \sqrt[3]{\epsilon}$. Thus,
\begin{equation*}
\begin{split}
\left|\frac{x^4y^3}{x^2+y^4}-0\right| &= \left|\frac{x^4y^3}{x^2+y^4}\right| \\
&= \frac{x^4|y|^3}{x^2+y^4} \\
&\leq \frac{x^4y^3}{x^4} \\
&= |y|^3 \\
&< \epsilon
\end{split}
\end{equation*}
and we are done.
Where have I gone wrong? Thanks.
| Yes your way is correct, more simply as $|x|<1$
$$\left|\frac{x^4y^3}{x^2+y^4}\right|\le\left|\frac{x^4y^3}{x^4+y^4}\right|=r^3\left|f(\theta)\right|\to 0$$
since $f(\theta)$ is bounded.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series.
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$
My try is as follows:
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?
| A quite elementary way is just using the two binomial formulas $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$ as follows:
\begin{eqnarray*}\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}
& = & (\sqrt[3]{x^{3}+3x^{2}} - x) + (x -\sqrt{x^{2}-2x})\\
& = & \frac{3x^2}{\sqrt[3]{(x^{3}+3x^{2})^2} + x\sqrt[3]{x^{3}+3x^{2}} + x^2} + \frac{2x}{x+\sqrt{x^{2}-2x}}\\
& = & \frac{3}{\sqrt[3]{(1+\frac{3}{x})^2} + \sqrt[3]{1+\frac{3}{x}} + 1} + \frac{2}{1+\sqrt{1-\frac{2}{x}}} \\
& \stackrel{x\to \infty}{\longrightarrow} & 1+1 = 2
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Can the sum of irrational square roots of two different rational numbers be another irrational square root of a rational number? Prove or refute that $\exists x, y, z \in$ with $\mathbb {Q}$ and $x\neq y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{z}$ and $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}\setminus\mathbb{Q}.$
If $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}$, we find that $\sqrt{1}+\sqrt{4}=\sqrt{9}$ or even $\sqrt{0}+\sqrt{1}=\sqrt{1}$ holds. Thus we are interested in the case where the roots are irrational numbers but the radicands are rational numbers.
After squaring both members of the equality we get $x+2\sqrt{xy}+y=z\implies\sqrt{xy}=\frac{z-x-y}{2}$. This means that the product of $\sqrt{x}$ by $\sqrt{y}$ is a rational number. We can only conclude that both $\sqrt{x}$ and $\sqrt{y}$ are rational or irrational.
I can't find any good aproach to this problem and I don't even know if this result is relevant. Thank you for advance.
| Letting $x = 2$, $y = 8$, and $z=18$ works:
$$\sqrt 2 + \sqrt 8 = \sqrt 2 + 2 \sqrt 2 = 3 \sqrt 2 = \sqrt 18.$$
More generally, for any $a, b, c \in \mathbb Q$ we have
$$ \sqrt{ab^2} + \sqrt{ac^2} = b\sqrt a + c \sqrt a = \sqrt{a(b+c)^2}.$$
This is an example whenever $b \ne \pm c$ and $\sqrt a$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number Theory (Polynomials) Find The Remainder? A polynomial $f(x) = x^{50} + 3x^{49} + 3x + 12$ when divided by $x - a$, it leaves remainder $3$ & when its quotient is further divided by $x - b$ it leaves remainder $5$, also when $f(x)$ is divided by $x^2 - ( b + a)x + ab$ the remainder is $x + 6$. Find $a$?
| So by Remainder theorem, $f(a)=3$ and $f(x)=g(x)(x-a)(x-b)+(x+6)$ for some polynomial $g(x)$. Using the first result in the second, you immediately get …
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3415131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which numbers can be realized in the form $a^x - b^x$ for $a, b$ of opposite parity and $x$ even?
How can we check whether a number can be represented in the form of $a^x - b^x$, where $a$ and $b$ are integers of opposite parity and $x$ is positive and even integer.
I thougth of using $$a^n - b^n = (a - b) (a^{n-1} + a^{n-2}b + \cdots + b^{n-2}a +b^{n-1}) .$$
| Hint For $x = 2 m$, we have
$$a^x - b^x = (a^m)^2 - (b^m)^2 ,$$
that is, any difference of even powers is a difference of squares, so we may as well restrict our attention to $x = 2$.
Now, the factorization of a difference of squares is $$a^2 - b^2 = (a + b) (a - b) ,$$ which specializes the factorization formula in the question statement.
Additional hint Since $a$ and $b$ have opposite parity, $a - b$ and $a + b$ are both odd, hence so is $a^2 - b^2$. Are all odd numbers expressible this way?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction.
Please, explain how to remove the radical from the denominator. Thanks, for your help.
| Multiply top and bottom by the denominator to get $$\frac{8}{ \sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}} } }=\frac{8 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} } {2-\sqrt{\frac{5+\sqrt 5}{2}} }.$$ Then multiply top and bottom by $$2+\sqrt{\frac{5+\sqrt 5}{2}},$$ the conjugate of the denominator, to get $$\frac{8 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)}{4-\frac{5+\sqrt 5}{2}}.$$ This simplifies to give $$\frac{16 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)}{3-\sqrt 5}.$$ Finally multiply by $3+\sqrt 5$ again to get $$\frac{16 {\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)(3+\sqrt 5)}{9-5},$$ which simplifies to give $$4{\sqrt{2-\sqrt{\frac{5+\sqrt 5}{2}}}} \left(2+\sqrt{\frac{5+\sqrt 5}{2}}\right)(3+\sqrt 5).$$
By the way what you did was not rationalise the expression (you can't, since it's not rational). You've only rationalised the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{n\to\infty} a_{n}$ when $a_{n+1}=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}$ If $a_{n+1}=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}$ for all positive integers $n$ and $a_{1}>0$, find the limit of $$\lim_{n\to\infty} a_{n}$$
| I show that,
for $n \ge 3$,
$\dfrac{6a_3}{n(n-1)}
\lt a_n
\lt \dfrac1{n-1}
$.
I don't know which bound is
close to the true value.
Since
$a_{n+1}
=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}
$,
$a_{n+1}
\lt\dfrac{a_{n}}{1+a_{n}}
=\dfrac{1}{1+1/a_{n}}
$,
so
$\dfrac1{a_{n+1}}
\gt 1+\dfrac1{a_n}
$.
Let
$b_n = \dfrac1{a_n}
$.
Then
$b_{n+1} > 1+b_n
$
so,
since $b_1 > 0$,
$b_n > n-1
$
so
$a_n \lt \dfrac1{n-1}
\to 0$.
To get an upper bound,
$\begin{array}\\
a_{n+1}
&=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}\\
&>\dfrac{a_{n}}{1+\frac{1}{n}+\frac1{n-1}+\frac1{(n-1)^3}}\\
&>\dfrac{a_{n}}{1+\frac{2}{n-1}}
\qquad\text{for } n \ge 3\\
\text{so}\\
\dfrac{a_{n+1}}{a_n}
&>\dfrac{1}{1+\frac{2}{n-1}}\\
&=\dfrac{n-1}{n+1}\\
\text{so}\\
\dfrac{a_{m}}{a_3}
&=\prod_{n=3}^{m-1}\dfrac{a_{n+1}}{a_n}\\
&>\prod_{n=3}^{m-1}\dfrac{n-1}{n+1}\\
&=\dfrac{\prod_{n=3}^{m-1}(n-1)}{\prod_{n=3}^{m-1}(n+1)}\\
&=\dfrac{\prod_{n=2}^{m-2}n}{\prod_{n=4}^{m}n}\\
&=\dfrac{2\cdot 3}{m(m-1)}\\
&=\dfrac{6}{m(m-1)}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $13^{({11}^{7})}\equiv x\pmod{10}$
Solve $x$, where $x\in[0,9]\cap\mathbb{Z}:$
$$13^{11^7}\equiv x\pmod{10}$$
Here is my thoughts:
Since $13\equiv3\pmod{10}$
Implies $13^{11^7}\equiv 3^{11^7}\pmod{10}$
$$3^0=1,3^1=3,3^2=9,3^3=27,3^4=81\cdots$$
For mod $10$, have reminder repeat between:
(they have to repeat, but i will skip the induction here) $$R_1=\{1,3,9,7\}$$
Then solve
$$11^7+1\equiv x\pmod{|R_1|}$$
$$11^0+1=2,11^1+1=12,11^2+1=122,11^3+1=1332$$
For mod $4$, have reminder index repeat between:
$$R_2=\{2,0\}$$
In another word, the reminder repeat between second and last term in $R_1$
Then solve $$7+1\equiv x\pmod{|R_2|},\text{ that }x=0$$
$x=0$, means the reminder is the last term in $R_1$ which is $7$, that implies
$$13^{11^7}\equiv7\pmod{10}$$
Therefore $x=7$
My qeustion: is there any theorem to apply so this calculation could be simpler$?$
Any help would be appreciated.
| $3^4=81\equiv1\mod 10$
$11^7\equiv(-1)^7=-1\equiv3\mod 4$
Therefore, $13^{11^7}\equiv3^{11^7}\equiv3^3=27\equiv7\mod 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$a^2b^2+a^2c^2+b^2c^2\leq 3$ Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$
Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.
My idea:
I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.
Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$
I have to show that $$y^2-2xz\leq 3$$
I tried to prove it with the sign of trinom, but it doesn't work.
| Another idea is to notice that:
$$2(a^2-a+1)^2=a^4+1+(a-1)^4\geq a^4+1$$
So, using this and Cauchy-Schwarz:
$$16 = 16(a^2-a+1)^2(b^2-b+1)^2(c^2-c+1)^2 \geq 2(a^4+1)(b^4+1)(c^4+1)=$$
$$=(a^4+a^4+1+1)(b^4+c^4+b^4c^4+1)\geq (a^2b^2+a^2c^2+b^2c^2+1)^2$$
It follows that:
$$a^2b^2+b^2c^2+c^2a^2 \leq 3$$
with equality when $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A peculiar Riemann Sum Im trying to express the following limit as a Riemann Sum
$$ \lim_{ n \to \infty} \sum_{i=1}^{n} \dfrac{ n^3 (2i-3) }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } $$
over the interval $[2,3]$
Clearly, we want $\dfrac{1}{n}$ out of the sum as to take $\Delta x$ in account. We can divide both sides by $\dfrac{1}{n^5}$ and we obtain the
$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{ \dfrac{(2i-3)}{n} }{ (n^2- \sqrt{2n^4 + i^2n^2} + in )(2n^2-i^2)^{3/2} } \cdot \frac{1}{\dfrac{1}{n^2 \cdot n^3}}$$
Which reduces to
$$ \lim_{ n \to \infty} \dfrac{1}{n}\sum_{i=1}^{n} \dfrac{ \dfrac{(2i-3)}{n} }{ \left( 1+ \dfrac{i}{n} - \sqrt{ 2 + \dfrac{i^2}{n^2} } \right)\left(2- \left( \dfrac{i}{n} \right)^2 \right) } $$
Now, this is almost done as we want our height x-value to be $x = 2 + \dfrac{i}{n}$. However, here is where I get stuck and I ask for some help. The term $\dfrac{2i-3}{n}$ in the numerator is the only term that is problematic here. How would we handle this?
| According to Riemann sum
$$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} \frac{b-a} {n} \sum_{i=1}^{n}f\left(a+i\cdot\frac {b-a} {n} \right)$$
we need to express the sum as a function of $\left(2+\frac{i}{n}\right)$ that is
$$\lim_{ n \to \infty} \frac{1}{n}\sum_{i=1}^{n} \frac{ 2\left(2+\frac{i}{n}\right)-4+\frac 3 n }{ \left( \left(2+ \frac{i}{n}\right)-1 - \sqrt{ \left(2 + \frac{i}{n}\right)^2-4\left(2+\frac in\right)+6 } \right)\left(-2+4\left(2+\frac i n\right)- \left(2+ \frac{i}{n} \right)^2 \right) }=$$
$$=\int_2^3\frac{2x-4}{(x-1-\sqrt{x^2-4x+6})(-2+4x-x^2)}dx$$
which is an improper integral since $x-1-\sqrt{x^2-4x+6}=0$ for $x=\frac 5 2$.
Refer also to
*
*Perfect understanding of Riemann Sums
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$$x\in[-1,0]$$
Finding the range
$$f(x)=\sin^{-1}\left(\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left([-1,0]+\frac{1}{2}\right)^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{-1}{2},\frac{1}{2}\right]^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[0,\frac{1}{4}\right]+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{3}{4},1\right]}\right)$$
$$f(x)=\sin^{-1}\left({\left[\frac{\sqrt{3}}{2},1\right]}\right)$$
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
but actual answer is $\left[\dfrac{\pi}{3},\dfrac{\pi}{2}\right]$
| The range of the function $\sin^{-1}$ is only $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Hence with $f(x) = \sin^{-1} \left( \left[ \frac{\sqrt{3}}{2}, 1\right] \right)$, the $\sin^{-1}$ functions maps these points to the range $\left[ \frac{\pi}{3}, \frac{\pi}{2}\right]$.
| {
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"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Show that $x_k = (\sin( k\pi h), \sin(2 k\pi h), \sin(3 k\pi h), ... )$ is an eigenvector of a tridiagonal A. I'm looking at problem 3 from Gilbert Strang's Linear Algebra and Its Applications, 4e, section 7.4:
Multiplying $Ax_k$ as written gives me:
$$ Ax_k = \begin{bmatrix}2\sin(k\pi h) &- \sin(2 k\pi h) \\ - \sin( k\pi h) &+ 2\sin(2k\pi h) &- \sin(3 k\pi h) \\ & - \sin(2 k\pi h) & + 2\sin(3k\pi h) & - \sin(4 k\pi h) \\ &&...\end{bmatrix} $$
(Note that this is a column vector, but I spaced it out to indicate the pattern.)
I'm struggling to see this as a multiple of $(\sin( k\pi h), \sin(2 k\pi h), \sin(3 k\pi h), ... ) $. I thought that maybe I am supposed to assume that $k$ and $h$ are integers so that I can use the periodicity of sine to simplify $Ax_k$. But that just makes $x_k$ = 0.
So I looked at the solution, which says that the corresponding eigenvalue is $2 - 2\cos(k\pi h)$, and tried to work backwards. But computing $2 - 2\cos(k\pi h) x_k$ also doesn't seem to get me anywhere close to $Ax_k$. Even the first entry is clearly different:
$$\begin{align}[2 - 2\cos(k\pi h) x_k]_1 & = 2\sin(k\pi h) - 2\cos(k\pi h)\sin(2k \pi h)\\ &\ne 2\sin(k\pi h) - \sin(2 k\pi h) \end{align}$$
(unless $k$ and $h$ are both integers, as above)
I would appreciate a hint.
| We can write
$$Ax_k = 2 \begin{bmatrix} \sin(k\pi h) \\ \sin(2k\pi h) \\ \vdots \\ \sin(nk \pi h) \end{bmatrix} -\begin{bmatrix} \sin(2k \pi h) \\ \sin(k\pi h) + \sin(3k \pi h) \\ \vdots \\ \sin((n-1) k \pi h) \end{bmatrix}$$
A helpful identity is that
$$\sin(x) + \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$$
From this we get that
$$\sin((\ell-1) k\pi h) + \sin((\ell+1)k\pi h) = 2 \sin(\ell k \pi h)\cos( k\pi h)$$
Using this, we have
$$Ax_k = 2 \begin{bmatrix} \sin(k\pi h) \\ \sin(2k\pi h) \\ \vdots \\ \sin(nk \pi h) \end{bmatrix} - \begin{bmatrix} \sin(2k \pi h) \\ 2\sin(2k\pi h)\cos(k\pi h) \\ \vdots \\ 2\sin((\ell+1) k\pi h) \cos(k\pi h) \\ \vdots \\ \sin((n-1)k \pi h) \end{bmatrix}$$
Note that
$$\sin(2k\pi h) = 2 \sin(k\pi h) \cos(k \pi h)$$
also, $\sin((n+1) k \pi h) = 0$ so
$$\sin((n-1) k \pi h) = \sin((n-1) k \pi h) + \sin((n+1) k \pi h) = 2\sin(n k \pi h) \cos(k \pi h)$$
so
$$Ax_k = 2 \begin{bmatrix} \sin(k\pi h) \\ \sin(2k\pi h) \\ \vdots \\ \sin(nk \pi h) \end{bmatrix} - \begin{bmatrix} 2\sin(k\pi h)\cos(k \pi h) \\ 2\sin(2k\pi h)\cos(k \pi h) \\ \vdots \\ 2\sin(nk \pi h)\cos(k \pi h) \end{bmatrix} = (2 - 2\cos(k \pi h))\begin{bmatrix} \sin(k\pi h) \\ \sin(2k\pi h) \\ \vdots \\ \sin(nk \pi h) \end{bmatrix}$$
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Stuck on this generating functions and partial fraction expansion step I'm having a hard time figuring out the thought process after the partial fractions are determined. The picture is below:
So they plug in x=0 to find 2x. How do they determine the other expansions?
| The idea is to add the geometric series that come up after partial fractions.
Let $A(x)=a_0+a_1x^1+a_2x^2+\cdots+a_nx^n=\frac{x}{(1-x)(1-2x)}$.
As a formal expression, we have the following expansion for the geometric series $$1+rx+r^2x^2+r^3x^3+\cdots=\frac{1}{1-rx},\quad r\in \mathbb{C}$$ Then
\begin{align}
A(x)=x\left(\frac{2}{1-2x}-\frac{1}{1-x}\right)&=2x(1+2x+2^2x^2+2^3x^3+\cdots)-x(1+x+x^2+x^3+\cdots)\\
&=(2x+2^2x^2+2^3x^3+2^4x^4+\cdots)-(x+x^2+x^3+x^4+\cdots)\\
&=(2-1)x+(2^2-1)x^2+(2^3-1)x^3+(2^4-1)x^4+\cdots+(2^n-1)x^n+\cdots\\
&=0+(2-1)x+(2^2-1)x^2+(2^3-1)x^3+(2^4-1)x^4+\cdots+(2^n-1)x^n+\cdots
\end{align}
Now, we have the equality
$$a_0+a_1x^1+a_2x^2+\cdots+a_nx^n+\cdots=(2^0-1)+(2-1)x+(2^2-1)x^2+\cdots+(2^n-1)x^n+\cdots
$$
Equating the coefficients of $x^n$, we have that $a_n=2^n-1$ for $n\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $a_{n+2} = \sqrt{7-\sqrt{7+a_n}}$, with $a_1=\sqrt{7}$, $a_2 = \sqrt{7-\sqrt{7}}$? How to prove the convergence of the real sequence $\{a_n\}$, which is defined by $a_{n+2} = \sqrt{7-\sqrt{7+a_n}}$, with $a_1=\sqrt{7}$, $a_2 = \sqrt{7-\sqrt{7}}$? Furthermore, how to verifty that 2 is the limit?
| Easy to see that $0 \le a_n \le \sqrt{7}$. Then $$|a_{n+2}-a_{m+2}|=\frac{|\sqrt{7+a_n}-\sqrt{7+a_m}|}{\sqrt{7-\sqrt{7+a_n}}+\sqrt{7-\sqrt{7+a_m}}} \le \frac{|\sqrt{7+a_n}-\sqrt{7+a_m}|}{\sqrt{7-\sqrt{7+\sqrt{7}}}+\sqrt{7-\sqrt{7+\sqrt{7}}}} < $$ $$ <|\sqrt{7+a_n}-\sqrt{7+a_m}|=\frac{|a_n-a_m|}{ \sqrt{7+a_n}+\sqrt{7+a_m}} \le \frac{|a_n-a_m|}{ \sqrt{7}+\sqrt{7}} <\frac{|a_n-a_m|}{5}$$ Since $|a_{n+2}-a_{m+2}|<\frac{|a_n-a_m|}{5}$ then $|a_{n+2N}-a_{m+2N}|<\frac{|a_n-a_m|}{5^N} \le \frac{\sqrt{7}}{5^N}$. Since $|a_{n+2N}-a_{m+2N}|< \frac{\sqrt{7}}{5^N}$ then sequence $a_n$ is Cauchy sequence. Then by Cauchy Convergence Criterion (http://mathonline.wikidot.com/the-cauchy-convergence-criterion) the sequence $a_n$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find Ratio of Integrals $I:J$ Given:
$$I=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{12}$$
and
$$J=\int_{0}^{1}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{(x+3)^8}$$
Find Value of $\frac{I}{J}$
My attempt:The Integral $I$ is easy to evaluate using Beta Function.
So i was trying to Manipulate $J$ to convert it to $I$ as follows:
We can write $J$ as:
$$J=\int_{0}^{1}\frac{x^{6}\left ( \frac{1}{x} -1\right )^\frac{7}{2}}{x^{8}\left ( 1+\frac{3}{x} \right )^8}$$
Now put $\frac{1}{x}=t$ we get:
$$J=\int_{1}^{\infty}\frac{(t-1)^{\frac{7}{2}}}{(1+3t)^8}$$
Using integration by Parts taking $u=(t-1)^{3.5}$ and $v=\frac{1}{(1+3t)^8}$ we get:
$$J=\frac{1}{6}\times \int_{1}^{\infty}\frac{(t-1)^{\frac{5}{2}}}{(1+3t)^7}$$
Repeating Parts again and again:
$$J=\frac{1}{432}\times \int_{1}^{\infty}\frac{\sqrt{t-1}}{(1+3t)^5}\:dt$$
Any way to proceed from here?
| As you noted, $$I=\frac 1 {12}B\left(\frac 7 2, \frac 9 2\right)$$
Also, if $|z|<1$ and $\alpha \in \mathbb R$
$$\frac 1 {(1+z)^\alpha} = \sum_{n\geq 0}\frac{(-\alpha)(-\alpha-1)...(-\alpha-n+1)}{n!}z^n\tag{1}$$
So for $\alpha =8$
$$\begin{split}
J &= \frac 1 {3^8}\int_0^1\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}\:dx}{\left(1+\frac x 3\right)^8}\\
&=\frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}\int_0^1x^{\frac{5+n}{2}}(1-x)^{\frac{7}{2}}dx\\
&= \frac 1 {3^8}\sum_{n\geq 0}\frac{(-8)(-9)...(-8-n+1)}{n!}\frac 1 {3^n}B \left(\frac 7 2 +n,\frac 9 2\right)
\end{split}$$
Now, because $$B(a+1, b) = B(a,b)\frac a {a+b}$$
we also have
$$B(a+n, b)=B(a,b)\cdot\frac{a}{a+b}\cdot\frac{a+1}{a+b+1}...\frac{a+n-1}{a+b+n-1}$$
Also note that $\frac 7 2 +\frac 9 2 = 8$. Therefore
$$B\left(\frac 7 2+n, \frac 9 2 \right)=B\left(\frac 7 2,\frac 9 2\right)\cdot\frac{\frac 7 2}{8}\cdot\frac{\frac 7 2+1}{9}...\frac{\frac 7 2+n-1}{8+n-1}$$
We now have
$$\begin{split}
J &=B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^8}\sum_{n\geq 0}\frac{(-\frac 7 2)(-\frac 7 2 - 1)...(-\frac 7 2-n+1)}{n!}\frac {1} {3^n}\\
&= B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^8}\frac 1 {(1+\frac 1 3)^{\frac 7 2}} \,\,\,\,\text{ (using (1))}\\
&=B\left(\frac 7 2,\frac 9 2\right)\frac 1 {3^{\frac 9 2}4^{\frac 7 2}}
\end{split}$$
Conclusion:
$$\boxed{\frac I J = \frac {3^{\frac 9 2}4^{\frac 7 2}}{12}=3^{\frac 7 2}4^{\frac 5 2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum _{cyc}\sqrt{1-\frac{\left(a+b\right)^2}{4}}\ge \sqrt{6}$ Let $a,b,c\ge 0$ such that $a^2+b^2+c^2=1$. Prove that $$\sqrt{1-\frac{\left(a+b\right)^2}{4}}+\sqrt{1-\frac{\left(b+c\right)^2}{4}}+\sqrt{1-\frac{\left(a+c\right)^2}{4}}\ge \sqrt{6}$$
My try:
$$\sqrt{1-\frac{\left(a+b\right)^2}{4}}=\sqrt{a^2+b^2+c^2-\frac{\left(a+b\right)^2}{4}}=\frac{\sqrt{3a^2-2ab+3b^2+4c^2}}{2}\ge\frac{\sqrt{2a^2+2b^2+4c^2}}{2}$$$
Then i need to prove $$\sum \sqrt{2a^2+2b^2+4c^2}\ge 2\sqrt{6}$$
But that's failed. Without using $a^2+b^2\ge 2ab$ to remove the radical i have no more idea, i tried to use holder but failed. Help me.
| Also, Minkowski (triangle inequality) helps.
Indeed, let $a\geq b\geq c$.
Thus, $$\sum_{cyc}\sqrt{1-\frac{(a+b)^2}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{3a^2-2ab+3b^2+4c^2}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{(a-b)^2+2+2c^2}\geq\frac{1}{2}\sqrt{(a-b+a-c+b-c)^2+2\cdot3^2+2(a+b+c)^2}=$$
$$=\sqrt{(a-c)^2+5+ab+ac+bc}=\sqrt{6+(a-b)(b-c)}\geq\sqrt6.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$ we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
| We have that
$$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n=2^n\cdot \sqrt{\left(1+\frac1n\right)^n} \to \infty\cdot\sqrt e$$
and by your first method
$$\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{2}}}\ge 4^{\frac n2} \to \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why do these eigenvalues give the zero vector for 3 eigenvectors? Given the quadratic form $q(x_{1}, x_{2}, x_{3}, x_{4}) = x_{1}^2 + x_{2}^2 - x_{3}^2 - x_{4}^2 + 4x_{1}x_{2} - 2x_{1}x_{4} + 2x_{3}x_{4}$, the eigenvalues I am getting are $\lambda = -1, -2.27, .14, 3.13$. However, all of these eigenvalues except $\lambda = -1$ are resulting in the zero-vector for the their respective eigenvectors, which is not possible by definition. What is wrong here?
Edit: The characteristic equation I am getting is $\lambda^4 - 8\lambda^2-6\lambda + 1 = 0$.
| As far as the eigenvalues, the other answers are fine.
A milder thing to do is find out how to make your symmetric matrix "congruent" to a diagonal matrix, then invoke Sylvester's Law of Inertia. Below, $PQ=QP=I.$
Two eigenvalues are positive and two negative.
$$ P^T H P = D $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
- 2 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
- \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & 0 & - 1 \\
2 & 1 & 0 & 0 \\
0 & 0 & - 1 & 1 \\
- 1 & 0 & 1 & - 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & - 2 & 0 & - \frac{ 1 }{ 3 } \\
0 & 1 & 0 & \frac{ 2 }{ 3 } \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & - 3 & 0 & 0 \\
0 & 0 & - 1 & 0 \\
0 & 0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
- 1 & - \frac{ 2 }{ 3 } & - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & - 3 & 0 & 0 \\
0 & 0 & - 1 & 0 \\
0 & 0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 & 0 & - 1 \\
0 & 1 & 0 & - \frac{ 2 }{ 3 } \\
0 & 0 & 1 & - 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
1 & 2 & 0 & - 1 \\
2 & 1 & 0 & 0 \\
0 & 0 & - 1 & 1 \\
- 1 & 0 & 1 & - 1 \\
\end{array}
\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$
As $3>0$ and $\dfrac{1}{y}>0$
$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$
$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$
$$x=\dfrac{3y-1}{y+3}$$
$y+3\in[4,\infty)$ as $y\ge1$, $3y-1\in [2,\infty)$ as $y\ge1$
For $x$ to be positive integer, $3y-1$ should be multiple of $y+3$
$$3y-1=m(y+3) \text { where } m\in Z^{+}$$
$$3y-my=3m-1$$
$$(3-m)y=3m-1$$
Here R.H.S is positive, so L.H.S should also be positive.
So $3-m>0$, hence $m<3$
So possible values of $m$ are {$1$,$2$}.
For $m=1$, $$3y-1=y+3$$
$$2y=4$$
$$y=2$$
$$x=\dfrac{3\cdot2-1}{2+3}$$
$$x=1$$
For $m=2$, $$3y-1=2(y+3)$$
$$3y-1=2y+6$$
$$y=7$$
$$x=\dfrac{3\cdot7-1}{7+3}$$
$$x=\dfrac{20}{10}$$
$$x=2$$
Is there some other nicer way to solve this problem.
| Your method seems decent. Here I propose another way to ensure the integer-ness of $x$ and $y$.
Note from a rough sketch and limits that the range of $f:\mathbb{R^+} \mapsto X,~f(y)= {3y-1\over y+3}$ is $\left[-\frac 13, 3\right)$. The only positive integers that lie in this range are ${1,2}$. You may use the inverse process $y={3x+1\over 3-x}$ and,use $x\in \{1,2\}$ to see for which of these $y$, $x$ is also a positive integer.
For $x=1$, $y=\frac 42 = 2$, which is a positive integer. So one solution is $(x,y)\equiv (1,2)$. For $x=2$, $y=\frac 71 = 7$, also a positive integer. So another solution is $(2,7)$. We have exhausted all possibilities for $y$. Our solution set is $\left\{(1,2),(2,7)\right\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving limits with indeterminate form I received the following problem:$$\lim_{x\to\infty}\sqrt[4]{x^4+x^3}-x$$
I cannot figure out how to tackle it. Simple substitution seems to get $\infty-\infty$. This is where I got when trying to play with it: $$=\lim_{x\to\infty}\sqrt[4]{x^4+x^3}-\frac{1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\frac{1}{x}\sqrt[4]{x^4+x^3}-1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\frac{1}{x}x\sqrt[4]{1+\frac{1}{x}}-1}{\frac{1}{x}}$$$$=\lim_{x\to\infty}\frac{\sqrt[4]{1+\frac{1}{x}}-1}{\frac{1}{x}}$$
I did see the answer - its $\frac14$. But, how do you proceed from here?
| First multiply with $\sqrt[4]{x^4+x^3}+x$ to get a difference of squares and then repeat it:
\begin{align}
\sqrt[4]{x^4+x^3}-x&= \frac{\left(\sqrt[4]{x^4+x^3}-x\right)\left(\sqrt[4]{x^4+x^3}+x\right)}{\sqrt[4]{x^4+x^3}+x}\\
&= \frac{\sqrt{x^4+x^3}-x^2}{\sqrt[4]{x^4+x^3}+x}\\
&= \frac{\left(\sqrt{x^4+x^3}-x^2\right)\left(\sqrt{x^4+x^3}+x^2\right)}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\
&= \frac{x^4+x^3-x^4}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\
&= \frac{x^3}{\left(\sqrt[4]{x^4+x^3}+x\right)\left(\sqrt{x^4+x^3}+x^2\right)}\\
&= \frac{1}{\left(\sqrt[4]{1+\frac1x}+1\right)\left(\sqrt{1+\frac1x}+1\right)}\\
&\xrightarrow{x\to\infty} \frac14
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$ without L'Hopital's rule I'm currently struggling with this task:
$$\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$$
By now I have only come up with the idea to use the formula of sum of cubes, so, the numerator would become $\cos{4x}-\cos{5x}$ and the denominator would be $3(1-\cos{3x})$:
$$\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}} = \frac{(\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})}{(1-\cos{3x})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})} = \frac{\cos{4x}-\cos{5x}}{3(1-\cos{3x})}$$
In this expression it looks like $\cos{4x}-\cos{5x}$ and $(1-\cos{3x})$ increase with the same speed (if I can say it this way) and the limit is likely to be $\frac{1}{3}$.
Still I have no idea how to prove that without using L'Hopital's rule which is prohibited by the task.
| Adding and subtracting $1$ in the numerator we can rewrite the expression under limit as $$\frac{1-\sqrt [3]{\cos 5x}}{1-\cos 3x}-\frac{1-\sqrt[3]{\cos 4x}} {1-\cos 3x}$$ and the first fraction above can be further rewritten as $$\frac{1-\sqrt[3]{\cos 5x}}{1-\cos 5x}\cdot\frac{1-\cos 5x}{(5x)^2}\cdot\frac{5^2}{3^2}\cdot\frac{(3x)^2}{1-\cos 3x}$$ so that it tends to $(1/3)(1/2)(25/9)(2)=25/27$. Similarly second fraction tends to $16/27$ and hence the desired limit is $9/27=1/3$.
The following standard limits are used in the above approach $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1},\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$. Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$.
| I have got one way of demonstrating it:
Where $z$ is an odd number
Where $n=(z-1)/2$
1). $(10^z)^3=(3•10^{z+n})^2+(10^{z+n})^2$
This means that $10^{3z}= 9•10^{3z-1}+10^{3z-1}$ for $3z-1=2z+2n$ since $n=(z-1)/2$
Therefore $1•10^{3z}=(9/10)10^{3z}+(1/10)10^{3z}$ which is also $10^{3z}(9/10+1/10)$ which is also $1•10^{3z}$
Therefore there are an infinite number of values for z
now if one superimposes what is in the brackets of 1). with $a$, $b$ and $c$ you will get $a^3=b^2+c^2$
Therefore $a$ has an infinity of values
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I write $2+5i$ in polar form? Original problem:
Write $2+5i$ in polar form.
My attempt:
$|2+5i|=\sqrt{4+25}=\sqrt{29}$
$\arg(2+5i)=\arctan{\frac{5}{2}}=\theta$
$2+5i=\sqrt{29}(\cos(\theta)+i\sin(\theta))$
But how do I calculate $\arctan({\frac{5}{2}})$ in radians?
I know $\tan(\theta)=\frac{5}{2}$ and this implies that if we have a rectangle triangle $ABC$ then $\overline{AB}=5$ and $\overline{BC}=2$.
Here I'm stuck here.
| USE: $z = re^{i\theta}$
$2+5i$ = ?
Using modulus which is $r = \sqrt{x^2+y^2}$
$r = \sqrt{(2)^2+(5)^2}$ = $\sqrt{29}$
$\theta = \arctan({\frac{5}{2}})$
$2+5i = \sqrt{29}e^{i\arctan({\frac{5}{2}})}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation or approximation of a modified elliptic integral of the first kind In my research, I have come across an integral of the following form:
\begin{equation}
\int_0^{\pi/2} \frac{\cos\left(a \arcsin\left[kx\right]\right)dx}{\sqrt{1-k^2\sin ^2x}}.
\end{equation}
I would like to evaluate this integral or obtain an approximation for its value. Mathematica is not able to evaluate the integral and searching a table of integrals was not helpful either. The integral is similar to the complete elliptic integral of the first kind, but numeric evaluation shows its value can differ dramatically from that of the complete elliptic integral of the first kind.
Thank you!
| Given the definite integral:
$$
I(a,\,b) := \int_0^{\frac{\pi}{2}} \frac{\cos\left(a\,\arcsin\left(b\,x\right)\right)}{\sqrt{1 - \left(b\,\sin x\right)^2}}\,\text{d}x
$$
through a substitution of the type $t = \frac{4}{\pi}\,x - 1$ we go back to this other integral:
$$
I(a,\,b) = \int_{-1}^1 \underbrace{\frac{\cos\left(a\,\arcsin\left(b\,\frac{\pi}{4}\,(t + 1)\right)\right)}{\sqrt{1 - \left(b\,\sin \left(\frac{\pi}{4}\,(t + 1)\right)\right)^2}}\,\frac{\pi}{4}}_{:= f(a,\,b,\,t)}\text{d}t
$$
to which it's possible to apply the Legendre-Gauss quadrature.
In particular, opting for the two-point formula:
$$
I(a,\,b) \approx k_1\,f(a,\,b,\,t_1) + k_2\,f(a,\,b,\,t_2)
$$
where is it:
$$
\begin{cases}
k_1\,t_1^0 + k_2\,t_2^0 = \frac{1 + (-1)^0}{1 + 0} \\
k_1\,t_1^1 + k_2\,t_2^1 = \frac{1 + (-1)^1}{1 + 1} \\
k_1\,t_1^2 + k_2\,t_2^2 = \frac{1 + (-1)^2}{1 + 2} \\
k_1\,t_1^3 + k_2\,t_2^3 = \frac{1 + (-1)^3}{1 + 3}
\end{cases}
\; \; \; \Leftrightarrow \; \; \;
\begin{cases}
k_{1,2} = 1 \\
t_{1,2} = \pm \frac{1}{\sqrt{3}}
\end{cases}
$$
it follows that:
$$
I(a,\,b) \approx f\left(a,\,b,\,-\frac{1}{\sqrt{3}}\right) + f\left(a,\,b,\,\frac{1}{\sqrt{3}}\right),
$$
approximation that involves at most an error equal to:
$$
\epsilon(a,\,b) = \frac{1}{135}\underset{-1 \le t \le 1}{\max} \left|\frac{\partial^4 f(a,\,b,\,t)}{\partial t^4}\right|.
$$
As an example:
$$
I\left(3,\,\frac{1}{10}\right) \approx f\left(3,\,\frac{1}{10},\,-\frac{1}{\sqrt{3}}\right) + f\left(3,\,\frac{1}{10},\,\frac{1}{\sqrt{3}}\right) = 1.51672
$$
with a maximum error equal to:
$$
\epsilon\left(3,\,\frac{1}{10}\right) = \frac{1}{135}\underset{-1 \le t \le 1}{\max} \left|\frac{\partial^4 f\left(3,\,\frac{1}{10},\,t\right)}{\partial t^4}\right| = 1.08341\cdot 10^{-4}\,.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}$
Solve that $$\lim_{x\to0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}$$
$$
\lim_{x\to 0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}=\lim_{x\to 0}\bigg[1+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}\\
[1^t+1^t+....+1^t]^{1/t}=n^{1/t}\leq[1^t+2^t+....+n^t]^{1/t}\leq [n.n^t]^{1/t}=n.n^{1/t}
$$
Can I use squeeze theorem here or is there a better way ?
Note: My reference gives the solution $n$
| Let $f(x)$ be the function taken with the limit, then
\begin{align*}
\log f(x)&=(\sin^{2}x)\log(1^{1/\sin^{2}x}+\cdots+n^{1/\sin^{2}x})\\
&=(\sin^{2}x)\log n^{1/\sin^{2}x}\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{n}{n}\right)^{1/\sin^{2}x}\right)\\
&=(\sin^{2}x)\log n^{1/\sin^{2}x}+(\sin^{2}x)\log\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{n}{n}\right)^{1/\sin^{2}x}\right)\\
&\leq\log n+(\sin^{2}x)\log(1^{1/\sin^{2}x}+\cdots+1^{1/\sin^{2}x})\\
&=\log n+(\sin^{2}x)\log n.
\end{align*}
On the other hand,
\begin{align*}
\log f(x)&\geq\log n+(\sin^{2}x)\log\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}\right)\\
&\geq\log n+(\sin^{2}x)\log\dfrac{1}{n^{(1/\sin^{2}x)-1}}.
\end{align*}
But we know that
\begin{align*}
(\sin^{2}x)\log\dfrac{1}{n^{(1/\sin^{2}x)-1}}\leq(\sin^{2}x)\cdot\dfrac{1}{n^{(1/\sin^{2}x)-1}}\rightarrow 0\cdot 0=0
\end{align*}
as $x\rightarrow 0$.
So by Squeeze Theorem $\log f(x)\rightarrow\log n$ and hence $f(x)\rightarrow n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can I say that this matrix is invertible? Let $A_{ij}=1$ if $i=j$ and let $A_{ij}=\frac{1}{2}$ if $i \neq j$. How can I formally argue that $A$ is an invertible matrix? I can made for particular examples, but I don't know how to argue for the general case.
| $$
\begin{bmatrix}
1&\frac12&\frac12&\cdots&\frac12\\
\frac12&1&\frac12&\cdots&\frac12\\
\frac12&\frac12&1&\cdots&\frac12\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\frac12&\frac12&\frac12&\cdots&1\\
\end{bmatrix}
$$
subtract $\frac1n\times$ the sum of the $n-1$ right columns from column $1$ and we get
$$
\begin{bmatrix}
\frac{n+1}{2n}&\frac12&\frac12&\cdots&\frac12\\
0&1&\frac12&\cdots&\frac12\\
0&\frac12&1&\cdots&\frac12\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&\frac12&\frac12&\cdots&1\\
\end{bmatrix}
$$
subtract $\frac1{n-1}\times$ the sum of the $n-2$ right columns from column $2$ and we get
$$
\begin{bmatrix}
\frac{n+1}{2n}&\frac1{2(n-1)}&\frac12&\cdots&\frac12\\
0&\frac{n}{2(n-1)}&\frac12&\cdots&\frac12\\
0&0&1&\cdots&\frac12\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0 &\frac12&\cdots&1\\
\end{bmatrix}
$$
Continuing for the rest of the columns, we end up with
$$
\begin{bmatrix}
\frac{n+1}{2n}&\frac1{2(n-1)}&\frac1{2(n-2)}&\cdots&\frac12\\
0&\frac{n}{2(n-1)}&\frac1{2(n-2)}&\cdots&\frac12\\
0&0&\frac{n-1}{2(n-2)}&\cdots&\frac12\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&1\\
\end{bmatrix}
$$
Computing the determinant of the upper triangular matrix gives
$$
\frac{n+1}{2^n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
what is 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 +1/9 - ...? I know that it is converging because it is alternating series with terms getting smaller to zero. but I do not know what it converges to value
| $ S = 1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n} - 2 ( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2n} - 2 \frac{1}{2} ( 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{n} ) $
Call $ H_n = 1 + \frac{1}{2} + ... \frac{1}{n} $ the harmonic series
We know that $ H_n - \ln n\,$ converges to the Euler-Mascheroni constant
$\gamma = 0.57...$
$S = H_{2n} - H_n = H_{2n} - \ln{2n} + H_n + \ln n + \ln 2n - \ln n = (H_{2n} - \ln 2n ) - ( H_n - \ln n ) + ln ( \frac{2n}{n} ) \to \ln( \frac{2n}{n} ) = \ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove trigonometric an identity $\frac{2\cos\alpha -1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$
Prove trigonometric an identity
$\frac{2\cos\alpha
-1}{\sqrt3-2\sin\alpha}=\tan\left(\frac{\alpha}{2}+\frac{\pi}{6}\right)$
My proof:
$\frac{2\cos\alpha
-1}{\sqrt3-2\sin\alpha}=\frac{\cos\alpha-\frac{1}{2}}{\frac{\sqrt3}{2}-\sin\alpha}=\frac{\cos\alpha-\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}-\sin\alpha}$
I have no idea what to do now
| Using Prosthaphaeresis Formulas
$\dfrac12=\sin\dfrac\pi6=\cos\dfrac\pi3$
$\dfrac{\sqrt3}2=\cos\dfrac\pi6=\sin\dfrac\pi3$
$$\cos\alpha-\cos\dfrac\pi3=2\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\sin\left(\dfrac\pi6+\dfrac\alpha2\right)$$
$$\sin\dfrac\pi3-\sin\alpha=2\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\cos\left(\dfrac\pi6+\dfrac\alpha2\right)$$
Remember we need $\sin\left(\dfrac\pi6-\dfrac\alpha2\right)\ne0$ to reach at the identity required
Alternatively, we can use Weierstrass Substitution in the left hand side and use $\tan(A+B)$ formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $.
*
*For $\int_0^{1}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I prove $\ln(1+x^2)< x^2$, using that I can prove $\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ converges, but it take too long so is there shorter way to do this problem?
*For $\int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I have no idea how to do this problem.
| There is a short way: use equivalents to remove unnecessary details:
*
*For $\displaystyle \int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,\mathrm d x$, the integrand is equivalent near $0$ to
$$\frac{x^2}{\sqrt{2x^5}}=\frac1{\sqrt{2x}},$$
and the integral of the latter converges on $(0,1]$.
*For $\displaystyle \int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,\mathrm dx $, the integrand is equivalent, when $x\to\infty$, to
$$\frac{\ln (x^2)}{\sqrt {x^6}}=\frac{2\ln x}{x^3}= \frac{\ln x}{x}\frac1{x^2}=o\Bigl(\frac1{x^2}\Bigr),$$
and the integral of $1/x^2$ converges on $[1,+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| Also, the Tangent Line method helps: $$x^2+y^2+z^2-27=\sum_{cyc}(x^2-9)=$$
$$=\sum_{cyc}\left(x^2-9-18\left(\frac{1}{x}-\frac{1}{3}\right)\right)=\sum_{cyc}\frac{(x-3)^2(x+6)}{x}\geq0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| Let me give you a much simpler solution, using the general fact that if $B$ is a square matrix of rank $1$, then $\det({\rm Id}_n+B) = 1+{\rm tr}(B)$. Let $A$ be the matrix you want to compute the determinant of. Then if $v = [a ~ b ~c]^\top$, we have that $A = \lambda{\rm Id}_3 + vv^\top$. This means that $$\begin{align}\det(A) &= \det(\lambda{\rm Id}_3+vv^\top) = \det\left(\lambda\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3 \det\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right) = \lambda^3\left(1+ {\rm tr}\left(\frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3\left(1+ \frac{\|v\|^2}{\lambda}\right) = \lambda^3 + \lambda^2\|v\|^2 \\ &= \lambda^2(\lambda + \|v\|^2).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Show equality of sets : Is there a better way? Let \begin{equation*}S:=\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+\left \{a\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\mid a\in \mathbb{R}\right \} \ \text{ and } \ T:=\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix}+\left \{b\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\mid b\in \mathbb{R}\right \}\end{equation*}
I want to show that $S=T$. I have done the following:
Let $x\in S$. Then \begin{equation*}x=\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+a\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\end{equation*} with $a\in \mathbb{R}$.
We have that:
\begin{align*}x&=\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}-2\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}+2\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}+a\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\\ & =\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}-\begin{pmatrix}4 \\ -2 \\ 2\end{pmatrix}+(2+a)\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix} \\ & =\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix}+(2+a)\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\in T\end{align*}
So $S\subseteq T$.
Let $x\in h$. Then \begin{equation*}x=\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix}+b\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\end{equation*} with $b\in \mathbb{R}$.
We have that:
\begin{align*}x&=\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix}+2\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}-2\begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}+b\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\\ & =\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix}+\begin{pmatrix}4 \\ -2 \\ 2\end{pmatrix}+(-2+b)\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix} \\ & =\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+(-2+b)\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}\in S\end{align*}
So $T\subseteq S$.
From the two relations we get $S=T$.
$$$$
Is there maybe a better way to show that?
| It suffices to note that the direction vectors are the same (we need in general they are parallel) and solve for
$$\begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}+x\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}=\begin{pmatrix}-3 \\ 2 \\ 0\end{pmatrix} \iff x\cdot \begin{pmatrix}2 \\ -1 \\ 1\end{pmatrix}=\begin{pmatrix}-4 \\ 2 \\ -2\end{pmatrix} \iff x=-2$$
indeed from this it follows that by $a=(b-2)$ they represents exactly the same set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(\cos x-1)^2(4\cos^2x+2\cos x-1)^2=16(1-\cos^2x)(2\cos^3x-\cos x)^2$
We obviously get $\cos x=1$ (which indeed is a solution of original equation) and it remains
$(1-\cos x)(4\cos^2x+2\cos x-1)^2=16(1+\cos x)(2\cos^3x-\cos x)^2$
Setting $\cos x=y$, this is :
$(1-y)(4y^2+2y-1)^2=16(1+y)(2y^3-y)^2$
Which is $64y^7+64y^6-48y^5-64y^4-4y^3+16y^2+5y-1=0$
I found no clever way to solve this degree-7 polynomial
| By tangent half angle identities we obtain by $t=\tan \frac x 2$
*
*$\cos 2x =2\cos^2 x-1=\frac{2(1-t^2)^2}{(1+t^2)^2}-1$
*$\cos 3x=4\cos^3 x-3\cos x=4\frac{(1-t^2)^3}{(1+t^2)^3}-3\frac{1-t^2}{1+t^2}$
*$\sin (4x)=2\sin (2x)\cos(2x)=4\sin x\cos x(2\cos^2 x-1)=4\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}\left(\frac{2(1-t^2)^2}{(1+t^2)^2}-1\right)$
then
$$\cos 2x-\cos 3x+\sin 4x=0$$
$$\frac{2(1-t^2)^2}{(1+t^2)^2}-1-4\frac{(1-t^2)^3}{(1+t^2)^3}+3\frac{1-t^2}{1+t^2}+4\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}\left(\frac{2(1-t^2)^2}{(1+t^2)^2}-1\right)=0$$
$$2(1-t^2)^2(1+t^2)^2-(1+t^2)^4-4(1-t^2)^3(1+t^2)+$$$$+3(1-t^2)(1+t^2)^3+8t(1-t^2)(t^4-6t^2+1)=0$$
that is
$$t (t^7 - 4 t^6 - 9 t^5 + 28 t^4 - 5 t^3 - 28 t^2 + 5 t + 4) = 0$$
which confirms that $t=0$ is a solution then
$$t^7 - 4 t^6 - 9 t^5 + 28 t^4 - 5 t^3 - 28 t^2 + 5 t + 4=0$$
which seems to have others $5$ not trivial solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$
So the general formula for this series is
$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$
Is there a closed form for this series?
| Start with a geometric series
\begin{eqnarray*}
\sum_{n=0}^{\infty} (-1)^n y^n = \frac{1}{1+y}.
\end{eqnarray*}
Integerate
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n y^{n+1}}{n+1} = \ln(1+y).
\end{eqnarray*}
Integerate again
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n y^{n+2}}{(n+1)(n+2)} = (1+y)\ln(1+y)-y.
\end{eqnarray*}
Now divide by $y^2$ and let $y=x^2$ and we have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(n+1)(n+2)} = \frac{(1+x^2)\ln(1+x^2)-x^2}{x^4}.
\end{eqnarray*}
| {
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Let $T$ be the set of all $3 × 3$ symmetric matrices all of whose entries are either $0$ or $1$. Answer following Let $T$ be the set of all $3 × 3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.
i) The number of matrices in $T$ is
ii)The number of matrices in $T$ for which the system of linear equations has unique solution where $A\in T$
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
iii) The number of matrices in $T$ for which the system of linear equations is inconsistent where $A\in T$
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
My attempt is as follows:-
i) $A=\begin{bmatrix}
a&g&h\\
g&b&i\\
h&i&c
\end{bmatrix}$
It is given that five of the entries are $1$ and four of them are $0$
Case $1$: Among elements $g$,$h$,$i$ only one element is $1$, so automatically we have $2$ elements in A which are $1$ as A is symmetric. Remaining $3$ ones will be along the diagonal. So $\dbinom{3}{1}\dbinom{3}{3}$ matrices will be there for case $1$
Case $2$: Among elements $g$,$h$,$i$ only $2$ elements are $1$, so automatically we have $4$ elements in A which are $1$ as A is symmetric. Remaining $1$ one will be along the diagonal. So $\dbinom{3}{2}\dbinom{3}{1}$ matrices will be there for case $1$
So in total we would have $\dbinom{3}{1}\dbinom{3}{3}+\dbinom{3}{2}\dbinom{3}{1}=12$ matrices
ii) For given system of linear equations to have a unique solution, we need to count the matrices which have determinant non-zero.
Now here I was not getting any clever way to count the matrices which have determinant non-zero. So I decided to write all $12$ matrices and check their determinant.
I got $6$ matrices which had their determinant non-zero. But I am not satisfied with my way. Any other way?
iii) Now here we first need to know the number of matrices which have determinant zero, so with the help of computation in part ii), we can say there are $6$ matrices with determinant zero.
$$A\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
Taking $adj(A)$ on both sides
$$adj(A)(A)\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=adj(A)\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$$
L.H.S would be zero matrix as $|A|=0$, so R.H.S should be non-zero matrix as we have to count the matrices with inconsistent solutions. For matrix $A$ to have inconsistent solution, at-least one of $C_{11},C_{12},C_{13}$ should be non-zero or in other words we can say first column of $adj(A)$ should have one non-zero element.
So I was getting $4$ matrices out of $6$ which had at-least one of $C_{11},C_{12},C_{13}$ non-zero.
| There are so few matrices satisfying the conditions that writing them out explicitly is probably the fastest and most reliable way of solving this. Elegant solutions are great, but if there is any kind of time pressure, it is usually best to go with the first correct solution!
$A_1=\begin{pmatrix}1&1&1\\1&0&0\\1&0&0\end{pmatrix},$
$A_2=\begin{pmatrix}0&1&1\\1&1&0\\1&0&0\end{pmatrix},$
$A_3=\begin{pmatrix}0&1&1\\1&0&0\\1&0&1\end{pmatrix},$
$A_4=\begin{pmatrix}1&0&1\\0&0&1\\1&1&0\end{pmatrix},$
$A_5=\begin{pmatrix}0&0&1\\0&1&1\\1&1&0\end{pmatrix},$
$A_6=\begin{pmatrix}0&0&1\\0&0&1\\1&1&1\end{pmatrix},$
$A_7=\begin{pmatrix}1&1&0\\1&0&1\\0&1&0\end{pmatrix},$
$A_8=\begin{pmatrix}0&1&0\\1&1&1\\0&1&0\end{pmatrix},$
$A_9=\begin{pmatrix}0&1&0\\1&0&1\\0&1&1\end{pmatrix},$
$A_{10}=\begin{pmatrix}1&1&0\\1&1&0\\0&0&1\end{pmatrix},$
$A_{11}=\begin{pmatrix}1&0&1\\0&1&0\\1&0&1\end{pmatrix},$
$A_{12}=\begin{pmatrix}1&0&0\\0&1&1\\0&1&1\end{pmatrix},$
$A_1,A_{12}$ have many solutions; $A_6,A_8,A_{10},A_{11}$ have no solutions; $A_2,A_3,A_4,A_5,A_7,A_9$ each have a unique solution.
It is easy to see whether the zero-determinant cases have 0 or many solutions because they have a repeated row. Where the rows 2 and 3 are the same there are many solutions (because we want a value 0 in each case for that component of the vector). Where one of the repeat rows is 1 we have zero solutions.
| {
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What is the maximum possible value of a + 4b? Let $a$ and $b$ be real numbers so that the equation $x^2 − (a + b)x + 9(a^2 + b^2 ) = \frac{17}{18}$ has an integer solution. What is the maximum possible value of $a + 4b$?
I can't figure out if I need to use the AM-GM inequality or the quadratic equation. Any help would be appreciated.
| Let $c = a + 4b$. Then $a = c - 4b$ and the equation becomes:
$x^2 - (c - 4b + b)x + 9(c^2 - 8bc + 16b^2 + b^2) = x^2 + (3b - c)x + 9(17b^2 - 8bc + c^2) = \frac{17}{18}$
Rearranging, we get
$cx + 9c^2 - 8bc = -x^2 + 3bx - 17b^2 - \frac{17}{18}$
or
$f(c) = 9c^2 + (x - 8b)c = -x^2 + 3bx - 17b^2 - \frac{17}{18}$
A critical point of $f(c)$ occurs when the derivative of $f(c)$ with respect to $c$ is equal to zero (the right side doesn't matter because all constants with respect to the variable $c$ become zero with the derivative):
$\partial_c f(c) = \partial_c((9c^2 + (x - 8b)c) = 18c + (x - 8b) = 0$
$c = \frac{x - 8b}{18}$
Because the integer variable $x$ is not bounded above and the real number variable $b$ is not bounded below, the critical point $c$ has no maximum over the entire range of values for $x$ and $b$, regardless if it is a maximum, a minimum, or a saddle point with fixed $x$ and $b$. Since $c = a + 4b$, there is no maximum for $a + 4b$ either.
| {
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Maximum and minimum of $x^2+2y^2+3z^2$ subject to $x^2+y^2+z^2=100$
Find the maximum and minimum values of the function $f(x,y,z)=x^2+2y^2+3z^2$ subject to the constraint $x^2+y^2+z^2=100$.
I know to find the critical points I need to solve the system of equations
$$\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$$
I ended up with
$$\begin{aligned} 2x &= \lambda 2x\\ 4y &= \lambda 2y\\ 6z &= \lambda 2z\end{aligned}$$
I don't know to go from here since $\lambda$ ends up as $1$, $2$, and $3$.
| This is a sort of "dual" of the extremization problem that is often asked here [for example: Lagrange Multipliers to find the maximum and minimum values ] , with the function being $ \ x^2 + y^2 + z^2 \ $ under the constraint $ \ ax^2 + by^2 + cz^2 \ = \ d \ $ with specified coefficients. The geometrical interpretation is similar, though. The constraint surface $ \ x^2 + y^2 + z^2 \ = \ 100 \ $ represents a spherical surface of radius $ \ 10 \ $ centered on the origin. Level surfaces of $ \ x^2 + 2y^2 + 3z^2 \ $ are triaxial ellipsoids of varying sizes, with all of them being similar.
For a particular ellipsoid $ \ x^2 + 2y^2 + 3z^2 \ = \ C \ \ \rightarrow \ \ \frac{x^2}{C} \ + \ \frac{y^2}{C/2} \ + \ \frac{z^2}{C/3} \ = \ 1 \ \ , $ the semi-axis lengths are $ \ \sqrt{C} \ , \ \sqrt{C/2} \ $ and $ \ \sqrt{C/3} \ $ . The smallest of this "family" of ellipsoids which will be tangent to the spherical surface is the one for which $ \ \sqrt{C} \ = \ 10 \ \Rightarrow \ C \ = \ 100 \ \ ; $ this gives us an ellipsoid inscribed within the sphere. The largest ellipsoid that can be tangent to the spherical surface is the one for which its shortest axis makes contact (the ellipsoid circumscribes the sphere); for this, $ \ \sqrt{C/3} \ = \ 10 \ \Rightarrow \ C \ = \ 300 \ \ . $ So the constrained minimum and maximum values of $ \ x^2 + 2y^2 + 3z^2 \ $ are $ \ 100 \ $ and $ \ 300 \ $ , respectively, as also shown in the other answers. [The ellipsoid which has the endpoints of its "middle-length" axis tangent to the spherical surface has $ \ \sqrt{C/2} \ = \ 10 \ \Rightarrow \ C \ = \ 200 \ \ , $ which has no special importance here.]
| {
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Find factor $f(x)$ conmmon to two quartic equation Let $f(x) = x^2 + bx + c$, where $b, c ∈ R$. If f(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x + 5$, then find $f(x)$
My approach ,on dividing both quartic equation by $f(x)$ remainder is zero, but not getting the answer
| If is a factor of $x^4 + 6 x^2 + 25$ and $3 x^4 + 4 x^2 + 28 x + 5$
then it is a factor of
$$
3 (x^4 + 6 x^2 + 25) - (3 x^4 + 4 x^2 + 28 x + 5)=14 (5 - 2 x + x^2)
$$
| {
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Is this inequality true $xy(x^{2}+y^{2})\le\frac{(x+y)^{4}}{8}$? Is this inequality true ?
$$xy(x^{2}+y^{2})≤\dfrac{(x+y)^{4}}{8}$$
$x,y>0$
If true how ? And which inequality has use it ?
I know that : $xy≤\dfrac{(x+y)^{2}}{4}$ by Am-Gm
But is $x^{2}+y^{2}≤\dfrac{(x+y)^{2}}{2}$ ?
In first in this rule : $\dfrac{x^{n}+y^{n}}{2}≥\left(\dfrac{x+y}{2}\right)^{n}$
But we have $≤$ not $≥$
Also after simplified a get :
$$2x^{2}+2y^{2}=x^{2}+y^{2}+2xy$$
$$x^{2}+y^{2}-2xy=(x-y)^{2}≥0$$
Correct $\color{#2f0}{\checkmark}$
Now I need generalized $$x^{p}+y^{p}≤\dfrac{(x+y)^{p}}{p}$$
I'm correct or no ? And where deferent between this inequality and power mean inequality
| Noodling.
I see $xy$ and I see $x^2 +y^2$ and think somehow I want to convert $xy(x^2 + y^2) \to x^2 + 2xy + y^2$ but gob only knows how.
I've got a hammer called AM.GM so I whack $(x+y)^2 = x^2 + 2xy + y^2$ with it just to see what will happen.
$2xy + x^2 + y^2 \ge 2\sqrt{2xy(x^2 + y^2)}$ happens
and well.... I can work with that.
$(x+y)^2 \ge 2\sqrt{2xy(x^2 + y^2)}$ so
$(x+y)^4 \ge 2^2\cdot [2xy(x^2 + y^2)]=8xy(x^2 + y^2)$ and .... that does it.
.....
Alternatively $xy(x^2 + y^2) \le \frac {(x+y)^4}8\iff$
$(x+y)^4 - 8x^3y - 8xy^3 \ge 0 \iff$
$(x^4 + 4x^3y +6x^2y^2 + 4xy^3 + y^4) - 8x^3y - 8xy^3 \ge 0 \iff$
$x^4 - 4x^3y +6x^2y^2 - 4xy^3 + y^4 \ge 0 \iff$
$(x-y)^4 \ge 0$.
| {
"language": "en",
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The internal and external bisectors of $\angle A$ meet line $BC$ at $E$ and $F$. Show that the tangent at $A$ to $\bigcirc ABC$ bisects $EF$.
Suppose the internal and external bisector of $\angle A$ meet the side $BC$ (produced) at $E$ and $F$, respectively. If the tangent at $A$ to $\bigcirc ABC$ meets $BC$ (produced) at $D$, prove that $D$ bisects $EF$.
| Let $c>b$. Let $K$ be a midpoint of $EF$. We have $\frac{EC}{EB}=\frac{b}{c}$, $\frac{FC}{FB}=\frac{b}{c}$, $EC+EB=a$, $FB-FC=a$. Then $EC=\frac{ab}{b+c}$, $EB=\frac{ac}{b+c}$, $FC=\frac{ab}{c-b}$. Then $EF=EC+FC=\frac{2abc}{c^2-b^2}$. Since $AK$ is median of triange $AEF$ with $\angle EAF=\frac{\pi}{2}$ then $AK=\frac{1}{2}EF$. Then $AK=\frac{abc}{c^2-b^2}$. Also we have $KC=KE-EC=\frac{ab^2}{c^2-b^2}$, $KB=KE+EB=\frac{ac^2}{c^2-b^2}$. Since $KC=\frac{ab^2}{c^2-b^2}$, $KB=\frac{ac^2}{c^2-b^2}$ and $AK=\frac{abc}{c^2-b^2}$ then easy to see that $KA^2=KC \cdot KB$. Since $KA^2=KC \cdot KB$ then $AK$ is tangent at $A$ to the circumcircle of triangle $ABC$. Then $K=D$. Then $D$ is midpoint of $EF$.
| {
"language": "en",
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"source": "stackexchange",
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Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?
Here it shows that the correct answer is $\frac{4\pi}{3}$.
| Not an answer to the question but a quick note: we can clean up your computation by working in terms of $u$ rather than in terms of $x$. With the substitution of $u = \tan(x/2)$, we find that
$$
du=\frac{1}{2}\sec^2\frac{x}{2}dx = \frac 12 (1 + u^2)\,dx
$$
Now, we have
$$
\int \frac{1}{5 + 4\cos x}dx =
\int \frac{1}{5 + 4\frac{1-u^2}{1+u^2}}dx =
\int \frac{(1+u^2)}{5(1+u^2) + 4(1-u^2)}dx =
\int \frac{(1+[u(x)]^2)}{3^2 + [u(x)]^2}\,dx.
$$
From here, substitution gives us
$$
2\int \frac{1}{3^2 + [u(x)]^2}\cdot\frac{1+[u(x)]^2}{2} dx = 2\int\frac{1}{3^2 + u^2}\,du.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An inequality on the roots of a transcendental equation
Denote the two real roots of $x-\ln x=a(a> 1)$ as $x_1,x_2$. Show that
$x_1+x_2\leq a+\sqrt{a}$.
If we introduce Lambert W function, we can express the roots as
$$x_1=-W_{-1}(-e^{-a}),~~~~x_2=-W(-e^{-a}).\\$$
Thus we only need to show
$$f(a)=-W_{-1}(-e^{-a})-W(-e^{-a})-a-\sqrt{a}\leq 0.$$
By graphing the function, we can see $f(a)$ is a decreasing function, thus $f(a)\leq f(1)=0,$ which is what we want. But how to prove this rigorously?
| Update
Remark: I came up with an ugly proof. Some proofs are not hard and thus omitted.
Hint for Facts 3 and 4: Use Euler's substitution $a = \frac{14+t^2}{4t+10}$ ($t>2$)
to obtain $\sqrt{4a^2+10a-14} = \frac{(t+7)(t-2)}{2t+5}$.
Let $f(x) = x - \ln x - a$. Let
\begin{align}
x_3 &= \frac{1+3a}{3+a} + a - \frac{1}{3}(2a+1+\sqrt{4a^2+10a-14}), \\
x_4 &= \frac{1}{3}(2a+1+\sqrt{4a^2+10a-14}).
\end{align}
We first give some auxiliary results (Facts 1 through 4).
Fact 1: $f(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Also, $\lim_{x\to 0} f(x) = +\infty$, $f(1) < 0$
and $\lim_{x\to +\infty} f(x) = +\infty $.
Fact 2: $f(\sqrt{a} + a - \tfrac{1}{5}) > 0$ for $a > \frac{121}{64}$.
Fact 3: $x_3 \in (0, 1)$ and $f(x_3) < 0$ for $1 < a \le \frac{121}{64}$.
Fact 4: $x_4\in (1, \infty)$ and $f(x_4) > 0$ for $1 < a \le \frac{121}{64}$.
Now we proceed. From Fact 1, we have $x_1\in (0, 1)$ and $x_2 \in (1, \infty)$.
We split into two cases:
1) $a > \frac{121}{64}$: Note that $f(\frac{1}{5}) = \frac{1}{5} - \ln \frac{1}{5} - a < 0$.
From Fact 1, we have $x_1 < \frac{1}{5}$. Clearly, $\sqrt{a} + a - \tfrac{1}{5} > 1$.
From Fact 1 and Fact 2, we have $x_2 < \sqrt{a} + a - \tfrac{1}{5}$. Thus, we have $x_1 + x_2 < \sqrt{a} + a$.
2) $1 < a \le \frac{121}{64}$: From Facts 1, 3 and 4, we have $x_1 < x_3$ and $x_2 < x_4$. Thus, we have $x_1 + x_2 < x_3 + x_4
= \frac{1+3a}{3+a} + a \le \sqrt{a} + a$ where we have used $\sqrt{a} \ge \frac{1+3a}{3+a}$ for $a > 1$.
We are done.
| {
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How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?
| Hint:
Use Partial Fraction Decomposition before differentiation
$$f(x)=\dfrac{x(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}=x+\dfrac A{x+1}+\dfrac B{x+2}+\dfrac C{x+3}$$
$$(x+1)(x+2)(x+3)f(x)=?$$
Set $x=-1,-2,-3$ to find $A,B,C$
$$f'(0)=1-\dfrac A{1^2}-\dfrac B{2^2}-\dfrac C{3^2}$$
| {
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Sum of squares of products of subsets without neighboring elements equals $(N+1)! -1$ Question: Let $n$ be any natural number. Consider all nonempty subsets of the set $\{1,2,...,n\}$, which do not contain any neighboring elements. Prove that the sum of the squares of the products of all numbers in these subsets is $$(n + 1)! - 1.$$ For example, if $n = 3$, then such subsets of $\{1,2,3\}$ are $\{1\}$, $\{2\}$, $\{3\}$, and $\{1,3\}$, and $$1^2 + 2^2 + 3^2 + (1\cdot3)^2 = 23 = 4! -1.$$
This question can be proved by induction, as shown here:
induction (sum of squares of products of elements of certain subsets of $\{1,\dots,n\}$)
This seems to me like something that is really out of the blue. So my question is, is there another way to see why this is true? Such as a combinatorial argument. In particular, does the quantity "sum of squares of products of numbers in subsets without neighboring elements" arise in some natural way?
| This is a little bit sketchy.
Consider the easier problem of finding the sum of the squares of the products of all elements for each subset of $\{1,2,3,...,n\}$ (so I am dropping the neighboring condition).
The answer is
$$(1+1^2)(1+2^2)(1+3^2)...(1+n^2)$$
This product can be written as a sum where each term is obtained by choosing $1$ or $j^2$ for each $(1+j^2)$ and then muliplying everything.
In fact, each term obtained this way is the square of the products of all elements of a particular subset.
For example, if we pick $1^2$ in $(1+1^2)$, $5^2$ in $(1+5^2)$, $11^2$ in $(1+11^2)$ and $1$ in all other terms of the product we get the square of the products of all elements of $\{1,5,11\}$.
Now I try to adapt this approach to the actual problem.
Consider the following sum:
$$\left(1+\frac{n^2}{n}\right)\left(1+\frac{(n-1)^2}{n-1}\right)\left(1+\frac{(n-2)^2}{n-2}\right)...\left(1+\frac{2^2}{2}\right)(1+1^2) \tag{1}$$
We start from the term $\left(1+\frac{n^2}{n}\right)$. We can decide to not include $n$ in our subset, hence we pick $1$ in the term $\left(1+\frac{n^2}{n}\right)$ and the next term we are going to consider is $\left(1+\frac{(n-1)^2}{n-1}\right)$.
Or, we can include $n$, hence we pick $\frac{n^2}{n}$ in the term $\left(1+\frac{n^2}{n}\right)$. In this case, we are multypling by $n^2$ (for the same reason as before) and we are dividing by $n$; notice that the term $\left(1+\frac{(n-1)^2}{n-1}\right)$ is exactly equal to $n$, so by dividing by $n$ we are cancelling $\left(1+\frac{(n-1)^2}{n-1}\right)$; hence, the next term we are going to consider next is $\left(1+\frac{(n-2)^2}{n-2}\right)$.
The idea is that if we include $n$ in our subset we cannot include $n-1$ (beacuse otherwise we would have two neighboring elements); cancelling the term $\left(1+\frac{(n-1)^2}{n-1}\right)$ is the same as not picking $n-1$.
We then repeat and the final result should be the square of the product of all elements for a subset of $\{1,2,3,...,n\}$ with no neighboring elements.
For example, consider the subset $\{13,5,3\}$; this corresponds to the term:
$$\frac{13^2}{13}\left(1+\frac{12^2}{12}\right)\frac{5^2}{5}\left(1+\frac{4^2}{4}\right)\frac{3^2}{3}\left(1+\frac{2^2}{2}\right)$$
$$=\frac{13^2}{13}(13)\frac{5^2}{5}\left(5\right)\frac{3^2}{3}\left(3\right)$$
$$=13*5*3$$
Notice that (1) is equal to $(n+1)!$ so this should be the value of the required sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complete solution of $\int x\sqrt{\frac{2\sin(x^2+1)-\sin2(x^2+1)}{2\sin(x^2+1)+\sin2(x^2+1)}}dx$
Let $x^2\neq n\pi-1, n\in\mathcal{N}$, Then
$$
\int x\sqrt{\frac{2\sin(x^2+1)-\sin2(x^2+1)}{2\sin(x^2+1)+\sin2(x^2+1)}}dx
$$
$$
\int x\sqrt{\frac{2\sin(x^2+1)-\sin2(x^2+1)}{2\sin(x^2+1)+\sin2(x^2+1)}}dx=\int x\sqrt{\frac{1-\cos(x^2+1)}{1+\cos(x^2+1)}}dx\\
=\int x\Big|\tan\frac{(x^2+1)}{2}\Big|dx\\
\text{Set }t=\frac{(x^2+1)}{2}\implies dt=xdx\\
I=\pm\int \tan tdt=\pm\log|\sec(x^2+1)|+C
$$
But my reference gives the solution $\log|\sec(x^2+1)|+C$, anyone confirm $\pm$ sign is relevant here ?
Do I not consider the domain and range of the functions in such problems ?
| Although positive sign appears before the simple odd integrand function
$$ \int 3 x\sqrt{1+x^2} dx $$
and its integral $ (1+x^2)^\frac32, $ the presence of $\sqrt{ }$ includes and implies the negative sign , so that we can use $\pm$ sign in front of the integral as well as the integrand.
The same is the case for integrands of general polynomial,trigonometric functions of such given integrands in general irrespective of function domain and range.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it out.
| Let $\tilde c=-c$. Therefore we have
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{\tilde c}=0\Rightarrow ab+a\tilde c+b\tilde c=0$$
Now we can use the following identity:
$$(a+b+\tilde c)^2=a^2+b^2+\tilde c^2+2\underbrace{(ab+a\tilde c+b\tilde c)}_{=0}=a^2+b^2+\tilde c^2$$
Therefore
$$a^2+b^2+c^2=a^2+b^2+ (-\tilde c)^2=a^2+b^2+\tilde c^2=(a+b+\tilde c)^2=(a+b-c)^2\Rightarrow $$
$$a^2+b^2+c^2=(a+b-c)^2$$
With the last equation we proved what we wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many real solutions $(x,y)$ for $x + y^2 = y^3 $ and $y + x^2 = x^3$ I think the answer is that there are $3$ real solutions.
*
*When $x = y$
*When $x = -y$
If $x\neq y$ is there a real solution? If so, how do we find the real solutions?
| Subtracting one equation from the other, we obtain $$(x-y)+\left(y^2-x^2\right)=\left(y^3-x^3\right)\Rightarrow$$ $$(x-y)\left(1+(x+y)+\left(x^2+xy+y^2\right)\right)=0\Rightarrow$$ $$(x-y)\left(\frac14 (1 + 2 x + y)^2+\frac1{12}(3y+1)^2+\frac23\right).$$ The second factor is strictly positive, so the only possibility is that $x=y$. In this case, our initial condition becomes $$x+x^2=x^3,$$ which gives the $\boxed{3}$ solutions $$(0,0),\left(\frac{1-\sqrt5}2, \frac{1-\sqrt5}2\right),\left(\frac{1+\sqrt5}2, \frac{1+\sqrt5}2\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
I tried solving but got stuck.
Show that it is true for n=1
$$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
I'm stuck and don't know how to proceed
| If induction is not mandatory,
I believe this is how the problem came into being
$$2(1+6)^n+3(1+4)^n\equiv2(1+6n)+3(1+4n)\pmod{(72,48)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Expected value of rolling 3 dice if we take square number
Questions:
$(1)$ Suppose we play a game. I roll a die up to three times. Each time I roll, you can either take the square of the number showing as dollars, or roll again. How much should you pay to play this game?
$(2)$ What effect will it have if we are charged for $\$1$ for each additional roll?
Question $(1)$ above is very common if we replace 'square of the number' with just 'number. For example, this post in MSE.
For $(1),$ assuming that we have $1$ die. Then the expected value is
$$\sum_{k=1}^6 \frac{k^2}{6} = \frac{91}{6}.$$
Now, assume that we have $2$ dice. Then
$$\frac{91}{6} \times \frac{1}{6} + \frac{91}{6} \times \frac{1}{6} + \frac{91}{6} \times \frac{1}{6} + \frac{4^2}{6} + \frac{5^2}{6} + \frac{6^2}{6} = \frac{245}{12}.$$
Similarly, if we have $3$ dice, then
$$\frac{245}{12}\times\frac{1}{6} + \frac{245}{12}\times\frac{1}{6} + \frac{245}{12}\times\frac{1}{6} + \frac{245}{12}\times\frac{1}{6} + \frac{5^2}{6} + \frac{6^2}{6} = \frac{428}{18}.$$
Are my calculations above correct?
For $(2),$ I think we should pay less. But I do not how much lesser is it.
| I agree with your calculations for the first part.
For the second part, your calculation for one roll is still correct. But for the second roll, your expected value for rerolling would be $\frac{91}6-1=\frac{85}6$
$$\frac{85}{6} \times \frac{1}{6} + \frac{85}{6} \times \frac{1}{6} + \frac{85}{6} \times \frac{1}{6} + \frac{4^2}{6} + \frac{5^2}{6} + \frac{6^2}{6} = \frac{239}{12}$$ and similarly for the third part except with $\frac{227}{12}$ instead of $\frac{245}{12}$. None of these subtractions change the decisions about when you would hold and when you would re-roll with just a \$1 charge, but you would want to stay aware of that in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488879",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \cos 2\theta \ln\left(\frac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$ $$\int \cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$$
My attempt is as follows:-
$$\ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)=t\tag{1}$$
$$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\left(\cos\theta-\sin\theta\right)^2-(-\sin\theta-\cos\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}=\dfrac{dt}{d\theta}$$
$$\dfrac{2}{\cos2\theta}=\dfrac{dt}{d\theta}$$
Let's calculate $\cos2\theta$ from equation $1$
$$\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}=e^t$$
$$\dfrac{1+\tan\theta}{1-\tan\theta}=e^t$$
Applying componendo and dividendo
$$\dfrac{2}{2\tan\theta}=\dfrac{e^t+1}{e^t-1}$$
$$\dfrac{e^t-1}{e^t+1}=\tan\theta$$
$$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\cos2\theta=\dfrac{(e^t+1)^2-(e^t-1)^2}{(e^t+1)^2+(e^t-1)^2}$$
$$\cos2\theta=\dfrac{4e^t}{2(e^{2t}+1)}$$
$$\cos2\theta=\dfrac{2e^t}{e^{2t}+1}\tag{2}$$
So integral will be
$$\dfrac{1}{2}\cdot\int \left(\dfrac{2e^t}{e^{2t}+1}\right)^2dt$$
$$\dfrac{1}{2}\cdot\int \dfrac{4e^{2t}}{(1+e^{2t})^2}$$
$$e^{2t}+1=y$$
$$2e^{2t}=\dfrac{dy}{dt}$$
$$2e^{2t}dt=dy
$$\int \dfrac{dy}{y^2}$$
$$-\dfrac{1}{y}+C$$
$$-\dfrac{1}{1+e^{2t}}+C$$
$$-\dfrac{1}{1+e^{\ln\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2}}+C$$
$$-\dfrac{1}{1+\dfrac{1+\sin2\theta}{1-\sin2\theta}}+C$$
$$-\dfrac{1-\sin2\theta}{2}+C$$
$$\dfrac{\sin2\theta}{2}+C'$$
And this should be actually wrong because if we differentiate the result, it will give $\cos2\theta$, but integrand is $\cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)$
What am I missing here, checked multiple times, but not able to get the mistake. Any directions?
| $$
\displaystyle \begin{array}{rl}
& \displaystyle \int\cos 2 \theta\ln\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right) d \theta \\
= & \displaystyle \frac{1}{2} \int \cos 2 \theta \ln \left(\frac{1+\sin 2 \theta}{1-\sin 2 \theta}\right) d \theta \\
= & \displaystyle \frac{1}{4} \int \ln \left(\frac{1+x}{1-x}\right) d x \\
= & \displaystyle \frac{1}{4}\left[\ln (1+x) d x-\int \ln (1-x) d x\right] \text { where } x=\sin 2 \theta\\
= & \displaystyle \frac{1}{4}[(1+x) \ln (1+x)-(1-x) \ln (1-x)]+C \\
= & \displaystyle \frac{1}{4}[(1+\sin 2 \theta) \ln (1+\sin 2 \theta)-(1-\sin 2 \theta) \ln (1-\sin 2\theta)+C
\end{array}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$ Find the remainder when $f(x) = x^{2016}+2x^{2015}-3x+4$ is divided by $g(x)=x^2+3x+2$
I try to factor them, but I failed. I know that it's impossible to divide it
| Notice: $x^2+3x+2=(x+1)(x+2)$ and let the remainder be $ax+b$ (must be at most linear since we divide by quadratic polynomial).
Write $$x^{2016}+2x^{2015}-3x+4=k(x)(x+2)(x+1)+ax+b$$
Put $x=-1$ we get $$6=-a+b$$ and put $x=-2 $ we get $$10 = -2a+b$$ Now solve this system and you get an answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a limit in order to find an asymptote I need to draw the graph of the function $y=\sqrt[3]{(x^2)(x+9)}$. Suppose $y=k\cdot x + b$ is the equation for the asymptote.
I managed to find $$k=\lim_{x\to\infty}\frac{y(x)}{x}=\lim_{x\to\infty}\sqrt[3]{\frac{x+9}{x}}=1$$
But then I got stuck finding $$b=\lim_{x\to\infty}y(x)-k\cdot x$$
Any ideas on how to deal with it?
| You can write $y(x)-x = \frac{y(x)^3-x^3}{y(x)^2+y(x)x+x^2}$. Let us work with this expression
\begin{align*}
y(x)-x &= \frac{x^2(x+9) - x^3}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\
&= \frac{9x^2}{(x^2(x+9))^{2/3} + x\sqrt[3]{x^2(x+9)} + x^2} \\
&= \frac{9}{\frac{1}{x^2}(x^2(x+9))^{2/3} + \frac{1}{x}\sqrt[3]{x^2(x+9)} + 1} \\
&= \frac{9}{(\frac{x+9}{x})^{3/2} + \sqrt[3]{\frac{x+9}{x}} + 1 }
\end{align*}
Taking a limit yields $\frac{9}{1+1+1}=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\tan x>x+\frac {x^3}3$ for $x\in(0,\frac\pi2)$
$$\tan x>x+\frac {x^3}3~\text{ for }~x\in\left(0,\frac\pi2\right)$$
My solution:
Both functions are monotone, increasing, equal at $x=0$. If i could show that the derivative of the first is greater than then derivative of second function that would be it. Taking the derivatives:
$$\frac1{\cos^2 x}>1+x^2$$
Applying the same reasoning on the derivatives and taking their derivatives we get
$$\frac{\tan x}{\cos^2 x}>x$$
Doing the same thing (it is possible to stop here if we use $\tan x>x$)
$$\frac{1+\tan x\sin 2x}{\cos^4 x}>1$$
It is easy to see that last inequality is true therefore all previous are also true.
Is my solution correct, i would be very disappointed if it weren't. Are there different ways to solve this ?
| Since $\tan(x)=-\frac{d}{dx}\log\cos x$ and
$$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) $$
we have
$$ \frac{\tan x}{x}=\sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4x^2}=\sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2}\sum_{m\geq 0}\left(\frac{4x^2}{\pi^2(2n+1)^2}\right)^m $$
and by exchanging $\sum_{n\geq 0}$ and $\sum_{m\geq 0}$ we get
$$ \color{red}{\frac{\tan x}{x}} = \sum_{m\geq 0}\frac{4^{m+1} x^{2m}}{\pi^{2m+2}}\cdot 2\sum_{n\geq 0}\frac{1}{(2n+1)^{2m+2}}=\color{red}{\sum_{m\geq 0}2(4^{m+1}-1)x^{2m}\cdot \frac{\zeta(2m+2)}{\pi^{2m+2}}} $$
and the given inequality follows from $\zeta(2)=\frac{\pi^2}{6}$, $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2m+2)\geq 1$ for any $m\in\mathbb{N}$.
This actually leads to
$$ \frac{\tan x}{x} \geq 1 + \frac{x^2}{3} + \frac{6 x^4 \left(21 \pi ^2-20 x^2\right)}{4 \pi ^4 x^4-5 \pi ^6 x^2+\pi ^8}$$
for any $x\in\left(0,\frac{\pi}{2}\right)$.
| {
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"url": "https://math.stackexchange.com/questions/3494024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.
Here are my proofs that $\sin \left(\dfrac{\pi}{2}-x\right)=\cos\left(x\right).$ First, angle addition says that $\sin\left( \dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\cos x-\sin x\cos\dfrac{\pi}{2}=(1)\cos x - \sin x (0) = \cos x.$ Second, a geometric proof. Consider a right triangle. We may assume WLOG that the hypotenuse is $1$. If not, then multiply all sides by the multiplicative inverse of the hypotenuse so that it is (the triangle obtained is similar, so the angles are preserved). Let one of the acute angles be $\alpha$. Then the side adjacent to that angle has length $\cos \alpha$. Now consider the angle $\dfrac{\pi}{2} - \alpha$. The side opposite to this angle has length $\cos \alpha$. But since $\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}}, \sin\left(\dfrac{\pi}{2} - \alpha\right)=\cos \alpha.$ Now, if $\alpha$ is not acute, we may add an integer multiple of $2\pi$ to $\alpha$ so that it is, without changing the value of $\sin \alpha$ since $\sin x$ is $2\pi$ periodic.
Now, since $0<x < \dfrac{\pi}{2}, \sin x,\cos x \in (0, 1)\Rightarrow \dfrac{3}{2}\sin x, \dfrac{3}{2}\cos x \in (0,\dfrac{3}{2})\subseteq (0,\dfrac{\pi}{2})$. So, using this fact, we have that $\cos \left(\dfrac{3}{2}\cos x\right)=\sin \left(\dfrac{3}{2}\sin x\right)\Leftrightarrow\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}.$ Using the fact that $\sin x = \sqrt{1-\cos ^2 x},$ I get that $\sin x = \dfrac{\frac{2\pi}{3}\pm \frac{2}{3}\sqrt{18-\pi^2}}{4},$ but I'm not sure how to use this to get $\sin \,(2x)$ into the desired form.
| Alternatively, a simpler approach involves realizing that $\begin{align}\sin x + \cos x = \dfrac{\pi}{3}&\Leftrightarrow (\sin x + \cos x)^2 = \dfrac{\pi^2}{9}\\
&\Leftrightarrow 2\sin x\cos x = \sin 2x = \dfrac{\pi^2-9}{9}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494116",
"timestamp": "2023-03-29T00:00:00",
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Derivation of the graph of $r=\frac{1}{\sin(\theta)}$ When playing around with desmos, I found a very interesting function:
$r=\frac{1}{\sin(\theta)}$
The graph of this function is a straight line with constant value $y=1$
I tried to prove this, however I failed:
Assuming $r=\sqrt {x^2+y^2}$ and $\theta = \arctan(\frac{y}{x})$ hold, we get:
$\sqrt {x^2+y^2} = \frac{1}{\sin(\arctan(\frac{y}{x}))}$
$\sqrt {x^2+y^2} = \frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$
$\sqrt {x^2+y^2} = \frac{1+\frac{y^2}{x^2}}{\frac{y}{x}}$ |$*\frac{y}{x}$
$\sqrt {\frac{y^2}{x^2}(x^2+y^2)} = 1+\frac{y^2}{x^2}$
$\sqrt {y^2 + \frac{y^4}{x^2}} = 1+\frac{y^2}{x^2}$ |$^2$
$y^2+\frac{y^4}{x^2} = 1+2\frac{y^2}{x^2}+\frac{y^4}{x^4}$ |$*x^4$
$x^4 y^2 + x^2 y^4 = x^4 + 2x^2 y^2 + y^4$
$(x^2 y^2)(x^2+y^2)=(x^2+y^2)^2$ |$:(x^2+y^2)$
$x^2 y^2 = x^2+y^2$ |$-y^2$
$x^2 y^2 - y^2 = x^2$
$(x^2 - 1) y^2 = x^2$ |$:(x^2-1)$
$y^2 = \frac{x^2}{x^2-1}$
$y = \sqrt{\frac{x^2}{x^2-1}}$
The graph of this function is obviously not a straight line with $y=1$.
Where did my calculation go wrong and what would be the actual derivation?
| A correct derivation is much simpler from the Cartesian-to-polar transformation, whereas you used the polar-to-Cartesian transformation and overcomplicated things:
$$r=\frac1{\sin\theta}\implies r\sin\theta=y=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| Hint: Imaginary part of $\frac {a+ib} {c+id}$ equals imaginary part of $\frac {(a+ib) (c-id)} {|c+id|^{2}}$ which is $\frac {bc-ad} {c^{2}+d^{2}}$ and this is $0$ iff $ad=bc$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int_0^{2\pi} \frac{\sin^2\theta}{1+\cos^2\theta}\,d\theta$ using residues? I have the following integral and I want to evaluate it using residues
$$I=\int_0^{2\pi} \frac{\sin^2\theta}{1+\cos^2\theta}d\theta$$
By using the transformation $\frac{1}{z}=e^{-i\theta}$, I got to show that
$$I=i\int_C\frac{z^4-2z^2+1}{z(z^4+6z^2+1)}dz$$
where $C$ is the centered unit circle.
I'm trying to use Cauchy's Residue Theorem, as we have 3 poles inside C: $z=0$, $z=i\sqrt{3-2\sqrt{2}}$ and $z=-i\sqrt{3-2\sqrt{2}}$. However, I'm really struggling to compute the residues of $z=i\sqrt{3-2\sqrt{2}}$ and $z=-i\sqrt{3-2\sqrt{2}}$ by hand.
Any tips or help to compute them?
| I guess, it may be a good idea to do few transformations before moving to complex plane.
First, you might note that
$$
I
= \int_0^{2\pi}\frac{\sin^2(\theta) d\theta}{1 + \cos^2(\theta)}
= \frac{1}{2}\int_0^{2\pi}\frac{1 - \cos(2\theta)}{3 + \cos(2\theta)}d(2\theta)
= \int_0^{2\pi}\frac{1 - \cos(x)}{3 + \cos(x)}dx.
$$
The latter transformation is based on the periodicity of $\cos$ -- we should double the range to $[0;4\pi]$, but instead leave it $[0;2\pi]$ and cancel $1/2$ quotient.
In complex plane this looks like
$$
I
= \int_0^{2\pi}\frac{2 - e^{ix} - e^{-ix}}{6 + e^{ix} + e^{-ix}}dx
= \oint_{|z|=1}\frac{i(z - 1)^2}{(z^2 + 6 z + 1)z} dz.
$$
The only poles within the integration contour are $z = 0$ and $z = -3 + 2\sqrt{2}$ that give residues $i$ and $-i\sqrt{2}$. Multiplying by $2\pi i$ and adding them we get the final result
$$
I = 2\pi(\sqrt{2}-1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: $\lim_{x\to \infty}\Bigg(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\Bigg)^{x\ln x}$
Find: $$\displaystyle\lim_{x\to
\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$
My attempt:
$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$
What I used:
$\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$
In the end, I got an indeterminate form: $\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$
Have I made a mistake anywhere? It seems suspicious.
added: replacement $\frac{x+1}{x^2+2x+3}$ by $\frac{1}{x}$ wasn't appealing either.
Would:$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$
be wrong?
//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$then$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$
| Whenever you see an expression of type $\{u(x) \} ^{v(x)} $ under limit it is best to take logarithms. This makes your expressions simpler and easier to type and write thereby allowing you to focus on the problem more efficiently.
Thus if $L$ is the desired limit then
\begin{align}
\log L&=\lim_{x\to\infty} (x\log x)\log\frac{\log(x^2+3x+4)}{\log(x^2+2x+3)}\notag\\
&=\lim_{x\to\infty} (x\log x) \log\left(1+\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\
&=\lim_{x\to \infty} (x\log x) \frac{\log(1+f(x))}{f(x)}\cdot f(x)\notag\\
&=\lim_{x\to \infty} x\log x\cdot 1\cdot f(x) \notag\\
&=\lim_{x\to\infty} x\log x\cdot\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}} {\log(x^2+2x+3)} \notag\\
&=\lim_{x\to\infty} \frac{x} {2}\log\left(1+\frac{x+1}{x^2+2x+3}\right)\frac{\log x^2}{\log(x^2+2x+3)}\notag\\
&=\lim_{x\to\infty} \frac{x} {2}\frac{\log(1+g(x))}{g(x)}\cdot g(x) \left(1+\dfrac{\log\dfrac{x^2}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\
&=\lim_{x\to \infty} \frac{x} {2}\cdot 1\cdot g(x)\cdot 1 \notag\\
&=\lim_{x\to\infty} \frac{x(x+1)}{2(x^2+2x+3)}\notag\\
&=\frac{1}{2}\notag
\end{align}
and hence $L=\sqrt{e}$. You should be able to notice that both $$f(x)=\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)} , g(x)=\frac{x+1}{x^2+2x+3} $$ clearly tend to $0$ as $x\to \infty $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Can we deduce the characteristic polynomial for this matrix? Given a square $n \times n$ matrix $A$ that satisfies $$\sum\limits_{k=0}^n a_k A^k = 0$$ for some coefficients $a_0, a_1, \dots, a_n,$ can we deduce that its characteristic polynomial is $\sum\limits_{k=0}^n a_k x^k$?
| Take matrix
$$
A =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2 \\
\end{array}
\right)
$$
Then
$$ A^4 - 11 A^3 + 41 A^2 - 61 A + 30 I = 0$$
However, the characteristic polynomial is
$$ x^4 - 6x^3 + 13x^2 - 12x + 4 $$
and the minimal polynomial is
$$ x^2 - 3 x + 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $\frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$.
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $$\large \frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$$
Well, I have provided my solution below, which is clearly not the most straightforward and convenient one. I would be greatly appreciated if you could come up with any other solutions.
| We have that $$\left\{ \begin{align} x + y &= m\\ xy &= n \end{align} \right. \implies m^2 - 3n - 1 = 0$$ and $$\frac{x}{x^2 + y} + \frac{y}{y^2 + x} = \frac{x(y^2 + x) + y(x^2 + y)}{(x^2 + y)(y^2 + x)} = \frac{mn + m^2 - 2n}{m^3 + n^2 - 3mn + n} = \frac{mn + n + 1}{n^2 + m + n}$$
It needs to be sufficient to prove that $$\frac{mn + n + 1}{n^2 + m + n} \le 1 \implies n^2 - mn + m - 1 \ge 0 \iff (xy)^2 - xy(x + y) + (x + y) + 1 \ge 0$$
$$\iff (xy - x - y + 1)(xy - 1) \ge 0 \implies (x - 1)(y - 1)(xy - 1) \ge 0$$
, which can be easily proven using Dirichlet's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Intuitive proof of commutativity of addition and multiplication? I have seen some "proofs" of the commutativity of addition and multiplication. But they don't really express how our mind thinks of this.
I "human" proof would go like this:
"If we have a number, $x$, of things on the left and a number, $y$, of things on the right the total number of things is $x+y$.
We can move the things on the left to the right and the things on the right to the left. There is the same number of things. But now we write this as $y+x$. Thus $x+y=y+x$."
This relies on the axiom that objects can be moved without disapearing. And the axiom that things can be moved from one place to another.
Similarly with multiplication. We could arrange things in a rectangular grid, and the "proof" would rely on the axiom that the number of things doesn't change under a rotation of the grid.
This seems like arithmetic might be derived from a set of physical axioms of geometry and movement rather than, say, abstract set theory.
Is there a way to make put these axioms into mathematical notation? Or is this just set theory in disguise?
| Here are $3$ sets of $5$ objects:
$$
\begin{array}{|c|}
\hline
\begin{array}{c}
\bullet \bullet\bullet \\ \bullet\bullet
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\bullet \\ \bullet & \bullet \\ \bullet & \bullet
\end{array} \\[10pt]
\hline
\begin{array}{c}
\bullet \bullet \\ \bullet\bullet\bullet
\end{array} \\[10pt]
\hline
\end{array}
$$
Grab one object from each set and move it to the other side of the room:
$$
\begin{array}{lcccr}
\begin{array}{|c|}
\hline
\begin{array}{c}
\bullet \bullet\phantom{\bullet} \\ \bullet\bullet
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\phantom{\bullet} \\ \bullet & \bullet \\ \bullet & \bullet
\end{array} \\[10pt]
\hline
\begin{array}{c}
\phantom{\bullet} \bullet \\ \bullet\bullet\bullet
\end{array} \\[10pt]
\hline
\end{array} & \qquad & \qquad & \qquad &
\begin{array}{c}
\bullet\bullet\bullet \\
\end{array}
\end{array}
$$
Now do it again:
$$
\begin{array}{lcccr}
\begin{array}{|c|}
\hline
\begin{array}{c}
\bullet \bullet\phantom{\bullet} \\ \bullet\phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\phantom{\bullet} \\ \bullet & \bullet \\ \bullet & \phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{c}
\phantom{\bullet} \bullet \\ \bullet\bullet\phantom{\bullet}
\end{array} \\[10pt]
\hline
\end{array} & \qquad & \qquad & \qquad &
\begin{array}{c}
\bullet\bullet\bullet \\[12pt] \bullet\bullet\bullet
\end{array}
\end{array}
$$
And again:
$$
\begin{array}{lcccr}
\begin{array}{|c|}
\hline
\begin{array}{c}
\phantom{\bullet} \bullet\phantom{\bullet} \\ \bullet\phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\phantom{\bullet} \\ \phantom{\bullet} & \bullet \\ \bullet & \phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{c}
\phantom{\bullet} \bullet \\ \phantom{\bullet}\bullet\phantom{\bullet}
\end{array} \\[10pt]
\hline
\end{array} & \qquad & \qquad & \qquad &
\begin{array}{c}
\bullet\bullet\bullet \\[12pt] \bullet\bullet\bullet \\[18pt] \bullet\bullet\bullet
\end{array}
\end{array}
$$
And again:
$$
\begin{array}{lcccr}
\begin{array}{|c|}
\hline
\begin{array}{c}
\phantom{\bullet} \phantom{\bullet}\phantom{\bullet} \\ \bullet\phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\phantom{\bullet} \\ \phantom{\bullet} & \bullet \\ \phantom{\bullet} & \phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{c}
\phantom{\bullet} \bullet \\ \phantom{\bullet\bullet\bullet}
\end{array} \\[10pt]
\hline
\end{array} & \qquad & \qquad & \qquad &
\begin{array}{c}
\bullet\bullet\bullet \\[12pt] \bullet\bullet\bullet \\[18pt] \bullet\bullet\bullet \\[10pt]
\bullet\bullet\bullet
\end{array}
\end{array}
$$
And again
$$
\begin{array}{lcccr}
\begin{array}{|c|}
\hline
\begin{array}{c}
\phantom{\bullet\bullet\bullet} \\ \phantom{\bullet\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{cc}
\phantom{\bullet} \\ \phantom{\bullet} & \phantom{\bullet} \\ \phantom{\bullet} & \phantom{\bullet}
\end{array} \\[10pt]
\hline
\begin{array}{c}
\phantom{\bullet\bullet} \\ \phantom{\bullet\bullet\bullet}
\end{array} \\[10pt]
\hline
\end{array} & \qquad & \qquad & \qquad &
\begin{array}{c}
\bullet\bullet\bullet \\[12pt] \bullet\bullet\bullet \\[18pt] \bullet\bullet\bullet \\[10pt]
\bullet\bullet\bullet \\[30pt] \bullet\bullet\bullet
\end{array}
\end{array}
$$
Three fives have become five threes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solution verification: $\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$
Find (without L'Hospital):$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos
x)}}$$
My attempt:
$$\lim_{x\to 0}\cosh x=1$$
$$\lim_{x\to 0}e^{2\sin x}\cosh x-1=\lim_{x\to 0}\frac{e^{2\sin x}-1}{2\sin x}\cdot2\sin x=2\lim_{x\to 0}\sin x$$
$$\lim_{x\to 0}(\cosh x-\cos x)=\lim_{x\to 0}(1-\cos x)=\lim_{x\to 0}\frac{1-\cos x}{x^2}\cdot x^2=\frac{1}{2}\lim_{x\to 0}x^2$$
$$\lim_{x\to 0}\sqrt[3]{x(\cosh x-\cos x)}=\lim_{x\to 0}\sqrt[3]{x\cdot\frac{1}{2}x^2}=\frac{1}{\sqrt[3]{2}}\lim_{x\to 0}x$$
$$2\sqrt[3]{2}\lim_{x\to 0}\frac{\sin x}{x}=2\sqrt[3]{2}$$
Is this legitimate?
| The key is to use laws of algebra of limits whose primary function is to help evaluate the limit of a complicated function which is composed of many simpler expressions connected by algebraic operations $+, -, \times, /$ given the limits of these simpler expressions. I have described these laws in a simple manner in this answer.
You should have in memory a table of well known standard limits which allow us to deal with the simpler expressions mentioned in last paragraph.
In this question we need the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1=\lim_{x\to 0}\frac{\sin x} {x} $$ and $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}=\lim_{x\to 0}\frac{\cosh x-1}{x^2}$$ First we deal with denominator. We can rewrite it as $$x\sqrt[3]{\frac{\cosh x-1}{x^2}+\frac{1-\cos x} {x^2}}$$ The expression under radical sign tends to $(1/2)+(1/2)=1$ and hence the denominator can be safely replaced by $x$.
Next we deal with numerator. It can be rewritten as $$e^{2\sin x} \cosh x-\cosh x +\cosh x - 1$$ or $$\cosh x (e^{2\sin x} - 1)+\cosh x - 1$$ Thus the desired limit is equal to the limit of $$\cosh x\cdot\frac{e^{2\sin x} - 1}{2\sin x} \cdot 2\cdot\frac{\sin x} {x} +x\cdot\frac{\cosh x - 1}{x^2}$$ which is $$1\cdot 1\cdot 2\cdot 1 +0\cdot\frac{1}{2}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| $30^\circ = \tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x}$
Take the tan of both sides
$\tan 30^\circ = \tan (\tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x})$
Angle addition - subtraction rule for tangent
$\tan(A+B) = \frac {\tan A+ \tan B}{1-\tan A\tan B}\\
\tan(A-B) = \frac {\tan A- \tan B}{1+\tan A\tan B}$
$\frac 1{\sqrt 3} = \frac {\tan (\tan^{-1} \frac {11}{x}) - \tan (\tan ^{-1}\frac {1}{x})}{1 + \tan (\tan^{-1} \frac {11}{x})\tan(\tan^{-1} \frac {1}{x})} $
$\tan (\tan^{-1} y) = y$
$\frac 1{\sqrt 3} = \frac {\frac {11}{x} - \frac{1}{x}}{1 + \frac {11}{x^2}} $
Multiply numerator and denominator by $x^2$ to kill the fractions.
$\frac 1{\sqrt 3} = \frac {10x}{x^2 + 11} $
$x^2 + 11 = 10\sqrt 3 x\\
x^2 - 10\sqrt3 x + 11 = 0$
Use the quadratic formula:
$x = 5\sqrt 3 \pm \sqrt {75 - 11}\\
x = 5\sqrt 3 \pm 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$ for the following given data
Let $A, B, C$ be real numbers such that
(i) $(\sin A, \cos B)$ lies on a unit circle centered at origin.
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$.
My multiple attempts are as follows:-
Attempt $1$:
$$\sin^2A+\cos^2B=1$$
$$\tan^2C+\sin^2A-2\sin A\tan C+\cot^2C+\cos^2 B-2\cot C\cos B$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin C}{\cos C}+\dfrac{\cos C\cos B}{\sin C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos B}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin^2C(\sin A-\cos B)+\cos B}{\sin C\cos C}\right)\tag{1}$$
Now from here how to proceed further.
Attempt $2$:
$$\sin^2A+\cos^2B=1$$
$$\sin^2A=\sin^2B$$
$$A=n\pi\pm B$$
Considering only the principal range, $A=B$, $A=-B$, $A=n\pi-B$, $A=n\pi+B$
Case $1$: $A=B,A=-B$
Put $B=A$ or $B=-A$ in equation $(1)$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos A}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A+\alpha)}{\sin C\cos C}$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot \sin(A+\alpha)$$
So minimum value will be $3-2\sqrt{2}$
Case $1$: $A=n\pi-B,A=n\pi+B$
Put $B=n\pi-A$ or $B=A-n\pi$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C-\cos^2 C\cos A}{\sin C\cos C}\right)$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A-\alpha)}{\sin C\cos C}$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot\sin(A-\alpha)$$
So minimum value will be $3-2\sqrt{2}$
Any other way to solve this question?
| If $x=\tan C$, and we have $\cos^2 A=\cos ^2B$, we need to minimize $$(x-\sin A)^2+\left(\dfrac{1}{x}-\cos A\right)^2=x^2+\dfrac{1}{x^2}+1-2\left(x\sin A+\dfrac{\cos A}{x}\right)$$
So, we need to maximize $\left(x\sin A+\dfrac{\cos A}{x}\right)$.
It is known that maximum of $\alpha \sin \theta+\beta \cos \theta$ is $\sqrt{\alpha^2+\beta^2}$ , so it's maximum is $\sqrt{x^2+\dfrac{1}{x^2}}$, so our expression becomes $$x^2+\dfrac{1}{x^2}-2\sqrt{x^2+\dfrac{1}{x^2}}+1=\left(\sqrt{x^2+\dfrac{1}{x^2}}-1\right)^2$$
And clearly by A.M-G.M inequality, $x^2+\dfrac{1}{x^2}\geq 2$, so the minimum is $(\sqrt{2}-1)^2$.
Equality occurs at $x=1=\tan C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.
| $$
\frac{a+b}{\csc x}=\frac{a-b}{\cot x}=k \qquad\implies\qquad
\begin{align}
a+b&=k\csc x \\
a-b&=k\cot x
\end{align}$$
Therefore,
$$\begin{align}
a^2 - b^2 &= ( a+ b )( a - b ) &&= k^2 \cdot \csc x \cot x \\
4 a b &= ( a + b )^2 - ( a - b )^2 = k^2 (\csc^2x-\cot^2x)&&=k^2\cdot 1
\end{align}$$
The result follows. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
$a_{n+1} = \sqrt{2 a_{n}}$, for any choice of $a_{0} > 0$. This sequence is monotone and bounded. And what its limit? Hello guys any help here?
I have this sequence: $a_{n+1} = \sqrt{2 a_{n}}$. This sequence is monotone and bounded for any choice of $a_{0} > 0$.
And what its limit?
| From the recurrence formula, we have:
$$a_1 = 2^{\frac{1}{2}} \cdot a_0^{\frac{1}{2}}$$
$$a_2 = 2^{\frac{1}{2}}\cdot a_1^{\frac{1}{2}} = 2^{\frac{1}{2}} \cdot 2^{\frac{1}{4}} \cdot a_0^{\frac{1}{4}} = 2^{\frac{1}{2}+\frac{1}{4}}\cdot a_0^{\frac{1}{4}}$$
$$a_3 = 2^{\frac{1}{2}} \cdot a_2^{\frac{1}{2}} = 2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}+\frac{1}{8}}\cdot a_0^{\frac{1}{8}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}} \cdot a_{0}^{\frac{1}{8}}$$
So in general
$$a_n = 2^{\frac{1}{2}+\frac{1}{4}+\ldots + \frac{1}{2^n}}\cdot a_0^{\frac{1}{2^n}}$$
Since
$$\lim_{n\to \infty} \sum_{k=1}^{n}\frac{1}{2^k} = \lim_{n\to \infty} \left(1-\frac{1}{2^{n}}\right) = 1$$
and
$$\lim_{n \to \infty} a_0^{\frac{1}{2^n}} = a_0^0 = 1$$
we can deduce that
$$\lim_{n\to \infty} a_n = 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Calculate matrix by using Cayley-Hamilton theorem Calculate matrix $B = A^{10}-3A^9-A^2+4A$ using Cayley-Hamilton theorem on $A$.
$$A = \begin{pmatrix}
2 & 2 & 2 & 5 \\
-1 & -1 & -1 & -5 \\
-2 & -2 & -1 & 0 \\
1 & 1 & 3 & 3
\end{pmatrix}$$
Now, I've calculated the characteristic polynomial of $A$:
$P_A(\lambda) = \lambda^4-3\lambda^3+\lambda^2-3\lambda$
So I know that $P(A) = 0 \rightarrow A^4-3A^3+A^2-3A = 0$, hereby $0$ is a $4 \times 4$ matrix.
$B = A^{10}-3A^9-A^2+4A = A^4 \cdot A^6 - 3A^3 \cdot A^6 + A^2 \cdot (-1) -3A + 7A $
Can I go further from here?
I tried doing polynomial division $B/P(A)$, but i stopped halfway since the numbers were getting too big and it didn't seem to get me to the right solution.
How do I transform $B$, so that I can use $P(A)=0$ and calculate B.
| Hint: Since $A^4-3A^3=-(A^2-3A)$
$A^{10}-3A^9=A^6(A^4-3A^3)=-A^6(A^2-3A)=-A^4(A^4-3A^3)=A^4(A^2-3A)=A^2(A^4-3A^2)=-A^2(A^2-3A)=-(A^4-3A^2)=A^2-3A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $I_n=\int_0 ^1{\frac{x^{n+1}}{x+3}}dx$, prove $\lim_{n \to \infty} n I_n=\frac{1}{4}$
Given $I_n=\displaystyle\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$ for $n\in\mathbb N$, prove that $$\lim_{n \to \infty} nI_n=\frac{1}{4}$$
Here is what the limit looks like:
$$\lim_{n \to \infty} n\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$$
I am not sure how to go about this exercise, but I tried solving the integral with no success. I integrated by parts once choosing $u = x^{n+1}$ and $v = \frac{1}{x+3}$ and this is what I got.
$$I_n=\ln 4-(n+1)\int_0^1{x^n\ln(x+3)}\,dx$$
I don't know how to proceed. I was asked previously to prove that $I_{n+1}+3I_n=\frac{1}{n+2}$, do you think this can be used somehow? I tried solving for $I_n$ but I don't see how it can help because of the $I_{n+1}$ that is left over? Help me out, please!
| By writing $n I_n = \frac{n}{n+2} \cdot (n+2)I_n$ and noting that $\frac{n}{n+2} \to 1$, it suffices to show that $(n+2)I_n$ converges to $\frac{1}{4}$. However, by integration by parts,
\begin{align*}
(n+2) I_n
&= \int_{0}^{1} (n+2)x^{n+1} \cdot \frac{1}{x+3} \, \mathrm{d}x \\
&= \underbrace{\left[ x^{n+2} \cdot \frac{1}{x+3} \right]_{0}^{1}}_{=\frac{1}{4}} + \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x.
\end{align*}
Moreover, from the inequality $0\leq \frac{1}{(x+3)^2} \leq \frac{1}{9}$ which holds true for $x \in [0, 1]$, we get
$$ 0 \leq \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x \leq \int_{0}^{1} \frac{x^{n+2}}{9} \, \mathrm{d}x = \frac{1}{9(n+3)}. $$
Combining altogether, we get
$$ \frac{1}{4} \leq (n+2)I_n \leq \frac{1}{4} + \frac{1}{9(n+3)} $$
and therefore $(n+2) I_n \to \frac{1}{4}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Minimal polynomial of $\sqrt{3+2\sqrt{3}}$ Let $a=\sqrt{3+2\sqrt{3}}$. Then
\begin{align*}
&a=\sqrt{3+2\sqrt{3}}\\
&\implies a^2=3+2\sqrt{3}\\
&\implies a^2-3 = 2\sqrt{3}\\
&\implies (a^2-3)^2 = 4\cdot 3=12\\
&\implies (a^2-3)^2-12=0.
\end{align*}
So, $a$ a root of
$$(x^2-3)^2-12=x^4-6x^2-3.$$
By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.
However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.
What is wrong with my attempt?
| I think you are right.
Another way:
$$\sqrt{3+2\sqrt3}=\sqrt[4]3\sqrt{2+\sqrt3}=\frac{\sqrt[4]3}{\sqrt2}\sqrt{4+2\sqrt3}=$$ $$=\frac{\sqrt[4]3}{\sqrt2}\sqrt{(1+\sqrt3)^2}=\frac{\sqrt[4]3}{\sqrt2}(1+\sqrt3),$$ which gives the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the volume given by $(x^2+y^2+z^2)^2 = x^2+y^2$ I am working on the following exercise:
Calculate the volume of the body bounded by the following surface:
$$(x^2+y^2+z^2)^2 = x^2+y^2$$
I would solve this with a triple integral, but I do not see how I can derive the boundaries from $(x^2+y^2+z^2)^2 = x^2+y^2$. Could you help me?
|
Graphed above is the cross section of the volume in a vertical plane using the cylindrical coordinate $r^2 = x^2+y^2$. Note that
$$(r^2+z^2)^2 = r^2\implies z^2=r-r^2$$
And, $r$ has the lower and upper limits $r=0$ and $r = 1$, respectively, when $z=0$.
Thus, express $z$ in terms of $r$ and integrate the volume as follows,
$$z_1 = -\sqrt{r-r^2},\>\>\>\>\>\>\>z_2 = \sqrt{r-r^2}$$
$$V = \int_0^{2\pi}\int_0^1 (z_2-z_1) rdrd\theta= 4\pi\int_0^1 \sqrt{r-r^2} rdr=\frac{\pi^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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