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Find a value b for the average value of $f(x)=-2x^2+3x+4$ over [0, b] Find a real number b, if it exists, for which the average value of $f(x)=-2x^2+3x+4$ over [0, b] is equal to 5. In other words, on the interval [0,b], there is an average value = 5 for the function $f(x)=-2x^2+3x+4$. I'm mainly checking my answer here. After using the average value theorem to find the integral (this is a rather trivial integral), I get the equation $5=\frac{1}{b}(\frac{-2}{3}x^3+\frac{3}{2}x^2+4x)|^{b}_{a}$. Now how do I show whether $b$ exists or not?
|
Since your interval is of the form $[0,b]$, this means that if the average value of $f(x)$ over said interval is $5$, then you get the equation
$$
5 = \frac{1}{b-0}\int_{0}^{b} -2x^2 +3x+4 \ dx
$$
Which, as you pointed out, gives you
\begin{align*}
5 = \frac{1}{b} \left(-\frac{2}{3}x^3 + \frac{3}{2} x^2 + 4x \right)\Bigg\vert_{0}^{b} = -\frac{2}{3}b^2 + \frac{3}{2} b + 4
\end{align*}
And rewriting the above equation we find that
$$
-\frac{2}{3}b^2 + \frac{3}{2} b -1 = 0 \implies 4\color{blue}{b^2}-9\color{blue}{b}+6 =0
$$
which is a quadratic equation in terms of $b$. We can use the quadratic formula to solve for $b$ by doing
$$
b = \frac{9 \pm \sqrt{(-9)^2 -4(4)(6)}}{2(4)}
$$
but since
$$
\sqrt{(-9)^2 -4(4)(6)} = \sqrt{-15}
$$
we find that the solutions for $b$ are complex since they involve the square root of a negative number! This means that there doesn't exist a real number $b$ such that the average value of $f(x) = -2x^2 +3x+4$ is $5$.
|
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|
The Infinitely Nested Radicals Problem and Ramanujan's wondrous formula In mathematics, a nested-radical is any expression where a radical (or root sign) is nested inside another radical, e.g. $\sqrt{2 + \sqrt{3}}$.
By extension, an infinitely nested radical (aka, a continued root) is an expression where infinitely many radical expressions are nested within each other. A famous example would be:
$\sqrt{x\sqrt{x\sqrt{x \dots}}} = \sqrt{x x} = x, \forall x \in \mathbb{R}$.
Another common (and easy to solve) variety occurs when the pattern of numbers within the radicals follows a regular repeating pattern.
A simple example of this type is $y = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}} = \sqrt{2 + y} = 2$. The key is simply to notice that the expression after one repeated cycle is the same as the original and solve the resulting polynomial equation. (A couple of examples here.
However, while the above examples were simple, nested roots can actually be quite important and satisfying beyond the surface level.
a) The Golden Ratio :
The Golden Ratio, $\phi = \frac{1 + \sqrt{5}}{2}$ is well-known for its regularity and recurring appearance is diverse areas of nature, art and mathematics. The famous continued fraction expression $\phi = 1 + \cfrac{1}{1+ \cfrac{1}{1 + \cdots}}$ is however, also equivalent to the folowing continued square root $\phi = \sqrt{1 + \sqrt{1+ \sqrt{1 + \dots}}}$
b) Viète's formula: An approximation for $\pi$ that has been called the starting point of mathematical analysis and even "the dawn of modern mathematics", the formula is:
$$\frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \cdot \dots$$ which already contains nested radicals.
Interestingly, this can also be expressed as:
$$\pi = \lim_{k\to\infty} 2^k\cdot\sqrt{2-\sqrt{2+\sqrt{2+... (\text{k - 1 times})}}}$$
which not only contains an infinitely nested radical, but it also contains the same elementary example $\sqrt{2+\sqrt{2 + \dots}}$ from above in slightly modified form!
(c) All real numbers in the range $[0,2]$ can be expressed as the infinitely nested radical $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$
Examples: $2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$, $1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$, $\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$ (The Golden Ratio again!)
(d) Infinitely nested equations are related to polynomial equations and can be sued to find their roots
There can be many more examples of such problems.
However, the difficult problem here is to solve non-repeating continued-roots such as $\sqrt{1 + \sqrt{2 + \sqrt{3 + \dots}}}$. From a cursory look, one might even expect that expression might be divergent! However, it is actually the Nested Radical constant $1.757932 \dots$ (OEIS A072449), for which no closed-form expression is known.
Thus, one can have an arbitrary continued-root, say $\sqrt{a_1 + b_1 \sqrt{a_2 + b_2 \sqrt{a_3 + b_3 \sqrt{a_4 + \dots}}}}$, where the constants might be determined by some complicated set of rules of formulae. And even for the simplest non-repeating cases, one's ability to solve will be entirely dependent on the ability for algebraic manipulation and knowledge of diverse theorems which might sometimes help.
Or so I believed ...
Enter Srinivasa Ramanujan...
In 1911, around 2 years before he first contacted Hardy, Ramanujan published the following problem in the Journal of the Indian Mathematical Society:
$$\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \dots}}}$$
After waiting 6 months without any solutions, he supplied the following formula from page 105 of his first notebook:
$$x + n + a = \sqrt{ax + (n+a)^2 + x \sqrt{a(x+n) + (n+a)^2 + (x+n) \sqrt{\dots}}}$$
And setting $x=2$, $n=1$ and $a=0$, we get the solution $$\sqrt{1^2 + (2) \sqrt{1^2+ (2+1)\sqrt{1^2 + \dots}}} = 2+1+0 = 3$$.
(Pg 86-87, The Man Who Knew Infinity: A Life of the Genius Ramanujan, Robert Kanigel, 5th ed. 1991)
Right off the bat, we can notice some interesting things here:
*
*While we cannot use this formula for the Nested Radical Constant above, it does allow us to find the values of infinitely many continued-roots. In particular we can immediately solve any fraction where the terms are in an Arithmetic Progression $\sqrt{a + b \sqrt{(a+d) + (a+d) \sqrt{(a+2d) + (a+2d) \sqrt{\dots}}}} = a+d + 1$.
*We can generalize this further:
$\sqrt{c.a + b \sqrt{(a+d) + c.(a+d) \sqrt{c.(a+2d) + (a+2d) \sqrt{\dots}}}} = a+d + c$.
*For any given natural number $n$, we can use this formula to find a number of unique nested fraction expressions equal to $q(n)$, the number of partitions of that number.
And so on...
My question:
*
*How can we prove this formula? While Ramanujan himself usually derived results with great insight via intuition, I expect that the formula itself should be provable with usual mathematical techniques, especially with the amount of influence Ramanujan had on subsequent mathematics, and the high volume of work on continued fractions which are related to nested-radicals. So, can anyone find any proof for this - either on their own or in existing literature?
*If we cannot prove this, is it possible to find some motivation that makes it more intuitively clear. This should be possible since that was the way it was originally obtained.
*Any other interesting observations about this formula? Is it related to any other mathematical results? Perhaps it is related to some series, constants or continued fractions?
|
Method 1:- (More General)
As $$(x+a)^2=x^2+a^2+2ax$$
Replacing $a$ by $n+a$ in above equation where $x,n,a\in R$
$$(x+(n+a))^2=x^2+(a+n)^2+2x(a+n)$$
$\ \implies(x+(n+a))^2=x^2+(a+n)^2+2ax+2nx$
$\ \implies(x+(n+a))^2=ax+(a+n)^2+x^2+2nx+ax$
$\ \implies(x+(n+a))^2=ax+(a+n)^2+x(x+2n+a)$
Taking positive square roots on both sides of above equation
$(x+n+a)=\sqrt{ax+(a+n)^2+x(x+2n+a)}$
Now replacing $x$ by $x+n$
$x+2n+a=\sqrt{a(x+n)+(a+n)^2+(x+n)(x+3n+a)}$
$x+3n+a=\sqrt{a(x+2n)+(a+n)^2+(x+2n)(x+4n+a)}$
$x+4n+a=\sqrt{a(x+3n)+(a+n)^2+(x+3n)(x+5n+a)}$
Inducing these results to $k\in N$ we have,
$x+kn+a=\sqrt{a(x+(k-1)n)+(a+n)^2+(x+(k-1)n)(x+(k+1)n+a)}$
Putting the value for $x+2n+a$ in square root of $x+n+a$
$$(x+n+a)=\sqrt{ax+(a+n)^2+x \sqrt{a(x+n)+(a+n)^2+(x+n)(x+3n+a)}}$$
Now Doing this Process continuously we have our orignal infinite nested radical for $x+n+a$.
On setting $x=2,a=0,n=1$ we get
$$\sqrt{1^2 + (2) \sqrt{1^2+ (2+1)\sqrt{1^2 + \dots}}} = 2+1+0 = 3$$
Method 2:-(slightly less general)
Let $f(n)=\sqrt {1+(n+1)\sqrt{1+(n+2)\sqrt{1+...}}}$
Then $f(n)=\sqrt{1+(n+1)f(n+1)}$
$\implies f(n+1)=\frac{(f(n))^2-1}{n+1}$
Asymptotically we have $(f(n))^2\sim(n+1)f(n+1)$
If we assume $f(n)=an+b$(By equating degrees on both sides) and put in equation for $f(n+1)$
we get,
$n^2(a^2-a)+n(2ab-2a-b)+(b^2-1-a-b)=0$
This gives $a=1$ and $b=2$
so $f(n)=n+2$
For $n=1$ we have
$$\sqrt {1+2\sqrt{1+3\sqrt{1+...}}}=3$$
|
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Finding the Tree for Prufer Code This is more of a fun question that I came up with. How would you go about building the tree for the prufer code:
$P(t) = 122333444455555666666777777788888888999999999(10)(10)(10)(10)(10)(10)(10)(10)(10)(10)$
|
I would take the easy way out and use Wolfram Alpha:
ResourceFunction["PruferCodeToLabeledTree"][{1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10}]
Resulting in
|
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|
In simplifying $\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}}$ to $\frac{x^2 +x +3}{|1-2q|}$, why use the absolute value? I have seen in a question
$$\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}}$$ was given to be
$$\frac{x^2 +x +3}{|1-2q|}$$
Why was absolute value given to $1-2q$?
|
Well, by definition you have that:
$$
\sqrt{\frac{(x^2 +x +3)^2}{(1-2q)^2}} = \frac{\sqrt{(x^2 +x +3)^2}}{\sqrt{(1-2q)^2}}
$$
and you got that $\forall x \in \mathbb{R}\ \sqrt{x^2} = |x|$
so you get that:
$$
\frac{\sqrt{(x^2 +x +3)^2}}{\sqrt{(1-2q)^2}} = \frac{|(x^2 +x +3)|}{|(1-2q)|}
$$
Clearly $x^2 \geq 0 \ \forall x \in \mathbb{R}$, and in particular you have that $x^2 \geq x $ while $|x| > 1$, if $x \in [-1,1]$ you can easily see that $ x^2 + x < 3$ so you have that $x^2+x+3$ is always greater than 0 so you can get rid of $|.|$ (as well when you consider $x^2-x+3$)
Now if you have $q > 0$ you can see that $1-2q$ is not necessarily positive, so you need to use your absolute value.
If you still have doubts I suggest you to graph the functions and watch that both $x^2-x+3$ and $x^2+x+3$ are positive in all its codomain! :)
|
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If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$
But the actual answer is $2.$
Am I missing something ?
|
The derivative, $$\frac{2}{3}-\frac{3}{2 (x+1)^2}$$ is zero when $x=1/2$:
|
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How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$? I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is there a way to rigorously show that the limit does not exist?
Note. $[[x]]$ is the greatest integer less than or equal to $x$ for all $x \in \mathbf{R}$.
Find the limits or state that they do not exist.
Problem 1.1.1 (b) $\lim_{x \to 0}x\cdot[[\frac{1}{x}]]$
Proof.
Consider the sequences $a_n:=\frac{1}{n}$ and $b_n=-\frac{1}{n}$. Both these sequences converge to zero.
The corresponding image sequences are,
\begin{align*}
(f(a_n)) &= \frac{1}{1}\cdot 1, \frac{1}{2} \cdot 2, \frac{1}{3}\cdot 3, \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
\begin{align*}
(f(b_n)) &= -\frac{1}{1}\cdot (-1), -\frac{1}{2} \cdot (-2), -\frac{1}{3}\cdot (-3), \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}
Also, consider the sequence $c_n:=\frac{1}{\sqrt{n}}$. The sequence $(c_n)$ also converges to zero.
\begin{align*}
(f(c_n)) &= \frac{1}{\sqrt{1}}\cdot [[\sqrt{1}]], \frac{1}{\sqrt{2}} \cdot [[\sqrt{2}]], \frac{1}{\sqrt{3}}\cdot [[\sqrt{3}]], \frac{1}{\sqrt{4}} \cdot [[\sqrt{4}]], \ldots \\
&= 1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},1,\frac{2}{\sqrt{5}},\frac{2}{\sqrt{6}},\frac{2}{\sqrt{7}},\frac{2}{\sqrt{8}},1,\ldots
\end{align*}
Thus, the image sequence oscillates between $0$ and $1$ and is not convergent. This violates the definition of functional limits. We require that, for all sequences $(x_n)$ in the domain of the function $f$, such that $(x_n) \to a$, $x_n \ne a$ for all $n \in \mathbf{N}$, the image sequence $f(x_n)$ converges to $L$. Then, $\lim_{x \to a}f(x) = L$. So, the above limit does not exist.
|
I'll make a try to show that $\lim\limits_{x\to0}x[[\frac{1}{x}]]=1$. For $x\not=0$ it is $[[\frac{1}{x}]]\leq \frac{1}{x}<[[\frac{1}{x}]]+1$ hence $\frac{1}{x}-1<[[\frac{1}{x}]]\leq \frac{1}{x}$ which means $1-x<x[[\frac{1}{x}]]\leq 1$ if $x>0$ and $1-x>x[[\frac{1}{x}]]\ge 1$ if $x<0$. So $\lim\limits_{x\to0}x[[\frac{1}{x}]]=\lim\limits_{x\to0^+}x[[\frac{1}{x}]]=\lim\limits_{x\to0^-}x[[\frac{1}{x}]]=1$
|
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Prove area of Koch snowflake by mathematical induction Here is an interesting construction of a geometric object known as the Koch snowflake.
Define a sequence of polygons $S_0$, $S_1$ recursively, starting with $S_0$ equal to an equilateral triangle with unit sides. We construct $S_{n+1}$ by removing the middle third of each edge of $S_n$ and replacing it with two line segments of the same length.
Let $a_n$ be the area of $S_n$. Observe that $a_0$ is just the area of the unit equilateral triangle which by elementary geometry is $\frac{\sqrt{3}}{4}$
Prove by induction that for all $n \geq 0$, the area of the nth snowflake is given by:
$$a_n = a_0\left(\frac{8}{5} - \frac{3}{5} \left(\frac{4}{9}\right)^n\right).$$
|
I want to post the solution, since someone might need it.
Proof. by ordinary induction. Let Induction hypothesis $P(n)$ be $$a_n = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^n\right).$$
Base case $(n=0):$ $a_0=a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^0\right) = a_0.$ holds.
Inductive step: Assume $P(n)$ holds for some $n \geq 0$. Show that $P(n) \Rightarrow P(n+1).$ We need to show $$a_{n+1} = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^{n+1}\right).$$ We can write $$a_{n+1} = a_n + e_n t_{n+1}$$ where $e_n = 3\cdot4^n$ and $t_{n+1} = \frac{a_0}{9^{n+1}}.$ Replacing $e_n$ and $t_{n+1}$ in the equation of $a_{n+1}$ gives $$a_{n+1} = a_n + 3\cdot4^n\cdot\left(\frac{a_0}{9^{n+1}}\right)=$$ $$a_n + 3\cdot\left(\frac{1}{9}\right)\cdot a_0\cdot\left(\frac{4^n}{9^{n}}\right)=$$ $$a_n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ Since P(n) holds, we can replace $a_n$ with $a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right).$ $$a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right) + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot \left(\frac{3}{5}\right)\cdot\left(\frac{4}{9}\right)^n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \left(\frac{3}{5} - \frac{1}{3}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{15}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{9} \cdot \frac{3}{5}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^{n+1} \cdot \frac{3}{5}=$$ $$a_0\cdot \left(\frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^{n+1}\right)$$ which proves $P(n+1).$
We can conclude, by induction principle, $\forall n \geq 0: P(n)$.
|
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$ \int _ {- 1} ^ {3} [| x ^ 2-6x | + \text{sgn} (x-2)]\, \text {d} x $
Calculate the integral $$\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x .$$
We know that the two functions are defined as follows,
\begin{align*}
f (x) &= | x ^ 2-6x |\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x^2-6x>0; \\
6x-x^2, & \text{ if } x^2-6x<0.
\end{cases}\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x\in(-\infty, 0)\cup (6,+\infty); \\
6x-x^2, & \text{ if } x\in(0,6).
\end{cases}.
\end{align*}
And, $$\text{sgn}(x-2)=\begin{cases}
1, & \text{ if } x>2; \\
-1, & \text{ if } x<2.
\end{cases}$$
I know that the integral gives $ 58 / $ 3. But someone can explain to me the step by step, how to find that result.
I considered doing it with the graph of the function which is the following,
Calculating, the area of the triangle on the left from $ -1 $ to $ 0 $, and then calculating the two on the right and adding it, but I don't know how to calculate the area on the right.
|
Let's first work on getting a piecewise equation for the whole thing. There are three points where the pieces can change, namely $x = 0, 2, 6$. So, we will consider each interval separately when writing down the piecewise equation. For example, for $x<0$, we need to consider the following:
$$\begin{cases}
\color{red}{x^2-6x}, & \color{red}{x\in(-\infty, 0)\cup (6,+\infty)} \\
6x-x^2, & x\in(0,6)
\end{cases} + \begin{cases}
1, & x>2; \\
\color{red}{-1}, & \color{red}{x<2}.
\end{cases} = \boxed{x^2 - 6x - 1,\quad x<0 \ }$$
In the same way we can obtain the rest of the cases:
$$\boxed{| x ^ 2-6x | + \operatorname{sgn} (x-2) = \begin{cases} x^2 - 6x -1, & x<0\\ 6x - x^2 -1, &0<x<2\\ 6x - x^2 + 1, &x>2 \end{cases}}$$
So, this tells us that the integral can be split up as
$$\boxed{\int_{-1}^{0}\bigr[ x^2 - 6x - 1\bigl]\, \text {d}x + \int_{0}^{2}\bigl[6x - x^2 -1\bigr]\, \text {d}x + \int_{2}^{3}\bigl[6x - x^2 + 1\bigr] \, \text {d}x \ }$$
|
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|
Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the example give here we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.
1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$
Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?
2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?
I appreciate your help! :)
|
For $0\le x\le1$,
*
*$\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\cdots\le x^3\left(\dfrac1{3!}+\dfrac1{4!}+\dfrac1{5!}+\cdots\right)$ and the sum inside the parenthesis is finite.
*Similarly, $\dfrac{x^2}2+\mathcal O(x^3)\le \dfrac{x^2}2+cx^3\le x^2\left(\dfrac12+c\right).$
|
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|
Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.
Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 - 4p
\\
-1 - p = - 16 - 4p
\\
3p = -15
\\
p = -5
\\
q = -1 - (-5)
\\
q = 4
$$
According to the book the answer should be $-3$ & $-4$
|
From $x^2 + px + q = a(x+1)(x-4) = x^2 -3x -4$ and from comparison of the coefficients follows $a = 1$, $p = -3$ and $q = -4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
|
In base $3$, we can say
$$N = \frac {1}{10} + \frac{2}{100} + \frac{1}{1000} + \frac{2}{10000} + \cdots
=0.121212\cdots$$
Staying in base 3, we get
\begin{array}{rcr}
N &= &0.121212\cdots \\
100N &= &12.121212 \cdots \\
-N &= &-0.121212\cdots \\
\hline
22N &= &12.000000\cdots \\
\end{array}
So $N = \dfrac{12}{22}$
In base $10$, this becomes $N = \dfrac 58$
Added 3/19/2021
You can also do this directly
\begin{align}
N &= \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}}
+ \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\
9N &= 3 + 2 + \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}}
+ \frac{2}{3^{4}} + \frac{1}{3^{5}} + … \\
9N &= 5 + N \\
N &= \frac 58
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
|
Here is an easy way to compute it :
Let's substitute $ \left\lbrace\begin{matrix}y=\frac{1}{x}\ \ \\ \mathrm{d}x=-\frac{\mathrm{d}y}{y^{2}}\end{matrix}\right. $, we get : \begin{aligned}\int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\int_{0}^{+\infty}{\frac{y^{2}\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}\\ &=\left[\frac{y\ln{y}}{2\left(1+y^{2}\right)}\right]_{0}^{+\infty}-\frac{1}{2}\int_{0}^{+\infty}{\frac{1+\ln{y}}{1+y^{2}}\,\mathrm{d}y}\\ &=-\frac{1}{2}\int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}-\frac{1}{2}\int_{0}^{+\infty}{\frac{\mathrm{d}y}{1+y^{2}}}\\ &=-\frac{1}{2}\left[\arctan{y}\right]_{0}^{+\infty}\\ \int_{0}^{+\infty}{\frac{\ln{x}}{\left(x^{2}+1\right)^{2}}\,\mathrm{d}x}&=-\frac{\pi}{4}\end{aligned}
$\textbf{Note :}$ We first integrated by parts, setting $ u':y\mapsto -\frac{y}{\left(1+y^{2}\right)^{2}} $, and $ v:y\mapsto y\ln{y} $. From the third to the fourth line, we used the fact that $ \int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}=0 $, which can be proven substituting $ y=\frac{1}{x} $.
$\textbf{Bonus :}$ Using the same method, prove that $ \left(\forall x>0 \right) $, we have : $$ \int_{\frac{1}{x}}^{x}{\frac{\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}=\frac{1}{2}\left(\frac{\pi}{2}-2\arctan{x}\right)+\frac{x\ln{x}}{1+x^{2}} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
|
Your solution is incorrect. The good solution is that $\sum_{k=1}^n \frac1k \sim \log n,$ so that
$$ 1\leq T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n} \leq \mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n$$
For any $k,$ there exists an $N,$ such that $\frac{\log n}{n^2} \leq \frac{1}{k n}$ for $n>N,$ so
$$\mathop {\lim }\limits_{n \to \infty }\left(1 + \frac{\log n}{n^2} \right)^n \leq \exp(\frac1k).$$ It follows that $T=1.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What will happen to $\theta$ when $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}$ and i >4? Some follow up questions related to Two mysterious missing angles in the sine values of acute angle list?
.After finding two missing angles, we have a list of 9 sine values for special angles between 0$^\circ$ to $90^\circ$: $\sin \theta =\frac{\sqrt {{2{\pm}\sqrt i}}}{2}, i=0,1,2,3,4 $. (A side note:This formula is easy to remember: start with
$\sin 45^\circ=$$\frac{\sqrt2}{2}$, let $\pm \sqrt i$ jump under the radical sign and follow behind 2. Thanks @some guy to remind this trick!)
Now, I have some curious thoughts:
*
*Why is the formula symmetric about $\theta = 45^\circ$?
*One would naturally want to continue this list with i beyond 4,.. in which case |$\sin \theta$| will not be bounded by [-1,1]. What does this mean? What are those angles? I expect we step into complex numbers, but do not have a clear picture. Can you please help?
\begin{align}\sin 0^\circ =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2},\sin 15^\circ =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2},\sin 22.5^\circ =\frac{\sqrt {\color{green}{2-\sqrt2}}}{2} \end{align}
\begin{align}\sin 30^\circ =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}
, \sin 45^\circ =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}
, \sin 60^\circ =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}\end{align}
\begin{align}\sin 67.5^\circ =\frac{\sqrt {\color{green}{2+\sqrt2}}}{2},\sin 75^\circ =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}, \sin 90^\circ =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}\end{align}
|
About your first question about why they are symmetric about $45^\circ$. Let's say that you have an angle that is $x^\circ$ from the $45^\circ$ angle to make an angle of $45-x^\circ$. Thus, the symmetric angle must be $45+x^\circ$. So, if you add these angles up, you get $90^\circ$, which means that they are complementary. You should know that cosine is equal to the sine of the complementary angle. So, if we take a pair of symmetric angles $45-x^\circ, 45+x^\circ$, their sines will be $sin(45-x),sin(45+x)$ or $sin(45-x),cos(45-x)$, if we use our identity. Now, we will use one final identity, and that is that $sin(\theta)^2+cos(\theta)^2=1$. So, we get that $sin(45-x)^2+cos(45-x)^2=1$. Now, we will prove that if $sin$ of one of the angles, $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$. Using our cosine identity, we know that $sin(45+x) = cos(45-x) = \frac{\sqrt{2 + \sqrt{m}}}2$. If we plug in our values into $sin(45-x)^2+cos(45-x)^2=1$, we get $(\frac{\sqrt{2 - \sqrt{m}}}2)^2+(\frac{\sqrt{2 + \sqrt{m}}}2)=1$ and if you simplify, you can see that this is true. Thus, our original assumptions that if $sin(45-x) = \frac{\sqrt{2 - \sqrt{m}}}2$, then the other angle should be $sin(45+x) = \frac{\sqrt{2 + \sqrt{m}}}2$ is true. If there is any confusion, let me know.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why do we multiply by $-1$ here? WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I've tried without multiplying by $-1$ and haven't managed to got $\frac56 π$ which is the answer (in $0<x≤2π$)
|
$\sin(x)$ is an odd function, so $\sin(x) = -\sin(-x)$. This means that
\begin{align*}
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} &\iff -\sin\left(-x+\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2}\\
&\iff\sin\left(-x+\frac{\pi}{4}\right)=-\frac{1+\sqrt3}{2\sqrt2}.
\end{align*}
In other words, both equations are equivalent. I'm not sure why WolframAlpha makes this transformation first, it's probably related to whatever algorithm it is using to solve the equation. For solving on paper it doesn't matter which approach you take.
To solve the equation in its original form, we first note that the equation is in the form $\sin(\theta) = a > 0$, which means that $\theta$ is in the first or second quadrant. Since the domain of $\arcsin(x)$ is the first and fourth quadrants, we will need to consider the solutions
$$\theta = \arcsin(a)\quad \text{and}\quad \theta = \pi - \arcsin(a).$$
With that in mind, using the first equation, we have
$$
\sin\left(x-\frac{\pi}{4}\right)=\frac{1+\sqrt3}{2\sqrt2} \implies x-\frac{\pi}{4} =\begin{cases} \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right)\\
\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right).
\end{cases}
$$
Solving for $x$ gives us
$$x= \begin{cases}\arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{2\pi}{3}\\\pi - \arcsin\left(\frac{1+\sqrt3}{2\sqrt2}\right) + \frac{\pi}{4} &= \frac{5\pi}{6}.\end{cases}$$
|
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|
Separating scaling, rotation and sheer coefficients correctly in this case This question follows-on from Correct the Fourier transform of a shear-distorted image? ($x_{new} = x + \alpha y$). I have a group of distorted $x, y$ points and have found a matrix $A$ that corrects their positions:
$$
\begin{bmatrix}
x' \\
y' \\
\end{bmatrix} =
\begin{bmatrix}
a_{xx} & a_{xy} \\
a_{yx} & a_{yy} \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
\end{bmatrix}
$$
I now want to break this up into three separate terms, one for pure scaling or magnification $M$, one for pure rotation $R$, and one for shear $S$. I think they will look like this:
$$M = \begin{bmatrix}
m_x & 0\\
0 & m_y \\
\end{bmatrix}
$$
$$R = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}
$$
$$S = \begin{bmatrix}
1 & s_{xy} \\
s_{yx} & 1 \\
\end{bmatrix}
$$
I see there are answers to
*
*Separating Out Parts of a Matrix (Translation, Rotation, Scaling)
*Is it true that any matrix can be decomposed into product of rotation,reflection,shear,scaling and projection matrices?
and I recognize that those answers may apply to my problem, but they are beyond me; I need a simpler, more applied answer, or at least assistance getting there.
Q1: I can imagine setting $A$ equal to the product, but I'm worried if the order matters. Will this order work? Would any order work equally as well?
$$A = R \ S \ M $$
Q2: I now have five parameters instead of four. Since I've separated out rotation, should I now use only one shear, i.e. I have to choose only one of $s_xy$ or $s_yx$ and set the other to zero? Or just use $s_{xy} = s_{yx} = s$?
$$S = \begin{bmatrix}
1 & s \\
s & 1 \\
\end{bmatrix}
$$
|
You have already observed the bad effect if you use $-s$ and $s$ in the off-diagonal terms of your "shear" matrix (making an antisymmetric matrix) -- the matrix doesn't just look like a rotation matrix, it actually is a combined rotation and scaling.
Changing the form of the matrix to a symmetric matrix does not help, because
$$
\begin{bmatrix} 1 & s \\ s & 1 \end{bmatrix} =
\begin{bmatrix} \cos\left(-\frac\pi4\right) & -\sin\left(-\frac\pi4\right) \\
\sin\left(-\frac\pi4\right) & \cos\left(-\frac\pi4\right) \end{bmatrix}
\begin{bmatrix} 1 - s & 0 \\ 0 & 1 + s \end{bmatrix}
\begin{bmatrix} \cos\left(\frac\pi4\right) & -\sin\left(\frac\pi4\right) \\
\sin\left(\frac\pi4\right) & \cos\left(\frac\pi4\right) \end{bmatrix}.
$$
That is, the effect of the symmetric matrix is what we get by rotating $\frac\pi4$ radians counterclockwise, dilating by a factor $1-s$ in the direction of the $x$ axis and $1+s$ in the direction of the $y$ axis, and rotating back by $\frac\pi4$ radians clockwise.
Equivalently, it is a dilation by $1-s$ in the direction of the vector
$[1,-1]^T$ and a dilation by $1+s$ in the direction of the vector $[1,1]^T.$
As a more concrete way of looking at it, just observe what the transformation does to the unit square: it distorts it into a rhombus for any value of $s.$ This is just stretching, compressing, or even reflecting it along its two diagonals; there is no shear at all.
On the other hand, if you look at the effect on the unit square of a transformation by the matrix
$$ \begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix}, $$
it is clearly a shear transformation: the vertices $(0,0)$ and $(1,0)$ are unaffected, while the vertices $(0,1)$ and $(1,1)$ are both moved $s$ units to the right.
Now let's see how we can decompose an arbitrary $2\times2$ matrix into matrices for rotation, scaling, and shear. We start with
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. $$
(Since there are only four entries, I find it simple enough to keep track of them by name rather than by subscripts, and the formulas will be less "busy" this way.)
Note what happens if we apply a rotation of $\theta$ radians to this matrix:
$$
\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}
\begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} a \cos\theta - c \sin\theta & b \cos\theta - d \sin\theta \\
c \cos\theta + a \sin\theta & d \cos\theta + b \sin\theta \end{bmatrix}.
$$
Let $\theta = -\operatorname{atan2}(c, a)$ where $\operatorname{atan2}$ is the two-parameter arc tangent function
(see this answer for a definition).
Then $\sin\theta = -\frac{c}{\sqrt{a^2+c^2}}$ and
$\cos\theta = \frac{a}{\sqrt{a^2+c^2}}$,
so $c \cos\theta + a \sin\theta = 0.$
Define a rotation matrix $Q$ (not $R$; let's reserve the name $R$ for later) by
$$
Q = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}.
$$
Then
$$
Q A =
\begin{bmatrix} a \cos\theta - c \sin\theta & b \cos\theta - d \sin\theta \\
0 & d \cos\theta + b \sin\theta \end{bmatrix}.
$$
Suppose $a \cos\theta - c \sin\theta \neq 0$
and $d \cos\theta + b \sin\theta \neq 0.$
(The alternative is that the matrix $QA$ is a projection and therefore so is the original matrix $A$.)
Define a scaling matrix $N$ (saving the name $M$ for later) by
$$
N = \begin{bmatrix} \frac{1}{a \cos\theta - c \sin\theta} & 0 \\
0 & \frac{1}{d \cos\theta + b \sin\theta} \end{bmatrix}.
$$
Then
$$
NQA =
\begin{bmatrix} 1 & \frac{b\cos\theta - d\sin\theta}{a\cos\theta - c\sin\theta}\\
0 & 1 \end{bmatrix}.
$$
This is a shear. So let $NQA = S$ and multiply both sides of the equation by
$Q^{-1}N^{-1}$; then
$$
A = Q^{-1} N^{-1} S = RMS
$$
where
\begin{align}
R &= Q^{-1} =
\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} =
\begin{bmatrix} \frac{a}{\sqrt{a^2 + c^2}} & \frac{-c}{\sqrt{a^2 + c^2}} \\
\frac{c}{\sqrt{a^2 + c^2}} & \frac{a}{\sqrt{a^2 + c^2}} \end{bmatrix}, \\
M &= N^{-1} = \begin{bmatrix} a \cos\theta - c \sin\theta & 0 \\
0 & d \cos\theta + b \sin\theta \end{bmatrix}
= \begin{bmatrix} \sqrt{a^2 + c^2} & 0 \\
0 & \frac{ad-bc}{\sqrt{a^2 + c^2}} \end{bmatrix}, \\
S &=
\begin{bmatrix} 1 &
\frac{b\cos\theta - d\sin\theta}{a\cos\theta - c\sin\theta}\\
0 & 1 \end{bmatrix} =
\begin{bmatrix} 1 & \frac{ab + cd}{a^2 + c^2} \\ 0 & 1 \end{bmatrix}.
\end{align}
This is good for every matrix $A$ except the case where $a = c = 0,$
which is a projection and therefore presumably not what you would use to "correct" the positions of points.
For completeness, however, if $a = c = 0$ and $b^2 + d^2 \neq 0$ we can write
$$
A = \begin{bmatrix} 0 & b \\ 0 & d \end{bmatrix} =
\begin{bmatrix} \frac{d}{\sqrt{b^2 + d^2}} & \frac{b}{\sqrt{b^2 + d^2}} \\
\frac{-b}{\sqrt{b^2 + d^2}} & \frac{d}{\sqrt{b^2 + d^2}} \end{bmatrix}
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{b^2 + d^2} \end{bmatrix}
\begin{bmatrix} 1 & s \\ 0 & 1 \end{bmatrix}
$$
where $s$ is whatever you like, that is, $A=RMS$ where $R$ is a rotation, $M$ is a scaling matrix (where $m_x = 0$) and $S$ is a shear.
And in the final case, $a = b = c = d,$ obviously you can just set $M$ to the zero matrix (that is, $m_x = m_y = 0$) and use any shear and rotation you like.
So it is indeed always possible to decompose $A$ into a product of the form $RMS,$
even in the cases you probably don't care about.
This suggests that it is possible to use another sequence such as $RSM,$
and indeed that particular sequence is possible.
In the $RMS$ derivation, the next step after zeroing out the lower left corner (via rotation) was to scale so that the diagonals are $1$
(which immediately gives you a shear matrix);
for $RSM,$ let's use a shear to produce a diagonal matrix.
That is, constructing $Q$ as before we have
$$
Q A =
\begin{bmatrix} a \cos\theta - c \sin\theta & b \cos\theta - d \sin\theta \\
0 & d \cos\theta + b \sin\theta \end{bmatrix}.
$$
A shear parallel to the $x$ axis will have no affect on the first column of this matrix, but the effect on the second column can be seen in
$$
\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} b \cos\theta - d \sin\theta \\
d \cos\theta + b \sin\theta \end{bmatrix} =
\begin{bmatrix} t(d\cos\theta + b\sin\theta) + b\cos\theta - d\sin\theta \\
d \cos\theta + b \sin\theta \end{bmatrix}.
$$
We need
$t(d\cos\theta + b\sin\theta) + b\cos\theta - d\sin\theta = 0,$
which we get by setting
$t = \frac{d\sin\theta - b\cos\theta}{d\cos\theta + b\sin\theta}.$
but the derivation might be a little more difficult to find (I have not tried)
and we should expect the resulting formulas to be quite different.
Then
$$
TQA =
\begin{bmatrix} a \cos\theta - c \sin\theta & 0 \\
0 & d \cos\theta + b \sin\theta \end{bmatrix}
$$
where $T$ is the shear matrix
$$
T =
\begin{bmatrix} 1 &
\frac{d\sin\theta - b\cos\theta}{d\cos\theta + b\sin\theta}\\
0 & 1 \end{bmatrix} =
\begin{bmatrix} 1 & -\frac{ab + cd}{ad - bc} \\ 0 & 1 \end{bmatrix}
$$
We see that $M = TQA$ is a scaling matrix, so we have
$A = RSM$ where
\begin{align}
R &= Q^{-1} =
\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} =
\begin{bmatrix} \frac{a}{\sqrt{a^2 + c^2}} & \frac{-c}{\sqrt{a^2 + c^2}} \\
\frac{c}{\sqrt{a^2 + c^2}} & \frac{a}{\sqrt{a^2 + c^2}} \end{bmatrix}, \\
S &= T^{-1} =
\begin{bmatrix} 1 & \frac{ab + cd}{ad - bc} \\ 0 & 1 \end{bmatrix}, \\
M &=
\begin{bmatrix} a \cos\theta - c \sin\theta & 0 \\
0 & d \cos\theta + b \sin\theta \end{bmatrix} =
\begin{bmatrix} \sqrt{a^2 + c^2} & 0 \\
0 & \frac{ad - bc}{\sqrt{a^2 + c^2}} \end{bmatrix}.
\end{align}
That is, the $RSM$ decomposition is just like the $RMS$ decomposition except for the shear factor.
|
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"url": "https://math.stackexchange.com/questions/4081235",
"timestamp": "2023-03-29T00:00:00",
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|
Applying Gauss' Divergence Theorem to given integral We are given the following two integrals:
$$\iint\limits_S D_n f\:dS $$ and
$$\iiint\limits_B \nabla \cdot (\nabla f) \: dV$$
where $S$ is the portion of the sphere $x^2+y^2+z^2=a^2$ in the first octant, $n$ is the unit normal vector to $S$ at $(x,y,z)$ and $f(x,y,z) = ln(x^2+y^2+z^2)$
For the first integral, we have:
$D_n f = \nabla f \cdot n$ and $n = \frac{1}{a}<x,y,z>$
Since $\nabla f = \: <\frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2}>$, this gives $D_n f = \nabla f \cdot n = \frac{2}{a}$.
So the integral becomes
$$\iint\limits_S D_nf \: dS = \iint\limits_S \frac{2}{a} \: dS = \frac{2}{a} \iint\limits_S dS = \frac{2}{a} \cdot A(S) = \frac{2}{a} \cdot 4\pi a^2 \cdot \frac{1}{8} = \pi a$$ ($\frac{1}{8}$ since the sphere is in the first octant)
For the second integral, we have
$\nabla \cdot (\nabla f) = \frac{2}{a^2}$ (since $x^2 + y^2 + z^2 = a^2$).
So the integral becomes
$$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \frac{2}{a^2} \iiint\limits_B dV = \frac{2}{a^2} \cdot V(B) = \frac{2}{a^2} \cdot \frac{4}{3} \pi a^3 \cdot \frac{1}{8} = \frac{1}{3} \pi a$$
However, according to Gauss' Divergence theorem, these two integrals should be equal to each other right? Since we have
$$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \iint\limits_S \nabla f \cdot dS = \iint\limits_S \nabla f \cdot n \: dS = \iint\limits_S D_n f \: dS$$
So how is it possible that I get two different answers, even though the Divergence theorem shows that the two integrals should be the same?
|
To apply divergence theorem, you must have a closed surface. So we close the surface by placing $3$ quarter disks in plane $x = 0, y = 0, z = 0$.
Please note that when you are doing volume integral, you cannot equate $\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{a^2}$. It should rather be,
$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{\rho^2}$
So the integral becomes,
$\displaystyle \int_0^{\pi/2} \int_0^{\pi/2} \int_0^a \frac{2}{\rho^2} \ \cdot\rho^2 \cdot \sin \phi \ d\rho \ d\phi \ d\theta = \pi a$.
Now to find flux through $S$, we must subtract flux through planar surfaces $x = 0, y = 0, z = 0$ but in this case, they are simply zero.
|
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|
Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
|
First, let us solve the equation $3^a = 2^b-1$, where $a, b \in \mathbb{N}$.
For $b = 0$ we have no solutions.
For $b = 1$ we obtain $a = 0$. For $b = 2$ we get $a=1$. If $b \ge 3$ then $2^b-1 \equiv -1(\text{mod } 8)$ and hence we have no other solutions, since $3^a \equiv 1,3(\text{mod } 8)$.
In conclusion, we get the solutions $(a,b) \in \{(0,1), (1,2) \}$.
Now, consider the equation $2^x + 3^y = z^2$, where $x,y,z \in \mathbb{N}$.
If $x = 0$, then $3^y = (z-1)(z+1)$ and hence $z-1 = 3^u$, $z+1 = 3^v$ for $u , v \in \mathbb{N}, u<v$. It follows that $2 = 3^v - 3^u$. Therefore, $u = 0$ and $v = 1$. In this case, we obtain the solution: $(x,y,z)= (0,1,2)$.
If $x=1$, then $2+3^y = z^2$. For $y = 0$ we have no solutions and if $y \ge1$, we have $z^2 \equiv 2(\text{mod } 3)$, which is impossible.
Assume $x \ge 2$. Then $y$ is even. Indeed, for $y$ odd we have $z^2 = 2^x+3^y \equiv 3(\text{mod } 4)$, which is false. So, $y = 2k$, for some $k \in \mathbb{N}$. It follows that $2^x = (z-3^k)(z+3^k)$. Hence $z-3^k = 2^u$ and $z+3^k = 2^v$, for $u , v \in \mathbb{N}, u<v$. Thus,
$$2\cdot 3^k = 2^v - 2^u \implies 3^k = 2^{v-1} - 2^{u-1}$$
It follows that $u-1 = 0$, i.e. $u=1$ and
$$3^k = 2^{v-1}-1$$
We have seen that the only solutions of this equation are $(k, v) \in \{ (0,2), (1,3) \}$.
Consequently, we obtain the other two solutions: $(x,y,z) = (3,0,3)$ and $(x,y,z) = (4,2,5)$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4082057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving $(\sqrt{2})^x+(\sqrt{2})^{x-1}=2(2\sqrt{2}+1)$ I'm in stuck with this simple equation.
$$(\sqrt{2})^x+(\sqrt{2})^{x-1}=2(2\sqrt{2}+1)$$
This is my solution:
$$\begin{align}(\sqrt{2})^x+(\sqrt{2})^x(\sqrt{2})^{-1} &=4\sqrt{2}+2 \tag{1}\\[4pt]
2^{x/2}+2^{(x-1)/2}&=2^2\cdot 2^{1/2}+2^1 \tag{2}\\[4pt]
{\frac x2}+{\frac {x-1} 2}&=2^{2+\frac 1 2}+2^1 \tag{3}\\[4pt]
{\frac {2x-1} 2}&={\frac 5 2}+1 \tag{4}\\[4pt]
{\frac {2x-1} 2}&={\frac 7 2} \tag{5}\\[4pt]
{\frac {2x-1} 2}-{\frac 7 2}&=0 \tag{6}\\[4pt]
{\frac {2x-8} 2}&=0 \tag{7}\\[4pt]
2x&=8 \tag{8}\\[4pt]
x&=4 \tag{9}
\end{align}$$
But the solution must be $x=4.33$. I can't find the error.
Is here someone who can help me? Thanks
|
Your incorrect step starts from here.
$$2^{\frac x2}+2^{\frac{x-1}{2}}≠\frac x2+\frac{x-1}{2}$$
Because, this is not a equality:
$$2^a+2^b ≠ a+b$$
You can do one of the correct solutions as follows:
$$(\sqrt 2)^{x-1}(\sqrt 2-1)=4\sqrt 2+2$$
$$(\sqrt 2)^{x-1}=\frac{4\sqrt 2+2}{\sqrt 2-1}$$
$$(\sqrt 2)^{x-1}=\frac{(4\sqrt 2+2)(\sqrt 2+1)}{(\sqrt 2-1)(\sqrt 2+1)}$$
$$(\sqrt 2)^{x-1}=(4\sqrt 2+2)(\sqrt 2+1)$$
$$(\sqrt 2)^{x-1}=6\sqrt 2+10$$
$$\frac {x-1}{2}=\log_2(6\sqrt 2+10)$$
$$\cdots \cdots \cdots$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Maximum possible value with 3 variables and fractional equations A cool problem I was trying to solve today but I got stuck on:
Find the maximum possible value of $x + y + z$ in the following system of equations:
$$\begin{align}
x^2 – (y– z)x – yz &= 0 \tag1 \\[4pt]
y^2 – \left(\frac8{z^2}– x\right)y – \frac{8x}{z^2}&= 0 \tag2\\[4pt]
z^2 – (x – y)z – xy &= 0 \tag3
\end{align}$$
I tried extending the first equation to
$$x^2 - xy + xz - yz = 0 \tag4$$
I then did the same thing for the second equation and got
$$y^2 - \frac{8y}{z^2} + xy - \frac{8x}{z^2} = 0 \tag5$$
For the 3rd equation:
$$z^2 - xz + yz - xy = 0 \tag6$$
I realized that adding the first and third equations got
$$x^2 + z^2 - 2xy = 0 \tag7$$
I also realized that the 2nd equation could be written as
$$y^2 - \frac{8}{z^2}(x + y) + xy = 0 \tag8$$
but I couldn't get much more. I was thinking that graphs could help me here but I don't have much of an idea.
It would be really helpful if someone could explain to me what I could do further to solve the problem.
|
From (1):
\begin{align*}
0 &= x^2 -(y-z)x-yz \\
&= (x+y+z)(x-y) -y(x-y) \text{,}
\end{align*}
so either
$$ x+y+z = y \qquad \text{ or } \qquad x-y = 0 \text{.} $$
We conclude either $x = -z$ or $x = y$.
From (2):
\begin{align*}
0 &= z^2 y^2 - (8-xz^2)y-8x \\
&= (x+y+z)(yz^2 - 8) - z(yz^2-8) \text{,}
\end{align*}
so either
$$ x+y+z = z \qquad \text{ or } \qquad yz^2 - 8 = 0 \text{.} $$
We conclude either $x = -y$ or $yz^2 = 8$.
From (3):
\begin{align*}
0 &= z^2 -(x-y)z-xy \\
&= (x+y+z)(z-x) - x(z-x) \text{,} \end{align*}
so either
$$ x+y+z = x \qquad \text{ or } \qquad z-x = 0 \text{.} $$
We conclude either $y = -z$ or $x = z$.
By cases:
*
*$x = -z$, $x = -y$, $y = -z$: The first two give $y = z$. With the third, this forces $y = z = 0$ and then $x = 0$, giving $x+y+z = 0$.
*$x = -z$, $x = -y$, $x = z$: The first and third require $x = z = 0$. With the second, $y = 0$, giving $x+y+z = 0$.
*$x = -z$, $yz^2 = 8$, $y = -z$: Using the third in the second, $-z^3 = 8$, which has no solutions in real numbers.
*$x = -z$, $yz^2 = 8$, $x = z$: The first and third require $x = z = 0$, making the second unsatisfiable.
*$x = y$, $x = -y$, $y = -z$: The first and second require $x = y = 0$. With the third, $z = 0$, giving $x+y+z = 0$.
*$x = y$, $x = -y$, $x = z$: Again, $x+y+z = 0$ in this case.
*$x = y$, $yz^2 = 8$, $y = -z$: Again, the second and third are unsatisfiable.
*$x = y$, $yz^2 = 8$, $x = z$: Eliminating $y$ between the first and second and $z$ between the result and third, $x^3 = 8$, so $x = 2$, so $x = y = z = 2$, giving $x+y+z = 6$.
The maximum of $x+y+z$ is $6$ occurring when $x = y = z = 2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove whether the following series is convergent or divergent Prove the convergence of the following series:
$$\sum\frac{1 + 2^n + 5^n}{3^n}$$
My idea was to write it as:
$$\frac{1}{3^n} + \frac{2^n}{3^n} + \frac{5^n}{3^n}$$
Which would mean that $\frac{1}{3^n}$ is a geometric series with a ratio of $\frac{1}{3}$, and since $\frac{1}{3} <1$,$\frac{1}{3^n}$ is convergent. However, I don't know what to do with the rest...
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Simply write $\frac{5^n}{3^n}$ as $\left(\frac53\right)^n$. Then,
$$\sum \frac{1+2^n+5^n}{3^n}>\sum \frac{5^n}{3^n}=\sum\left(\frac53\right)^n>\sum 1^n=\infty$$
So the entire sum diverges.
|
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|
How can I prove this formula $\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$? While solving this trigonometric integral :
$$\int_0^\infty \sin\left(a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x$$
I came across this formula :
$$\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$$
And I'm wondering if I can prove it, but I hadn't any idea, because it seems like it depends on the properties of the function $f$.
|
Assume that $a,b > 0$. If $\displaystyle{t = a x - \frac{b}{x}}$, then
$$t^2 + 2 a b = a^2 x^2 + \frac{b^2}{x^2} - 2 ab + 2 a b =
a^2 x^2 + \frac{b^2}{x^2}.$$
Also one has
$$dt = \left(a + \frac{b}{x^2} \right) dx$$
as $x$ varies from $0$ to $\infty$, we see that $t$ varies from $-\infty$ to $\infty$. Hence, as long as one is careful about convergence, you formally get
$$\int_{-\infty}^{\infty} f(t^2 + 2 a b) dt = \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) \left(a + \frac{b}{x^2} \right) dx$$
$$ = a \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx + \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) \frac{b \cdot dx}{x^2}.$$
In the second integral, make the substitution $x \mapsto (b/a) u^{-1}$.
This replaces $a^2 x^2$ by $b^2/u^2$ and $b^2/x^2$ to $a^2 u^2$,
it
changes the integrand from $[0,\infty)$ to $(\infty,0]$, and it changes
$$\frac{b \cdot dx}{x^2} = \frac{b \cdot d((b/a) u^{-1})}{(b/a)^2 u^{-2}} = \frac{- (b^2/a) u^{-2}}{b^2 a^{-2} u^{-2}}
= -a \cdot du,$$
and thus (with the minus sign accounting for reversing the integrand)
$$\int_{-\infty}^{\infty} f(t^2 + 2 a b) dt
=
a \int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx + a \int_{0}^{\infty} f \left(a^2 u^2 + \frac{b^2}{u^2} \right) du.$$
Since the LHS is symmetric it $t \mapsto -t$, replacing the integrand by $[0,\infty)$ divides both sides by $2$ giving
$$\frac{1}{a} \int_{0}^{\infty} f(t^2 + 2 a b) dt
=
\int_{0}^{\infty} f \left(a^2 x^2 + \frac{b^2}{x^2} \right) dx.$$
|
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|
Solution verification about complex multiplication in polar form I am trying to prove the process behind complex multiplication in polar form, which is similar to this:
The product of two complex numbers in polar form $r_{1}\,\angle\,\theta_{1}$ and $r_{2}\,\angle\,\theta_{2}$ is $r_{1}r_{2}\,\angle\,(\theta_{1} + \theta_{2})$.
The proof where the polar form is written into rectangular form which will then use sum-to-product formula is already established like this, so I want to try something new.
Can someone verify if my attempt is correct?
Define two complex numbers $z_{1} = a + bi$ and $z_{2} = c + di$. Writing it in polar form, we get $z _{1} = \sqrt{a^{2} + b^{2}}\,\angle\,\tan^{-1}\left(\frac{b}{a}\right)$ and $z_{2} = \sqrt{c^{2} + d^{2}}\,\angle\,\tan^{-1}\left(\frac{d}{c}\right)$, respectively. Also, the product of the two complex numbers $z_{1}z_{2}$ is equal to $(ac - bd) + (ad + bc)i$. Writing this in polar form, we have $z_{1}z_{2} = \sqrt{(ac - bd)^{2} + (ad + bc)^{2}}\,\angle\,\tan^{-1}\left(\frac{ad + bc}{ac - bd}\right)$. Let this product be $z_{3}$.
We then check if the radius of the product is equal to the product of the radii of the factors.
\begin{align*}r_{1}r_{2} &\overset{?}{=} r_{3} \\ \left(\sqrt{a^{2} + b^{2}}\right)\left(\sqrt{c^{2} + d^{2}}\right) &\overset{?}{=}\sqrt{(ac - bd)^{2} + (ad + bc)^{2}} \\ \sqrt{(a^{2} + b^{2})(c^{2} + d^{2})} &\overset{?}{=} \sqrt{(ac)^{2} - 2abcd + (bd)^{2} + (ad)^{2} + 2abcd + (bc)^{2}} \\ \sqrt{(ac)^{2} + (ad)^{2} + (bc)^{2} + (bd)^{2}} &\overset{?}{=} \sqrt{(ac)^{2} + (bd)^{2} + (ad)^{2} + (bc)^{2}} \\ \sqrt{(ac)^{2} + (ad)^{2} + (bc)^{2} + (bd)^{2}} &= \sqrt{(ac)^{2} + (ad)^{2} + (bc)^{2} + (bd)^{2}}\end{align*}
Next, we check if the angle of the product is the sum of the angles of the factors.
\begin{align*}\theta_{1} + \theta_{2} &\overset{?}{=} \theta_{3} \\\tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) &\overset{?}{=} \tan^{-1}\left(\frac{ad + bc}{ac - bd}\right).\end{align*}
Before proceeding, we must restrict $\theta_{i}$ on the range of the arctangent function which is $(-\frac{\pi}{2},\frac{\pi}{2})$ as the arctangent function. If, for instance, $\mathrm{Re}(z_{i}) = 0$, then the sign of $\mathrm{Im}(z_{i})$ will determine the sign of $\frac{\pi}{2}$. Also, if $\mathrm{Re}(z_{i}) < 0$ and $\mathrm{Im}(z_{i}) > 0$, we can just take the angle of a complex number $z$ where $\mathrm{Re}(z) = \mathrm{Im}(z_{i})$ and $\mathrm{Im}(z) = \mathrm{Re}(z_{i})$.
Rewrite the relation by letting $b$, $d$, $a$, and $c$ take the value of $\frac{b}{a}$, $\frac{d}{c}$, $1$, and $1$. Then, by simplifying,
\begin{align*}\tan^{-1}\left(\frac{(\frac{b}{a})}{1}\right) + \tan^{-1}\left(\frac{(\frac{d}{c})}{1}\right) &\overset{?}{=} \tan^{-1}\left(\frac{(1)(\frac{d}{c}) + (\frac{b}{a})(1)}{(1)(1) - (\frac{b}{a})(\frac{d}{c})}\right) \\ \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) &\overset{?}{=} \tan^{-1}\left(\frac{\frac{b}{a} + \frac{c}{d}}{1 - (\frac{b}{a})(\frac{d}{c})}\right)\end{align*}
Then, let $\tan u = \frac{b}{a}$ and $\tan v = \frac{d}{c}$. By substitution,
\begin{align*}\tan^{-1}\left(\tan u\right) + \tan^{-1}\left(\tan v\right) &\overset{?}{=} \tan^{-1}\left(\frac{\tan u + \tan v}{1 - (\tan u)(\tan v)}\right) \\ u + v &\overset{?}{=} \tan^{-1}\left(\frac{\tan u + \tan v}{1 - (\tan u)(\tan v)}\right) \\ \tan(u + v) &\overset{?}{=} \frac{\tan u + \tan v}{1 - (\tan u)(\tan v)}\end{align*}
The last relation is known to be true. Thus, $$\tan(u + v) = \frac{\tan u + \tan v}{1 - (\tan u)(\tan v)}$$
Since the product of the radii of the factors is equal to the radius of the product and the angle of the product is equal to the sum of the angles of the factors, therefore, $(r_{1}\,\angle\,\theta_{1})(r_{2}\,\angle\,\theta_{2}) = r_{1}r_{2}\,\angle\,(\theta_{1} + \theta_{2})$.
Edit: Tried to fix the angle problem. "Before proceeding...", in particular.
|
Your computations are correct, but you do not cover all possible cases. Problems are
*
*Your formulae based on $\tan^{-1} = \arctan$ only work if both $a \ne 0$ and $c \ne 0$. This reflects the fact $\tan x$ is not defined if $x = k \pi/2$ with odd $k \in \mathbb Z$. Thus you must separately treat a number of special cases: $a = 0, c = 0$ / $a \ne 0, c = 0$ / $a = 0, c \ne 0$.
*There are infinitely many branches of the arc tangent, let us denote them by $\arctan_k : \mathbb R \to ((2k - 1)\pi/2, (2k + 1)\pi/2)$ with $k \in \mathbb Z$. If $k$ is odd, then your formulae describe points $z$ with real part $\Re(z) > 0$, and if $k$ is even, then your formulae describe points $z$ with $\Re(z) < 0$. Usually one works with the principal value $\arctan = \arctan_0 : \mathbb R \to (-\pi/2, \pi/2)$. Anyway, it gives again many cases which have to be considered. As you do it, it only works if all three points $z_1, z_2, z_1z_2$ have real part $> 0$. And it seems that if for example $\Re(z_1) > 0 $ and $\Re(z_2) < 0$ you get really into troubles because you cannot use the same branch of $\arctan$ for both points.
*The polar form is not unique, angles are determined only up to a summand $2k\pi$. The formula $(r_{1}\,\angle\,\theta_{1})(r_{2}\,\angle\,\theta_{2}) = r_{1}r_{2}\,\angle\,(\theta_{1} + \theta_{2})$ is true for all choices of $\theta_i$. You can of course restrict to angles in $[0,2\pi)$, but the sum of two such values may by outside of this interval. This can easily be treated, but again it needs additional arguments.
|
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|
Taylor series for $e^{-x \ln x}$ What is a Taylor expansion for the following function?
$$ e^{-x \ln x} $$
I assume you can't do a Taylor expansion around $x=0$, since the function doesn't exist at that point. The next best choice seems to be $x=1$, since that's when $\ln(x)=0$. So, when I try to do an expansion around $x=1$, I get an answer that seems to have no clean closed-form (verified with wolfram alpha):
$$ 1 - (x - 1) + \frac{1}{2} (x - 1)^3 - \frac{1}{3} (x - 1)^4 + \frac{1}{12}(x - 1)^5 + \frac{1}{120}(x - 1)^6 - \dots$$
(Notice the above series has a zero coefficient for the $(x-1)^2$ term...which makes it even trickier to generate a closed-form expression for).
However, I believe this function should be easily expanded and neatly integrated term-by-term (as suggested by the solution to question 1A on this prelim exam question from UC Berkeley).
Where am I going wrong?
|
We have
\begin{align*}
\color{blue}{x^{-x}}&=1 - (x - 1) + \frac{1}{2} (x - 1)^3 - \frac{1}{3} (x - 1)^4 + \frac{1}{12}(x - 1)^5 + \frac{1}{120}(x - 1)^6 - \dots\\
&=\sum_{n=0}^\infty a_n\frac{(x-1)^n}{n!}
\end{align*}
when expanded at $x=1$,
where
\begin{align*}
\color{blue}{(a_n)_{n\geq 0}=(1, -1, 0, 3, -8, 10, 6, -42, -160,\ldots)}
\end{align*}
is stored as A176118 in OEIS.
|
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|
Does the system of two equation in two variables given below have a possible real solution? Consider the following two equations:
$$\sqrt {3x} \left(1 + \frac{1}{x+y} \right) = 2 $$
$$\sqrt{7y}\left(1 -\frac{1}{x+y}\right)= 4\sqrt{2}$$
.
It is expected to verify if the equations possess a solution and if yes what is the floor of $ y/x $ for that solution.
.
What have I tried:
.
Eliminating the $\frac {1}{x+y}$ term from the two equations leads to :
$$ \frac{2}{\sqrt{3x}} + \frac{4\sqrt{2}}{\sqrt{7y}} = 2 $$
Which gives
$$ y = \frac{24x}{7(1- \sqrt{3x})^2}$$
Also from the first equation value of y is :
$$ y = \frac{\sqrt{3x}}{2-\sqrt{3x}} - x $$
Equating the two values of y and simplifying resulted in a polynomial with fractional powers of x. This proved to be a dead end for me.
.
Note: some underlying inquisitions:
*
*Is there a method for simplifying such equations that I am unaware of?
*Are the given two equations representing hyperbola? (Checked the graph on desmos!). If yes how can one see that without referring to the plot.
|
Starting from @Moo's comment $$1323 x^4+3024 x^3+2610 x^2-7456 x+147 = 0$$ is the same as
$$\left(9 x^2+30 x+49\right) \left(147 x^2-154 x+3\right)=0$$ The first quadratic does not show real roots and for the second
$$x_\pm=\frac{11\pm4 \sqrt{7}}{21} $$
Similarly
$$3087 y^4-25284 y^3+44716 y^2-38816 y+4032 = 0$$is the same as
$$\left(63 y^2-120 y+112\right)\left(49 y^2-308 y+36\right)=0 $$
The first quadratic does not show real roots and for the second
$$y_\pm=\frac{2\left(11\pm 4 \sqrt{7}\right)}{7} =6 x_\pm$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Approximate result for $\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)$? What would be a quick way to approximately determine the value of
$$\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)=\phi\left(\frac{1}{2}\right) , $$
where $\phi(q)$ is the Euler function? By approximating, I mean determine the first few digits of the result starting from the product representation without additional knowledge of other properties of $\phi(q)$.
For instance, my non-programmable CASIO fx-991ES calculator comes with a summation function $\sum_n$, but no multiplication function $\prod_n$, so assuming that I have access to this low complexity piece of equipment, I got an approximation as
$$\begin{align}\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right) &= \exp\left(\ln\left(\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)\right)\right)\\
&\approx\exp\left(\sum_{n=1}^{100}\ln\left(1-\frac{1}{2^n}\right)\right)\\
&=0.2887880951\end{align}$$
by doing a finite sum and exponentiating the result instead. Curiously, all of the digits are correct, so maybe summing even fewer terms would be sufficient to get just a few digits right.
But what if all I have is paper and pen? Is there a way to get a good approximation in a few lines?
|
$$\prod_{n=1}^\infty\left(1-\frac{1}{2^n}\right)=\Bigg[\prod_{n=1}^p\left(1-\frac{1}{2^n}\right)\Bigg]\Bigg[\prod_{n=p+1}^\infty\left(1-\frac{1}{2^n}\right)\Bigg]$$
Fot the second product, take its logarithm
$$\log\Bigg[\prod_{n=p+1}^\infty\left(1-\frac{1}{2^n}\right)\Bigg]=-\sum_{n=p+1}^\infty \left(\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{3 x^3}+\frac{1}{4 x^4}+O\left(\frac{1}{x^5}\right)\right)$$ where $x=2^n$.
So, the approximation will write
$$\Pi_p=\frac {a_p}{b_p} \exp\left(- \frac {c_p}{d_p}\right) $$ Below are listed the various constants
$$\left(
\begin{array}{ccccc}
p & a_p & b_p & c_p & d_p \\
1 & 1 & 2 & 1229 & 2240 \\
2 & 3 & 8 & 28087 & 107520 \\
3 & 21 & 64 & 73229 & 573440 \\
4 & 315 & 1024 & 1738567 & 27525120 \\
5 & 9765 & 32768 & 4611629 & 146800640
\end{array}
\right)$$ and now the numerical values
$$\left(
\begin{array}{cc}
p & \text{result} \\
1 & \color{red}{0.288}8615143 \\
2 & \color{red}{0.2887}901234 \\
3 & \color{red}{0.288788}1550 \\
4 & \color{red}{0.28878809}69 \\
5 & \color{red}{0.2887880951 }
\end{array}
\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove $\{a_n\}$ converges.
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.
First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?
|
I'm not so sure about this one, but here's my attempt-
\begin{align}
\lVert a_{n+1}-a_n \rVert &= \left\lVert \frac{1}{a_n^2}-\frac{1}{a_{n-2}^2} \right\rVert\\
&\leq \frac{ \lVert {a_{n-2}^2}-a_n^2 \rVert }{16} \\
&= \frac{ \lVert {a_{n-2}+a_{n}} \rVert \lVert {a_{n-2}-a_{n}} \rVert }{16} \\
&\leq \frac{6}{16} \lVert {a_{n-2}-a_{n}} \rVert \\
&=\frac{6}{16} \left\lVert \frac{1}{a_{n-3}^2} + \frac{1}{a_{n-4}^2} - \frac{1}{a_{n-1}^2} - \frac{1}{a_{n-2}^2} \right\rVert \\
&=\frac{6}{16} \left\lVert \left(\frac{1}{a_{n-3}^2} - \frac{1}{a_{n-1}^2}\right) + \left(\frac{1}{a_{n-4}^2} - \frac{1}{a_{n-2}^2}\right) \right\rVert \\
&\leq \frac{6}{16} \cdot \frac{6}{16} \lVert (a_{n-1}-a_{n-3})+(a_{n-2}-a_{n-4}) \rVert\\
&\ \ \vdots \\
&\leq \left(\frac{6}{16}\right)^n max(|| a_4-a_2 ||,\lVert a_3-a_1||)2^n \\
&=\left(\frac{12}{16}\right)^n max(|| a_4-a_2 ||,\lVert a_3-a_1||)
\end{align}
(Original image: https://i.stack.imgur.com/uBBqe.jpg )
Does this seem correct? I tried using Cauchy's convergence test and the fact that the tail lies between 2 and 3. If not, I hope it provides at least some clue to the correct path. The last step is a result of the observation that each term of $a_k - a_{k-2}$ should produce two similar terms(as seen while expanding), whose index decreases linearly.
Edit: I have added the term $max(|| a_4-a_2 ||,\lVert a_3-a_1||)$ instead of the original single term to take care of the parity issue.
|
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|
irrational integral $ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$ I have to solve this irrational integral $$ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$$
It seems that the most convenient way to operate is doing the substitution
$$ x= \frac{t^2}{3-2t}$$
according to the rule,
obtaining the integral:
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Then $$\frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}= \frac{A}{t-2}+\frac{B}{t+3}+\frac{C}{3-2t}+\frac{D}{(3-2t)^2}$$
then I found the coefficient $A= \frac{12}{5},B= -\frac{12}{5}, C= \frac{15}{2},D= -\frac{9}{2}$
In my book the integral on which to operate is:
$$ \int -2\frac{(t^2-3t)(3+t-t^2)}{(t^2-7t+6)(3-2t)^2}\, dx$$
that is different from my
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Perhaps I made mistakes in the first passages and I checked lots of times my calculations. Can someone indicate where I'm making mistakes?
|
In your integral ,you can make break the numnerator in terms of denominator by first putting a -ve sign in front of the integral , then proceed .
|
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"timestamp": "2023-03-29T00:00:00",
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|
If x, y, w, z >0 and $x^4$+$y^4$+$w^4$+$z^4$ <=4 prove 1/$x^4$+1/$y^4$+1/$w^4$+1/$z^4$>=4 I would appreciate suggestions to solve:
If x, y, w, z > 0 and $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 prove the following:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
From plugging in numbers into Excel, it looks like x, y, w, z must be numbers near 1.
I tried to do a simple case:
If a, b, k > 0 and a + b < k prove the following: 1/a + 1/b > k
a + b < k so 1/(a+b) > 1/k
But 1/a + 1/b > 1/(a+b) so 1/a + 1/b > 1/k which is not what I want to prove.
It seems that I would have to impose a condition for the possible values of k.
|
Thank you very much for the hints. Here is the solution from the hint of achille hui:
We are told that $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 so ($x^4$+$y^4$ + $w^4$ + $z^4$)/4 <= 1
But AM >= HM (or HM <= AM) so:
4/(1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$) <= ($x^4$+$y^4$ + $w^4$ + $z^4$)/4 <= 1
Which means:
4/(1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$) <= 1
Re-arranging:
4 <= 1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$
In other words:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
|
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|
How can I solve $x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$? I have this right now:
$$x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$$
$$\frac{dx}{x\left(y^2+z\right)}=\frac{dy}{-y\left(x^2+z\right)}=\frac{dz}{\left(x^2-y^2\right)z}$$
I get the first first integral like this:
$$\frac{xdx}{x^2\left(y^2+z\right)}=\frac{ydy}{-y^2\left(x^2+z\right)}=\frac{-dz}{-\left(x^2-y^2\right)z}$$
$$\frac{d\left(\frac{1}{2}x^2\right)}{x^2y^2+x^2z}=\frac{d\left(\frac{1}{2}y^2\right)}{-y^2x^2-y^2z}=\frac{d\left(-z\right)}{-x^2z+y^2z}$$
$$\frac{d\left(\frac{1}{2}x^2+\frac{1}{2}y^2-z\right)}{0}=ds$$
$$d\left(\frac{1}{2}x^2+\frac{1}{2}y^2-z\right)=0$$
$$\frac{1}{2}x^2+\frac{1}{2}y^2-z=C$$
$$x^2+y^2-2z=C_1$$
$$\Psi _1\left(x,y,z\right)=x^2+y^2-2z$$
But I am not sure how to get the second first integral
I tried using $z=\frac{1}{2}x^2+\frac{1}{2}y^2-C$ when doing:
$$\frac{dx-dy}{x\left(y^2+z\right)+y\left(x^2+z\right)}=\frac{dz}{\left(x^2-y^2\right)z}$$
$$\frac{d\left(x-y\right)}{\frac{1}{2}\left(x+y\right)^3-C\left(x+y\right)}=\frac{dz}{\left(x+y\right)\left(x-y\right)z}$$
$$\frac{2d\left(x-y\right)}{\left(x+y\right)^2-2C}=\frac{dz}{\left(x-y\right)z}$$
$$\frac{2\left(x-y\right)d\left(x-y\right)}{\left(x+y\right)^2-2C}=\frac{dz}{z}$$
Let $w=\left(x-y\right)$ and $C=\frac{1}{2}x^2+\frac{1}{2}y^2-z$, then:
$$\frac{2w\:dw}{4z-w^2}=\frac{dz}{z}$$
Then we do for $v=w^2$:
$$\frac{\:dv}{4z-v}=\frac{dz}{z}$$
Which we then solve:
$$\frac{\:dv}{dz}=4-\frac{v}{z}$$
Using $p=\frac{v}{z}$ we get:
$$p+p'z=4-p$$
$$-\frac{1}{2}\frac{dp}{p-2}=\frac{dz}{z}$$
Integrating:
$$-\frac{1}{2}ln\left|p-2\right|+C_2=ln\left|z\right|$$
$$ln\left|\frac{1}{p+2}\right|+C_2=ln\left|z^2\right|$$
$$\frac{C_2}{p+2}=z^2$$
$$C_2=z^2\left(\left(x-y\right)^2+2\right)$$
Is there an easier way to calculate the second first integral? I don't see any mistake in my calculations but still it's very very long
|
Take the first ODE:
$$\frac{dx}{x\left(y^2+z\right)}=\frac{dy}{-y\left(x^2+z\right)}$$
$$y(x^2+z)dx+x(y^2+z)dy=0$$
$$\dfrac 12xy(2xdx+2ydy)+z(ydx+xdy)=0$$
$$\dfrac 12xy(dx^2+dy^2)+z(ydx+xdy)=0$$
$$\dfrac 12xyd(x^2+y^2)+zd(xy)=0$$
And eliminate the $z$ variable since you have:
$$x^2+y^2-2z=C_1$$
$$z=\dfrac {x^2+y^2-C_1}2$$
The ODE becomes separable.
$$xyd(x^2+y^2)+( {x^2+y^2-C_1})d(xy)=0$$
After integration we get:
$$\ln (x^2+y^2-C_1)+\ln (xy)=C$$
$$\ln (2z)+\ln (xy)=C$$
$$\boxed {zxy=C_2}$$
|
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|
Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
|
$$
I = \int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx = \int\frac{3x^2-1}{\sqrt{(x + \frac12)^2 - \frac54}}dx
$$
Replacing $x= y\cdot \sqrt{\frac54} - \frac12$ gives
$$
I = \int\frac{(15 y^2)/4 - (3 \sqrt 5 y)/2 - 1/4}{\sqrt{(y^2-1)}}dy
$$
Now let $y = \cosh (z)$ which gives
$$
I = \int [15 (\cosh (z))^2)/4 - (3 \sqrt 5 \cosh (z))/2 - 1/4 ] \; dz
$$
This can easily be performed, giving
$$I = \frac{13 z - 12 \sqrt 5 \sinh(z) + 15 \sinh(z) \cosh(z)}{8} + {\rm constant}$$
Replacing $z$ by $x$ (via $y$) gives
$$
I = \frac18 \Big[6 \sqrt{x^2 + x - 1} (2 x - 3) + 13 \tanh^{-1}\frac{2 x + 1}{2 \sqrt{x^2 + x - 1}} \Big] + {\rm constant}
$$
|
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|
$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$ The problem
Given that $a,b>0$ and $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$$
Find the value of $$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)$$
My attempt
We have from the given condition
$$2+\frac{\log a}{\log 2}=3+\frac{\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
$\implies$
$$\frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
By ratio and proportion we get each ratio equal to
$$\frac{2\log 2+3\log 3+\log(ab)-\log(a+b)}{0}$$
Thus we have
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
But i am unable to find $\log(ab)$
|
Set
\begin{align}\\ \frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}=y\\\to \log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108) \to 108ab=a+b\end{align}
This is where you are.
Let us find another equation to eliminate a or b.
From \begin{align}\\2\log 2\log 3+\log a\log 3=3 \log 3\log 2+\log b\log 2\end{align} \begin{align}a=2b^{\frac{\ln 2}{\ln 3}}, 108\cdot 2b^{\frac{\ln 2+\ln 3}{\ln 3}}=2b^{\frac{\ln 2}{\ln 3}}+b\end{align}
Therefore
\begin{align}b \approx 0.01, a\approx 0.11, a+b \approx 0.12, ab\approx 0.0011
\\log_{ab}(\frac{1}{a}+\frac{1}{b})=\log_{ab}{(a+b)} -1\approx -0.6883\end{align}
|
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|
Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the numerator and the denominator by $x$, then letting it go towards $-\infty$
$$\frac{\frac{\left(\sqrt{x^2+14}+x\right)}{x}}{\frac{\left(\sqrt{x^2-2}+x\right)}{x}}=\frac{\left(\sqrt{\frac{x^2}{x^2}+\frac{14}{x^2}}+\frac{x}{x}\right)}{\left(\sqrt{\frac{x^2}{x^2}-\frac{2}{x^2}}+\frac{x}{x}\right)}=\frac{\left(\sqrt{1+\frac{14}{x^2}}+1\right)}{\left(\sqrt{1-\frac{2}{x^2}}+1\right)}\:=\:\frac{\left(\sqrt{1}+1\right)}{\left(\sqrt{1}+1\right)}\:=\:\frac22 \ = \ 1$$
WolframAlpha's result: $-7$. It has a long, complicated 25 step solution.
Is the solution $1$, $-7$ or neither?
Edit: of course, I set it $x$ to go towards $-\infty$ in WolframAlpha too
|
You can divide by $x$ but not in that way. Since $x\to-\infty$ it is eventually negative i.e. $-x>0$ so you should have $$\frac{\sqrt{x^2+c}}x = \frac{\sqrt{x^2+c}}{-\sqrt{x^2}}= -\sqrt{1+\frac c{x^2}}=-1-\frac c{2x^2}+O(|x|^{-3})$$
This last inequality via the approximation $\sqrt{1+t}= 1 + \frac t2+O(t^2)$ which can be shown via Taylor’s theorem. Thus
$$\frac{\sqrt{x^2+14}+x} {\sqrt{x^2-2}+x}=\frac{-14x^{-2}/2 +O(|x|^{-3})}{2x^{-2}/2 + O(|x|^{-3})} \to -7 \qquad (x\to-\infty)$$
quick Desmos graph to verify:
|
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|
RMS value of trapeziodal fourier series I calculated RMS value of trapezoidal fourier series but the numerical results are not same with its normal formula.
$b_n = \frac{8\cdot A}{\pi \cdot u \cdot n^2}\cdot sin(\frac{n \cdot u}{ 2})$
$f_{rms} = \sqrt{ a_0^2 + \frac{a_1^2 + a_2^2 +a_3^2+..... + b_1^2 + b_2^2 + b_3^2}{2}}$
$V_{rms,f} = \sqrt{\frac{8 \cdot A }{\pi \cdot u \cdot 2}(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2}.... \frac{1}{(2n-1)^2})}$
$\sum(2n-1)^2 = \frac{\pi^2}{8}$
$V_{rms,f} = \sqrt{\frac{A \cdot \pi}{u \cdot 2}}$
if $A = 12$ and $u = pi/6$, $V_{rms,f} = 6$
Normal RMS formula
$V_{rms} = A \cdot \sqrt{1-\frac{2 \cdot u}{3 \cdot \pi}}$
if $A = 12$ and $u = pi/6$, $V_{rms,f} = 11.313$
Where did I mistake ?
|
Using Octave code similar to those in Shifted square wave Fourier series
format long;
A = 12;
u = pi / 6;
t = 0:0.0001:(2*pi);
n = (1:2:1000).';
fourier_bn = @(n) 8 .* A ./ (pi .* u .* n.^2) .* sin(n .* u ./ 2);
fourier_term = @(n) fourier_bn(n) .* sin(n .* t);
fourier = sum(cell2mat(arrayfun(fourier_term, n, 'UniformOutput', false)));
fourier_rms_discrete = sqrt(sum(fourier.^2) ./ length(t))
fourier_rms_1st = sqrt(sum(fourier_bn(n).^2) ./ 2)
We can see that both discrete approximation and the 1st formula gives us $V_{\text{rms}, f} = 11.313...$, which is the same as what you computed using the 2nd formula.
Therefore, the 2nd computation is correct, but the 1st one is wrong. It is wrong because
$$\begin{align}
f_{\text{rms}}
&= \sqrt{a_0^2 + \frac{a_1^2 + a_2^2 + a_3^2 + \dots + b_1^2 + b_2^2 + b_3^2 + \dots}{2}} \\
&= \sqrt{\frac{b_1^2 + b_2^2 + b_3^2 + \dots}{2}} \\
&= \sqrt{\frac{\left(\frac{8A}{\pi u 1^2} \sin \frac{1u}{2} \right)^2 + \left(\frac{8A}{\pi u 2^2} \sin \frac{2u}{2} \right)^2 + \left(\frac{8A}{\pi u 3^2} \sin \frac{3u}{2} \right)^2 + \dots}{2}} \\
&= \sqrt{\frac{\left(\frac{8A}{\pi u}\right)^2 \left(\left(\frac{\sin \frac{1u}{2}}{1^2}\right)^2 + \left(\frac{\sin \frac{2u}{2}}{2^2}\right)^2 + \left(\frac{\sin \frac{3u}{2}}{3^2}\right)^2 + \dots\right)}{2}} \\
&= \sqrt{\left(\frac{8A}{\sqrt{2} \pi u}\right)^2 \left(\left(\frac{\sin \frac{1u}{2}}{1^2}\right)^2 + \left(\frac{\sin \frac{2u}{2}}{2^2}\right)^2 + \left(\frac{\sin \frac{3u}{2}}{3^2}\right)^2 + \dots\right)} \\
&= \frac{8A}{\sqrt{2} \pi u} \sqrt{\frac{\sin^2 \frac{1u}{2}}{1^4} + \frac{\sin^2 \frac{2u}{2}}{2^4} + \frac{\sin^2 \frac{3u}{2}}{3^4} + \dots}
\end{align}$$
Clearly,
$$\frac{8A}{\sqrt{2} \pi u} \sqrt{\frac{\sin^2 \frac{1u}{2}}{1^4} + \frac{\sin^2 \frac{2u}{2}}{2^4} + \frac{\sin^2 \frac{3u}{2}}{3^4} + \dots} \ne \sqrt{\frac{A \pi}{u 2}}$$
because at $u = 2\pi, A = 1$ we have $\text{LHS} = 0$ but $\text{RHS} > 0$.
|
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|
Integral $\int_0^\infty \frac{x^{2k}}{x^2+1}dx$ I am little bit confused by the following integral:
$$\int_0^\infty \frac{x^{2k}}{x^2+1}dx,$$
which according to WA is equal to
$$\int_0^\infty \frac{x^{2k}}{x^2+1}dx=\frac{\pi}{2}\sec(\pi k),\quad \text{for}\ \operatorname{Re}(k)>-\frac{1}{2}.$$
However, by plugging $k=1,2,...$, to RHS, this should be equal to $\frac{\pi}{2}(-1)^k$.
On the other side, plugging $k=1,2,...$ to LHS, these integrals should not exist. Since I could not derive that result, I would like to know what is going on here? Maybe WA evaluates is wrongly? Thanks for any hint.
|
Letting $\displaystyle \frac{1}{t}=x^{m}+1$, then $
\displaystyle d x=\frac{1}{m} \left(\frac{1}{t}-1\right)^{\frac{1}{m}-1}\left(-\frac{1}{t^{2}}\right) d t.
$
Consequently
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x &=-\int_{1}^{0} \frac{\left(\frac{1}{t}-1\right)^{\frac{r}{m}}}{\frac{1}{t}} \frac{1}{m t^{2}}\left(\frac{1}{t}-1\right)^{\frac{1}{m}-1} d t \\
&=\frac{1}{m} \int_{0}^{1} \frac{(1-t)^{\frac{r}{m}}(1-t)^{\frac{1}{m}-1}}{t^{\frac{r+1}{m}}} d t \\
&=\frac{1}{m} \int_{0}^{1} t^{-\frac{r+1}{m}}(1-t)^{\frac{r+1}{m}-1} d t \\
&=\frac{1}{m} B\left(1-\frac{r+1}{m}, \frac{r+1}{m}\right)
\end{aligned}
$$
By the property of Beta function,
$$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
where $\operatorname{Re}(x)>0$ and $\operatorname{Re}(y)>0,$
we have
$$
\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\Gamma\left(1-\frac{r+1}{m}\right) \Gamma\left(\frac{r+1}{m}\right)}{m\Gamma(1)}
$$
Using the Euler’s Reflection Theorem,
$$
\Gamma(1-z) \Gamma(z)=\pi \csc (\pi z),
$$
where $z\notin Z$,
we can now conclude that
$$\boxed{\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m}}$$
where $m>r+1>0$.
In particular, when $r=2k$ and $m=2$,
$$\int_{0}^{\infty} \frac{x^{2k}}{x^{2}+1} d x=\frac{\pi}{2} \csc \frac{(2k+1) \pi}{2}=\frac{\pi}{2} \sec (k\pi)$$
However, the property of Beta function,
$$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
holds for $\operatorname{Re}(x)>0$ and $\operatorname{Re}(y)>0,$
if and only if $1-\frac{2 k+1}{2}>0 \Leftrightarrow k<\frac{1}{2}$.
Therefore for any $k<\frac{1}{2}$,
$$\int_{0}^{\infty} \frac{x^{2k}}{x^{2}+1} d x=\frac{\pi}{2} \csc \frac{(2k+1) \pi}{2}=\frac{\pi}{2} \sec (k\pi) \tag*{(*)} $$
but doesn’t hold for all $k\geq \frac{1}{2}, $ which explains why we can’t put $k=1,2,3, \cdots$ into $(*)$.
|
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|
Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$ Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$.
Where do I start with this integral? I can easily see that it is possible to fator $x^{2}$ out on the denominator and use partial fractions. The numerator is also factorable but it does not have any integer roots. Can someone show me how to integrate this (either using what I said about partial fractions), or another method? Thanks
|
Simplify step by step:
$$\begin{align}
\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx
&= \int 1+ \frac{3x^2+x-1}{x^3+x^2}dx \\
&= x + \int\left(\frac{3x^2 + 2x}{x^3+x^2} -\frac{x + 1}{x^3+x^2}\right)dx \\
&= x + \int\frac{3x^2 + 2x}{x^3+x^2}dx - \int\frac{x + 1}{x^3+x^2}dx \\
&= x + \ln|x^3+x^2| - \int\frac{1}{x^2}dx \\
&= x+\ln|x^3 + x^2| + \frac{1}{x}+C
\end{align}$$
|
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|
Find minimum of ${(a+3c)\over(a+2b+c)} + {4b\over(a+b+2c)} - {8c\over(a+b+3c)}$ for non-negative reals
Let $a, b, c \ge 0$, not all zero. Find the minimum value of $${(a+3c)\over(a+2b+c)} + {4b\over(a+b+2c)} - {8c\over(a+b+3c)}.$$
Here was my attempt:
If c tends to infinity and a and b are small we get 1/3.
Now , i took $b=0$ and we get $(a-c)^2\over(a+c)(a+3c)$ clearly at $a=c$ this is $0$. And hence i can conclude that the infimum is $0$. This looked optimal.
But I took $a=1, b=3, c=10$ where the result is negative.
I don't see any definitve approach to this problem now. Hence, requesting your help !
|
As Albus mentioned, we use the substitution:
$ x = a + 2b + c, y = a + b + 2c, z = a + b + 3c$.
This system gives us $ a = -x + 5y - 3z, b = x - 2y + z, c = 0x - y + z$.
The expression becomes:
$$\frac{ -x + 2y } { x} + \frac{ 4 x - 8 y + 4z } { y} + \frac{ 8y - 8z } { z} = \frac{ 2y}{x} + \frac{ 4x}{y} + \frac{4z}{y} + \frac{ 8y}{z} -17. $$
Since $ \frac{ 2y}{x} + \frac{ 4x}{y} \geq 2\sqrt{8} = 4 \sqrt{2}$ and $ \frac{ 4z}{y} + \frac{8y}{z} \geq 2 \sqrt{32} = 8 \sqrt{2}$, hence a lower bound is $ 12 \sqrt{2} - 17 \approx -0.0294$.
Equality is achieved when $ y = \sqrt{2} x, z = \sqrt{2} y$, or equivalently when
$ a : b : c = 5\sqrt{2} - 7 : 3 - 2 \sqrt{2} : 2 - \sqrt{2} $, which we have to verify have the same sign (If not, that can't be achieved.)
|
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|
$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
&= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{2+2 x+4 \sqrt{x+1}} \cdot d x\\&=\int \frac{1}{1+x+2 \sqrt{x+1}} \cdot d x\\
&\quad +\int \frac{\sqrt{x+1}}{2(1+x)+4 \sqrt{x+1}}\cdot d x \\
&\quad -\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot d x
\end{split}
$$
From this last line i can evaluate the first two antiderivatives but the last one $$\int\frac{\sqrt{3-x}}{2(1+x)+4 \sqrt{x+1}} \cdot dx$$ seems a bit hard for me two evaluate, i don't find the good substitution. I am opened to any suggestion. Thanks in advance!
|
Solution without trigonometry:
Let $u=\sqrt{x+1}$ and $v=\sqrt{3-x}$, then $u^2+v^2=4$ and $u^2-v^2=2x-2$, so $udu+vdv=0$ and $dx=udu-vdv=2udu=-2vdv$.
\begin{align*}\int\frac{dx}{2+u+v}&=\int\frac{(u+v-2)dx}{(u+v)^2-4}=
\int\frac{(u+v-2)dx}{2uv}=\int\left(\frac{dx}{2v}+\frac{dx}{2u}-\frac{dx}{uv}\right)\\&=\int\left(-dv+du-\frac{dx}{uv}\right)=-v+u-\int\frac{du}{v}+\int\frac{dv}{u}\\&=-v+u-\int\frac{(u^2+v^2)du}{4v}+\int\frac{(u^2+v^2)dv}{4u}\\&=-v+u+\int\frac{v(udv-vdu)}{4v}+\int\frac{u(udv-vdu)}{4u}\\&=-v+u+\frac12\int(udv-vdu)=-v+u+2\int\frac{udv-vdu}{u^2+v^2}\\&=-v+u+2\int\frac{u^2d(v/u)}{u^2+v^2}=-v+u+2\arctan(v/u)+C.\end{align*}
|
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|
Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can we do something else at the beginning? $$x-14=(x-13)-1\\x-15=(x-14)-1\\..?$$
|
Hint: Note that
$$\frac{x-13}{x-14} = \frac{x-14+1}{x-14} = 1 + \frac{1}{x-14} \tag{1}\label{eq1A}$$
$$\frac{x-15}{x-16} = \frac{x-16+1}{x-16} = 1 + \frac{1}{x-16} \tag{2}\label{eq2A}$$
|
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|
Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^2-6x+15)\\x^4-12x^3+15x^2-10x^3+120x-150x+15x^2-180x+225=\\=4x^3-24x^2+60x$$ which is an equation I can't solve. I tried to simplify the LHS by $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{(x^2-6x+15)-4x}{x^2-6x+15}=1-\dfrac{4x}{x^2-6x+15}$$ but this isn't helpful at all. Any help would be appreciated! :) Thank you in advance!
|
Componendo and dividendo yields:
$$\frac{x^2-6x+15}{x^2-10x+15} = \frac{x^2-12x+15}{4x}$$
$$\Rightarrow \frac{4x}{x^2 - 10x + 15} = \frac{x^2-16x+15}{4x} \tag{$\frac{a-b}{b} = \frac{c-d}{d}$}$$
$$\Rightarrow u=x^2-13x+15: \frac{4x}{u + 3x} = \frac{u - 3x}{4x}$$
$$\Rightarrow u^2 - 9x^2 = 16x^2$$
$$\Rightarrow (u - 5x)(u + 5x) = 0$$
$$\Rightarrow (x^2 - 18x + 15)(x^2 - 8x + 15) = 0$$
|
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|
How the right term of the derivative is gained? This deduction is one of the typical ones I think.
What I want to deduce is the right term from the left term of the below equation.
$$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{-a}{x\sqrt{a^{2}+x^{2}}}$$
$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\frac{d}{dx}\left(x^{-1}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((x^{-1})'\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((-1\cdot x^{-2})\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\sqrt{a^{2}+x^{2}}\right)\right)$
$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left(-x^{-2}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)\left(x^{-1}\frac{d}{dx}\left(\left(a^{2}+x^{2}\right)^{\frac{1}{2}}\right)\right)$
Anyone deduced it in someday?
ps. I have to go to work. Back after about 7hours.
|
Small error: a missing + sign in the fourth line from the product rule:
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right)\left((x^{-1})'\left(a+\sqrt{a^{2}+x^{2}}\right) ~~\mathbf{+}~~ x^{-1}\frac{d}{dx}\left(a+\sqrt{a^{2}+x^{2}}\right)\right)$$
Continuing on with your idea, the next step would be to use the chain rule on that square root; with the + sign in the proper place we get
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right) \left(-\frac{1}{x^2}\left(a+\sqrt{a^{2}+x^{2}}\right) ~~\mathbf{+}~~ \frac{1}{x}\frac{x}{\sqrt{a^{2}+x^{2}}}\right)$$
Then the main trick is to get a common denominator:
$$=\left(\frac{x}{a+\sqrt{a^{2}+x^{2}}}\right) \frac{-a\sqrt{a^2+x^2}-a^2-x^2 + x^2}{x^2\sqrt{a^2+x^2}}$$
and I'm sure you can take it from here.
|
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|
Pythagorean triples conditions Pythagorian triple is every triple of natural numbers $(x, y, z)$ such that $x, y, z$ are sides of a right triangle, where $z$ is the hypotenuse.
Now, Pythagorean theorem says:
$$x^2 + y^2 = z^2 \tag1$$
If we look just natural solutions to the equation $(1)$, without geometrical condition that a right triangle with sides $x, y, z$ do exist, do we have more solutions? And if yes, in which pattern?
Of course, condition that for natural triple satisfying $x^2 + y^2 = z^2$ there must exist right triangle with sides $x, y, z$, which is equivalent to add 3 conditions to the $(1)$: (triangle inequality)
*$x^2 + y^2 = z^2$ and
*$x + y > z$
*$x + z > y$
*$y + z > x$
(Because if 1., 2., 3. holds, we can construct right triangle with sides $x, y, z$.)
|
One other way to analyse the equation $x^2+y^2=z^2$ is to solve it in its quadratic form.
Assuming $x\ne y$ otherwise $z$ would be irrational with factor $\sqrt{2}$.
Lets take $(x<y<z)>0$ without loss of generality and define $x=y-m$ and $z=y+n$ where $m$ and $n$ are positive integers. Giving
$$(y-m)^2+y^2=(y+n)^2\tag{1}$$
with $(z-x)=m+n$, $(z-y)=n$ and $(y-x)=m$
Which on rearrangement becomes
$$y^2-2(m+n)y+(m^2-n^2)=0\tag{2}$$
which can be solved as a quadratic in the usual way, with the result:
$$y=(m+n)+\sqrt{2n(m+n)}$$
[ or alternatively $y=(z-x)+\sqrt{2(z-y)(z-x)}$ ]
With a positive only square root since $m$ and $n$ are always positive integers.
Therefore
$x=n+\sqrt{2n(m+n)}$, $y= (m+n)+\sqrt{2n(m+n)}$ and $z= (m+2n)+\sqrt{2n(m+n)}$
Then
$(1.)\;\;\;$ $x+y=(m+2n)+2\sqrt{2n(m+n)}>z\;(=(m+2n)+\sqrt{2n(m+n)})$,
$(2.)\;\;\;$ $y+z=(2m+3n)+2\sqrt{2n(m+n)}>x\;(=n+\sqrt{2n(m+n)})$
and
$(3.)\;\;\;$ $x+z=(m+3n)+2\sqrt{2n(m+n)}>y\;(=(m+n)+\sqrt{2n(m+n)})$.
(It is also true that $m\ge n$ or equivalently $3z \ge 5x$, but I'm not sure about the best way to prove that.)
(In the same manner $x^3+y^3=z^3$ can be converted to a cubic equation, $x^4+y^4=z^4$ to a quartic equation, $x^5+y^5=z^5$ to a quintic equation and so on)
Of course you don't need to go this far to prove statements (1.), (2.) and (3.).
Since it can be shown algebraically that $(x+y)^2$ is greater than $z^2$, this immediately implies $(x+y)>z$.
Since we assumed at the outset $z>y$, it trivially follows that $x+z>y$.
Since we assumed at the outset $z>x$, it trivially follows that $y+z>x$.
Recalling the Pythagorean solution $x = a^2-b^2, y = 2ab, z = a^2+b^2$, $y>x$ can't in general be assumed.
|
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|
Study the series of $\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$ with respect to $a>0$ I have to study the series $\sum a_n$ with $a_n=\int_0^{\frac{1}{n^a}}\sin{(\sqrt[3]{x})}\,dx$, with respect to $a>0$.
I have thought to use the asymptotic criterion for series.
In particular I can observe that:
$\sin{x}\sim x-\frac{x^3}{3!}$ and so: $\sin{(\sqrt[3]{x})}\sim \sqrt[3]{x}- \frac{x}{3!}$ for $x\to 0$.
Now this means that: $$\int_0^{\frac{1}{n^a}}\sqrt[3]{x}- \frac{x}{3!}\,dx=[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}]\rvert_0^{\frac{1}{n^a}}-[\frac{x^{2}}{2\cdot 3!}]\rvert_0^{\frac{1}{n^a}}=\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$$ Now since $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}$ it self is asymptotic to $\frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ then I can say:
$$a_n\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}-\frac{1}{12n^{2a}}\sim \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$$
Now $\sum_{n=1}^{\infty} \frac{1}{\frac{4}{3}n^{\frac{4}{3}a}}$ is convergent for $\frac{4}{3}a>1\iff a>\frac{3}{4}$ and so from asymptote criterion also the original series converges for this value of $a$.
I need a check of my attempt and in case there is something wrong can you tell me where and how can I correct this?
(I can't use the Lebesgue theore since it is not something that I know)
|
Here is a more rigourous version of what you did. Using the Lagrange form of the remainder term in Taylor's formula, we have
$$
\left| {\sin w - w} \right| \le \frac{{w^3 }}{6}
$$
for all $w\geq 0$. Consequently,
\begin{align*}
\left| {a_n - \frac{3}{{4n^{4a/3} }}} \right| &= \left| {\int_0^{1/n^a } {(\sin \sqrt[3]{x} - \sqrt[3]{x})dx} } \right| \\ & \le \int_0^{1/n^a } {\left| {\sin \sqrt[3]{x} - \sqrt[3]{x}} \right|dx} \le \int_0^{1/n^a } {\frac{x}{6}dx} = \frac{1}{{12n^{2a} }}.
\end{align*}
Thus,
$$
\left| {\frac{{a_n }}{{\frac{3}{{4n^{4a/3} }}}} - 1} \right| \le \frac{1}{{9n^{2a/3} }}
$$
showing that
$$
a_n \sim \frac{3}{{4n^{4a/3} }}
$$
as $n\to +\infty$.
Addendum. By L'Hôpital's rule
$$
\mathop {\lim }\limits_{w \to 0} \frac{{\int_0^w {\sin \sqrt[3]{x}dx} }}{{\frac{3}{4}w^{4/3} }} = \mathop {\lim }\limits_{w \to 0} \frac{{\sin \sqrt[3]{w}}}{{\sqrt[3]{w}}} = 1
$$
In particular, with $w=1/n^a$,
$$
a_n \sim \frac{3}{{4n^{4a/3} }}
$$
as $n\to +\infty$.
|
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|
Solve the following system of equations:$|x|+|y|=2$ and $y=x+1$.
Solve the following system of equations:
$|x|+|y|=2$ and $y=x+1$, where $x$ is a real number.
Approach:
I substituted $y$ in equation $1$, so:
$$ \ \ \ \ \ \ \ \ \ \ |x|+|x+1|=2$$
$$1 \ \ \ \ \ \ \ \ \ \ 1$$
$$2 \ \ \ \ \ \ \ \ \ \ 0$$
$$0 \ \ \ \ \ \ \ \ \ \ 2$$
These are the total possibilities I think, because $|x|, |x+1| \ge 0$.
Then I made cases,
Case $1$: $|x|=1$, so, $x = \pm 1$
and $|x+1| = 1$, so $x+1=\pm 1$ so, $x = 0,-2$.
But no value of $x$ is matching, so this case gets rejected.
Case $2$: $|x|=2$, so, $x = \pm 2$
and $|x+1| = 0$, so $x=-1$.
But no value of $x$ is matching, so this case also gets rejected.
Case $3$: $|x|=0$, so, $x = 0$
and $|x+1| = 2$, so $x+1 = \pm 2$ and so, $x=1,-3$.
Here also, no value of $x$ is matching. So no solution exists.
Is this solution correct? Please confirm. If there is a shorter method to approach the question, please share it.
|
HINT
*
*If $x≥0$, then $y=x+1>0$
$$\begin{cases} x+y=2 \\ y=x+1 \end{cases}$$
Then, we need
*
*If $-1≤x<0, ~y≥0 $
$$\begin{cases} y-x=2 \\ y=x+1 \end{cases}$$
*
*If $x<-1, ~y<0$
$$\begin{cases} -x-y=2 \\ y=x+1 \end{cases}$$
|
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"url": "https://math.stackexchange.com/questions/4124031",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Determine the limit using Taylor expansions: I struggle with this one—maybe someone could point me in the right direction.
$$\lim_{x\to 0} \frac{5^{(1+\tan^2x)} -5}{1-\cos^2x}$$
Getting the Taylor series expansion for $\tan^2x$ and $\sin^2x$ is no problem, but I struggle with getting further along at this step:
$$\frac{5^1 \cdot 5^{x^2} \cdot 5^{(2/3)x^4} -5}{x^2 - \frac{x^4}3}$$
Any help would be greatly appreciated!
|
$\lim\limits_{x\to0}\dfrac{5^{1+\tan^2(x)}-5}{1-\cos^2(x)}=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\sin^2(x)}=$
$=5\lim\limits_{x\to0}\left[\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\dfrac1{\cos^2(x)}\right]=$
$=5\lim\limits_{x\to0}\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}\underset{\overbrace{\text{by letting }y=\tan^2(x)}}{=}$
$=5\lim\limits_{y\to0}\dfrac{5^y-1}y\cdot\lim\limits_{x\to0}\dfrac1{\cos^2(x)}=$
$=5\cdot\ln5\cdot1=$
$=5\ln5\;.$
Addendum :
If you want to calculate the limit by using Taylor expansions, you can proceed in the following way :
$\dfrac{5^{1+\tan^2(x)}-5}{1-\cos^2(x)}=$
$=5\cdot\dfrac{5^{\tan^2(x)}-1}{\sin^2(x)}=$
$=5\cdot\dfrac{5^{\tan^2(x)}-1}{\tan^2(x)}\cdot\dfrac1{\cos^2(x)}\underset{\overbrace{\text{by letting }y=\tan^2(x)}}{=}$
$=5\cdot\dfrac{5^y-1}y\cdot\dfrac1{\cos^2(x)}=$
$=5\ln5\cdot\dfrac{e^{y\ln5}-1}{y\ln5}\cdot\dfrac1{\cos^2(x)}=$
$=5\ln5\cdot\dfrac{y\ln5+\frac{(y\ln5)^2}2+O(y^3)}{y\ln5}\cdot\dfrac1{\left(1-\frac{x^2}2+O(x^4)\right)^2}=$
$=5\ln5\cdot\dfrac{1+\frac{y\ln5}2+O(y^2)}{1-x^2+O(x^4)}\xrightarrow{\color{blue}{x\to0}}5\ln5\;.$
It is not necessary to write only Taylor expansions in powers of $x$, we can also write a Taylor expansion in powers of $y$ which is an infinitesimal function of $x$ as $x\to0\;.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\sin P+\sin Q=2\sin\frac{P+Q}2\cos\frac{P-Q}2$ using the sin addition formula and double angle formula This question asks me to prove the validity of the trig identity by using the compound angle identities for $\sin(A+B)$ and $\sin(A-B)$
$$\sin P +\sin Q =2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}$$
I didn't get it at first so looked at the mark scheme and they set $$A+B=P$$ $$A-B=Q$$ and then got $$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$ I get all the algebra, as in when they sub everything in to make it equal but I don't get why they suddenly got $A+B$ to equal $P$ and $Q$?
Also, if they got $\sin(A+B)+\sin(A-B)$ why would they not use the sin addition formula to actually prove it? How can they just say that $2\sin A\cos B=\sin(A+B)+\sin(A-B)$ and then sub $P$ and $Q$ back in to prove the question?
|
They did it because they could. $A$ and $B$ don't have any significance on their own. If you have any two numbers $x,y$ and can always find two other numbers where $x= A+B$ and $y= A-B$. (just let $A=\frac {x+y}2$ and $B=\frac {x-y}2$). This can be a useful manipultion to make as expression appear simpler.
But we could ignore it.
$P = \frac {P+Q}2 + \frac {P-Q}2$
And $Q = \frac {P+Q}2 -\frac {P-Q}2$.
So $\sin P = \sin (\frac {P+Q}2 + \frac {P-Q}2) = \sin \frac {P+Q}2\cos\frac {P-Q}2 + \cos \frac {P+Q}2\sin \frac {P-Q}2$.
(Interestingly that will true regardless of what $Q$ is. We could take it as a trig identity that $\sin \theta = \sin (\frac {\theta+x}2 + \frac {\theta-x}2) = \sin \frac {\theta+x}2\cos\frac {\theta-x}2 + \cos \frac {\theta+x}2\sin \frac {\theta-x}2$ for all $x$.)
And $\sin Q =\sin (\frac {P+Q}2 - \frac {P-Q}2) = \sin \frac {P+Q}2\cos\frac {P-Q}2 - \cos \frac {P+Q}2\sin \frac {P-Q}2$
And so adding them up you get.... $2 \sin \frac {P+Q}2\cos \frac {P-Q}2$.
....
The people who wrote the proof simply thought it'd be easier to write and to follow if we replace $P$ with $A+B$ and $Q$ with $A-B$ (by letting $A:= \frac {P+Q}2$ and $B:=\frac {P-Q}2$) and prove $\sin(A+B) - \sin(A-B) = 2\sin A\cos B$.
Was it easier to write and to follow? Well, I'll let you be the jury on that.
=====
tl;dr
I think what the writer of the proof wanted to express but failed to make the point clear was this:
Prove $\sin P + \sin Q = 2\sin \frac {P+Q}2 \cos \sin \frac{P-Q}2$.
Proof: Let's assign labels $A = \frac {P+Q}2$ and $B=\frac {P-Q}2$. If we make this substitution we see that $A+B = \frac {P+Q}2 + \frac {P-Q}2 = \frac {2P}2 = P$ and that $A-B = \frac {P+Q}2 -\frac {P-Q}2 = \frac {2Q}2 = Q$.
Using this substitution our statement to prove becomes simply
$\sin(A+B) + \sin (A-B) = 2\sin A \cos B$.
ANd that follows immediately from the sine addition rule $\sin(A\pm B)=\sin A\cos B \pm \cos A\sin B$
(so $\sin(A+B) + \sin (A-B)= $
$(\sin A\cos B + \cos A\sin B) + (\sin A\cos B -\cos A\sin B) =$
$ 2\sin A\cos B$.)
....
Had the proof been written in that wording would it have been clearer?
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sum\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ converges or diverges? The original question is to show that $\;\sum\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ either converges or diverges.
I know it diverges but I'm having difficulty arriving at something useful for $ a_n $.
Here's what I did:
$ a_n = \left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 + 1}{n^2 +2n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 - 1}{n^2 +2n + 1}\right)^{n^2} = \left(\dfrac{(n-1)(n+1)}{(n+1)^2}\right)^{n^2} = \left(\dfrac{n-1}{n+1}\right)^{n^2} = \left( 1 - \dfrac{2}{n+1}\right)^{n^2} $
Do you have idea how to continue from here? ( my idea was to arrive to some expression that goes to infinity and deduce by the comparison test that $ a_n $ diverges ). I feel like something is right infront of my eyes but I can't see it.
Edit: The series converges. However, how do you propose I should continue from where I've left?
|
Let's use the root test:
$$\sqrt[\Large n]{\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}} = \left( \frac{n^2+1}{n^2+n+1} \right)^{n} = \left[\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\right]^{\frac{-n^2}{n^2+n+1}}\stackrel{n\to\infty}{\longrightarrow} e^{-1} < 1,$$
and therefore the series converges.
We used:
*
*$\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\to e$, because $\frac{n}{n^2+n+1}\to 0,$
*$\frac{-n^2}{n^2+n+1}\to -1.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For what values of $a$ is the vector $(a ^ 2, a, 1)$ in $\langle (1,2,3), (1,1,1), (0,1,2) \rangle$? For what values of $a$ is the vector $(a ^ 2, a, 1)$ in $\langle (1,2,3), (1,1,1), (0,1,2) \rangle$?
My question is, should I find scalars $ \alpha, \beta, \gamma $ such that $ \alpha (1,2,3) + \beta (1,1,1) + \gamma (0,1,2) = (a ^ 2, a, 1) $?
How can I prove that a vector is in the generator set?
Could you give me an idea?
|
$ \alpha (1,2,3) + \beta (1,1,1) + \gamma (0,1,2) = (a ^ 2, a, 1) $
this becomes the following system of equations:
$\alpha + \beta +0 \gamma=a^2$
$2\alpha + \beta + \gamma=a$
$3\alpha + \beta +2 \gamma=1$
Solving by gauss elimination:
First write the system in augmented matrix form
$
\begin{bmatrix}
1 & 1 & 0 & a^2\\
2 & 1 & 1 & a \\
3 & 1 & 2 & 1
\end{bmatrix}$
$r_2 \to r_2-2r_1$ and $r_3 \to r_3-3r_1$
$
\begin{bmatrix}
1 & 1 & 0 & a^2\\
0 & -1 & 1 & a-2a^2 \\
0 & -2 & 2 & 1-3a^2
\end{bmatrix}$
$r_2 \to -r_2$
$
\begin{bmatrix}
1 & 1 & 0 & a^2\\
0 & 1 & -1 & 2a^2-a \\
0 & -2 & 2 & 1-3a^2
\end{bmatrix}$
$r_1 \to r_1-r_2$ and $r_3 \to r_3+2r_2$
$
\begin{bmatrix}
1 & 0 & 1 & a-a^2\\
0 & 1 & -1 & 2a^2-a \\
0 & 0 & 0 & a^2-2a+1
\end{bmatrix}$
$a^2-2a+1$ should equal zero to have infinite solutions or there is no solutions
so
$a^2-2a+1=0 \to (a-1)^2=0 \to a=1$
after substituting $a=1$ the matrix become:
$
\begin{bmatrix}
1 & 0 & 1 & 0\\
0 & 1 & -1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}$
$\alpha+\gamma=0$
$\beta-\gamma=1$
$\gamma=t \to \alpha=-t$ and $\beta=1+t $
$ t \in \mathbb{R}$
|
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|
How to solve this binomial summation Problem where I was stuck at:
$$\sum_{k=0}^{n} (-4)^k\cdot \binom{n+k}{ 2k },$$ what I tried was to get observe that inside term is coefficients in expansion of $(1+2x)^{n+k}$ but evaluating it afterwards is not working for me.
|
We seek to evaluate
$$\sum_{k=0}^n (-1)^k 2^{2k} {n+k\choose 2k}$$
which is
$$\sum_{k=0}^n (-1)^k 2^{2k} {n+k\choose n-k}
= [z^n] (1+z)^n \sum_{k=0}^n (-1)^k 2^{2k} (1+z)^k z^k.$$
Here the coefficient extractor enforces the upper limit of the sum and
we find
$$[z^n] (1+z)^n \frac{1}{1+4z(1+z)} =
[z^n] (1+z)^n \frac{1}{(1+2z)^2}.$$
This is
$$\frac{1}{4} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} (1+z)^n
\frac{1}{(z+1/2)^2}.$$
Residues sum to zero and the residue at infinity is zero by inspection.
Hence the residue at zero is minus the residue at $-1/2.$ We have
$$\frac{1}{4}
\left.\left( \frac{1}{z^{n+1}} (1+z)^n \right)'\right|_{z=-1/2}
= \frac{1}{4}
\left.\left( - \frac{n+1}{z^{n+2}} (1+z)^n
+ \frac{n}{z^{n+1}} (1+z)^{n-1}\right)\right|_{z=-1/2}
\\ = \frac{1}{4} \left( - (n+1) (-1)^n 2^{n+2} \frac{1}{2^n}
+ n (-1)^{n+1} 2^{n+1} \frac{1}{2^{n-1}} \right)
\\ = -(n+1) (-1)^n + n (-1)^{n+1}.$$
Taking into account the sign we get at last
$$\bbox[5px,border:2px solid #00A000]{
(-1)^n \times (2n+1).}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ can we conclude that $x^n+y^n+z^n\in\mathbb{Q}$ for all $n\in\mathbb{N}$. Is there an explicit form? I had seen a fun problem that is exactly the problem in the question, except it was a specific case of this. Turns out, if $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ then $x^5+y^5+z^5 = 6$. I put this equation into Wolfram Alpha for different values of $n$. For instance, $x^8+y^8+z^8=\frac{51}{72}$. $x^{11}+y^{11}+z^{11}=\frac{11117}{216}.$
Firstly, is there an explicit expression to evaluate $x^n+y^n+z^n$? And, if not, are we able to prove that this is always a rational number, at least? WolframAlpha is unable to calculate an explicit form for $x^n+y^n+z^n$.
|
Yes. In https://en.m.wikipedia.org/wiki/Symmetric_polynomial, look at Power-sum Symmetric Polynomials in the section Special kinds of symmetric polynomials
|
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|
Algebra trick clarification question I have the equation $x = \sqrt{3} + \sqrt{2} \,\,$ and the trick is that $\frac{1}{x} = \sqrt{3} - \sqrt{2}$. I don't see how the minus sign appears just by taking 1 over x?
|
If we are told that $x = \sqrt{3}+\sqrt{2}$ then it follows that $\dfrac{1}{x} = \dfrac{1}{\sqrt{3}+\sqrt{2}}$
$=\dfrac{1}{\sqrt{3}+\sqrt{2}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} = \dfrac{\sqrt{3}-\sqrt{2}}{3-2}$
$=\sqrt{3}-\sqrt{2}$
This same pattern can be extended to if $x=\sqrt{n+1}+\sqrt{n}$ then $\dfrac{1}{x}=\sqrt{n+1}-\sqrt{n}$
|
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|
How many triangles are there whose sides are prime numbers and have a perimeter equal to $29$?
Perimeter of a triangle is equal to $29$ and the values of all three sides are prime number. How many many incongruent triangles we have with this properties?
$1)2\qquad\qquad\qquad2)3\qquad\qquad\qquad3)4\qquad\qquad\qquad4)5$
First I realized that we can't have a side equal to $2$ because then perimeter will be even number. And we know that if $a,b,c$ are the values of sides we have:
$$a+b>c$$
Hence $29-c>c$ and $c<14.5$, similarly we have $a<14.5$ and $b<14.5$, so we have $a,b,c\in\{3,5,7,11,13\}$ and $a+b+c=29$
From here is it possible to proceed without Try and Error ?
|
Modulo $6$ the residues of the primes are $3,-1,1,-1,1$ respectively – and our target has residue $-1$ modulo $6$. If we use two $3$s the last side must be $23$, which is impossible. If we use one $3$ the other two sides must have residue $1$ each and sum to $26$ – the only possibility is $13$ and $13$. If no $3$s, there must be two residue $-1$ numbers and one residue $1$ number:
*
*If the residue $1$ number is $7$, we can only have $11$ and $11$
*If that number is $13$, we can only have $5$ and $11$
Hence there are $3$ incongruent admissible triangles: $\{3,13,13\},\{7,11,11\},\{5,11,13\}$.
|
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|
Find $x$ for $3^x = 5/3$ Why my answer is wrong? $3^x = 5/3$
$e^{\ln 3^x} = 5/3$
$e^{x \ln 3}=5/3$
$e^x \times e^{\ln 3} = 5/3$
$e^x \times 3 = 5/3$
$e^x = 5/9$
$x = \ln(5/9)$
|
We know that $a^x = e^{x \ln a}$
thus $3^x = e^{x \ln 3} = 5/3$
applying $\ln$ to both sides, we get $x \ln 3 = \ln (5/3) = \ln(5) - \ln(3)$
The rest is straightforward algebra:
$$x \ln 3 + \ln 3 - \ln 5 = 0$$
$$(x+1) \ln 3 = \ln 5$$
$$x + 1 = \ln 5 / \ln 3$$
$$x = \frac{\ln 5}{\ln 3} - 1$$
|
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"url": "https://math.stackexchange.com/questions/4154947",
"timestamp": "2023-03-29T00:00:00",
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|
Paradox of $-i$ seemingly equal to $1$ via the Wallis product for $\pi$ and the Euler sine product Assuming $x$ is a real variable throughout,$$\frac{\sinh(ix)}{i}=\sin(x)$$
$$\frac{\sinh(\pi ix)}{\pi ix} = \frac{\sin(\pi x)}{\pi x}$$
$$\frac{e^{\pi ix}-e^{-\pi ix}}{2\pi ix}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$
Letting $x=\frac{1}{2}$ yields:
$$\frac{2}{\pi i}=\prod_{n=1}^{\infty}\left(1-\frac{1}{4n^2}\right)$$
$$\frac{1}{i}=-i=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{4n^2-1}{4n^2}\right)=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)$$
The famous Wallis product for $\pi$ tells us:
$$\frac{\pi}{2}=\frac{2\cdot 2}{1\cdot 3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\frac{6\cdot 6}{5\cdot 7}\dots=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)$$
Therefore, in a multiplication which I hope to be legitimate, we seem to get:
$$-i=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)\cdot\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)=\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{(2n-1)(2n+1)}\right)=1$$
Clearly, $1\neq-i$, so what have I done wrong here? I randomly came up with this while half asleep, so there may be some trivial error here - perhaps the products cannot be combined like this - but I'm at a loss!
Thanks for any help.
|
You appear to be making a mistake in the left hand side of your fourth equation. $e^{\frac{1}{2}i \pi}=i$, and $e^{-\frac{1}{2}i \pi}=-i$. Therefore
$$\frac{e^{\frac{1}{2}i\pi}-e^{-\frac{1}{2}i\pi}}{2\pi i\frac{1}{2}}=\frac{2i}{2\pi i}=\frac{2}{\pi}$$
Carrying this correction through gives $1=1$ at the end.
|
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|
Can there be a triangle ABC if $\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}$? Can there be a triangle ABC if $$\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}\;?$$ Equating the ratios to $k$ we get $\cos A=k$, $\cos B=2k$, $\cos C=3k$.
Then the identity $$\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1 \implies 12k^3+14k^2-1=0$$
$f(k)=12k^3+14k^2-1$ being monotonic for $k>0$ can have at most one real positive root. Further, $f(0)=-1, f(1/3)=1>0$, so there will be one real root in $(0,1/3)$. Hence all three cosines will be positive and less that 1, for a unque triangle to be possible.
What can be other ways to solve this question?
|
\begin{align}
\frac{\cos A}{1}&
=\frac{\cos B}{2}=\frac{\cos C}{3}
=\frac{\cos A+\cos B+\cos C}{6}
\tag{1}\label{1}
.
\end{align}
\begin{align}
6\cos A&=\cos A+\cos B+\cos C
=\frac{r}{R}+1=
v+1
\tag{2}\label{2}
,
\end{align}
where $r$ and $R$ are inradius and circumradius
of the corresponding $\triangle ABC$,
$v=\tfrac{r}{R}\in(0,\tfrac12]$ for a valid triangle.
Hence,
\begin{align}
\cos A&=\frac{v+1}6\in(0,\tfrac14)
\tag{3}\label{3}
,\\
\cos B&=2\cos A =\frac{v+1}3 \in(0,\tfrac12)
\tag{4}\label{4}
,\\
\cos C&=3\cos A=\frac{v+1}2 \in(0,\tfrac34)
\tag{5}\label{5}
.
\end{align}
From known identity
\begin{align}
\cos A\cos B\cos C
&=\tfrac14(u^2-(v+2)^2)
\tag{6}\label{6}
,
\end{align}
where $u=\tfrac\rho{R}$,
and $\rho$ is semiperimeter of $\triangle ABC$,
we can express $u^2$ in terms of $v$:
\begin{align}
\tfrac1{36}(v+1)^3
&=\tfrac14(u^2-(v+2)^2)
,\\
u^2&=
\tfrac19v^3+\tfrac43v^2+\tfrac{13}3 v+\tfrac{37}9
\tag{7}\label{7}
,
\end{align}
and with the help of
\begin{align}
\cos A\cos B+\cos B\cos C+\cos C\cos A
&= \tfrac14(u^2+v^2)-1
\tag{8}\label{8}
\end{align}
we arrive at the cubic equation in $v$
\begin{align}
v^3+10v^2+17v-10&=0
,
\end{align}
which has only one positive solution
\begin{align}
v&=
\tfrac{14}3\cos\left(
\tfrac\pi3
-\tfrac13\arctan\left(\tfrac9{100}\sqrt{1329}\right)
\right)
-\tfrac{10}3
\approx 0.458757
.
\end{align}
Then
\begin{align}
u(v) &= \tfrac13\sqrt{v^3+12v^2+39v+37}
\approx 2.52792343
,\\
u_{\min}(v)&=
\sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}}
\approx 2.5158959
,\\
u_{\max}(v)&=
\sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}}
\approx 2.53465839
,
\end{align}
so indeed we have a unique valid triangle with given properties.
|
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|
About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$
Problem: Let $x > 0$. Prove that
$$x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12.$$
Remark 1: The problem was posted on MSE (now closed).
Remark 2: I have a proof (see below). My proof is not nice.
For example, we need to prove that
$\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$ for which my proof is not nice.
I want to know if there are some nice proofs. Also, I want my proof reviewed for its correctness.
Any comments and solutions are welcome and appreciated.
My proof (sketch):
We split into cases:
i) $x \ge 1$:
Clearly, $x^{x^{x^{x^{x^x}}}}\ge x^x$.
By Bernoulli's inequality, we have
$x^x = (1 + (x - 1))^x \ge 1 + (x - 1)x = x^2 - x + 1 \ge \frac12 x^2 + \frac12$. The inequality is true.
ii) $0 < x < 1$:
It suffices to prove that
$$x^{x^{x^{x^x}}}\ln x \ge \ln \frac{x^2 + 1}{2}$$
or
$$x^{x^{x^{x^x}}} \le \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ln x \le \ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}.$$
It suffices to prove that
$$x^{x^{x^x}}\ge \frac{7}{12} \ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}. \tag{1}$$
First, it is easy to prove that $$x^x \ge \mathrm{e}^{-1/\mathrm{e}}
\ge \frac{1}{\ln x}\ln\frac{\ln\frac{7}{12}}{\ln x}.$$
Thus, the left inequality in (1) is true.
Second, let $f(x) = x^{7/12}\ln x - \ln \frac{x^2 + 1}{2}$. We have
\begin{align*}
f'(x) &= \frac{7}{12x^{5/12}}
\left(\ln x + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
&\le \frac{7}{12x^{5/12}}
\left(\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
&\le 0 \tag{2}
\end{align*}
where we have used
$\ln x \le \frac{3x^2 - 3}{x^2 + 4x + 1}$ for all $x$ in $(0, 1]$.
Also, $f(1) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $(0, 1)$.
Thus, the right inequality in (1) is true.
Note: For the inequality $\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$,
we let $x = y^{12}$ and it suffices to prove that
$11y^{47} + \cdots + 3 \ge 0$ (a polynomial of degree $47$, a long expression) for all $0 < y < 1$.
We are done.
|
Remarks: @Erik Satie considered $^6 x \ge \lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$ for $(38/100, 1)$. I gave alternative
proof of $-\frac{W(-\ln x)}{\ln x} \ge \frac12 x^2 + \frac12$ for all $x$ in $(38/100, 1)$.
Case $x \in (38/100, 1)$:
According to Theorem in [1] (Page 240), we have
$\lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$
where $W(\cdot)$ is the principal branch of the Lambert W function. Also, we have $^6 x \ge {^8}x
\ge {^{10}}x \ge \cdots$ which results in
$^6 x \ge \lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$.
Let us prove that
$-\frac{W(-\ln x)}{\ln x} \ge \frac12 x^2 + \frac12$ for all $x$ in $(38/100, 1)$.
To this end, with the substitution $x = \mathrm{e}^{-y}$ for $y\in (0, -\ln\frac{38}{100})$, we need to prove that
$$\frac{W(y)}{y} \ge \frac12 \mathrm{e}^{-2y} + \frac12$$
or
$$W(y) \ge \frac12 y\mathrm{e}^{-2y} + \frac12 y.$$
Since $u \mapsto u\mathrm{e}^u$ is strictly increasing on $(0, \infty)$, it suffices to prove that
$$W(y)\mathrm{e}^{W(y)} \ge \left(\frac12 y\mathrm{e}^{-2y} + \frac12 y\right)
\mathrm{exp}\left({\frac12 y\mathrm{e}^{-2y} + \frac12 y}\right)$$
that is
$$y \ge \left(\frac12 y\mathrm{e}^{-2y} + \frac12 y\right)
\mathrm{exp}\left({\frac12 y\mathrm{e}^{-2y} + \frac12 y}\right)$$
where we have used the fact $W(y)\mathrm{e}^{W(y)} = y$ for all $y > 0$.
With the substitution $z = \mathrm{e}^{-2y}$, it suffices to prove that, for all $z$ in $(38^2/100^2, 1)$,
$$0 \ge \ln \frac{1 + z}{2} - \frac{1 + z}{4}\ln z.$$
The remaining is smooth.
Reference
[1] R. Arthur Knoebel, “Exponentials Reiterated,” The American Mathematical Monthly, No. 4, Vol. 88 (1981), pp. 235-252, Apr. 1981.
https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0
|
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|
A six digit number is formed randomly using digits $\{1,2,3\}$ with repetitions. Choose the correct option(s):
A six digit number is formed randomly using digits $\{1,2,3\}$ with repetitions. Choose the correct option(s):
*
*A) Probability that all digits are used at least once is $\dfrac{20}{27}$
*B) Probability that digit $1$ is used odd number of times and $2$ is used even number of times is $\dfrac{1-3^{-6}}4$
*C) Probability that all digits are used as well as odd digits are used odd number of times and even digit is used even number of times is $\dfrac{3^6-2^7+1}{4\cdot3^6}$
Total cases = $3^6$
*
*A) All digits used at least once = Total cases$-$Cases when one digit is not used
Or, would it be
All digits used at least once = Total cases$-$Cases when one digit is not used$-$Cases when two digits are not used
In any case, I am not getting $\frac{20}{27}$
*
*B) $1$ can be used $1$ or $3$ or $5$ times. $2$ can be used $0$ or $2$ or $4$ times.
So, favorable cases=$^6C_1(^5C_0+^5C_2+^5C_4)+^6C_3(^3C_0+^3C_2)+^6C_5=182$, and it matches with the answer.
But the answer given in the option is of different format $\dfrac{3^6-1}{4\cdot3^6}$. Looks like they are subtacting $1$ from the total cases and diving by $4$ to get the favorable case. Why?
*
*C) $1$ can be used $1$ or $3$ or $5$ times. $2$ can be used $2$ or $4$ times.
So, favorable cases=$^6C_1(^5C_2+^5C_4)+^6C_3(^3C_2)+^6C_5=156$
But as per the option, favorable cases= $150.5$. Also, looks like they are subtracting $2^7-1$ from total cases and then diving by $4$, what could be the motivation for this, even if this is wrong?
|
+1 to your answer, for showing your work. I agree that your analysis for Part B is correct. My analysis is:
Part A
Use Inclusion-Exclusion.
$\displaystyle \frac{N\text{(umerator)}}{D\text{(enominator)}},~$ with $\displaystyle D = \frac{1}{3^6}.$
$\displaystyle N = 3^6 - \left[\binom{3}{1}2^6\right] + \left[\binom{3}{2}1^6\right] = 729 - 192 + 3 = 540.$
$\displaystyle \frac{540}{729} = \frac{20}{27}$
So, the answer given for part (A) is correct.
Part B
Again, $\displaystyle D = 3^6.$
$\displaystyle N = \sum_{k=1}^6 f(k),~$ as described below:
$f(1) : (1)$ "1", $(0)$ 2's : $\binom{6}{1} = 6.$
$f(2) : (1)$ "1", $(2)$ 2's : $\binom{6}{1}\binom{5}{2} = 60.$
$f(3) : (1)$ "1", $(4)$ 2's : $\binom{6}{1}\binom{5}{4} = 30.$
$f(4) : (3)$ "1"'s, $(0)$ 2's : $\binom{6}{3}\ = 20.$
$f(5) : (3)$ "1"'s, $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$
$f(6) : (5)$ "1"'s, $(0)$ 2's : $\binom{6}{5} = 6.$
$\displaystyle N = 182.$
$\displaystyle \frac{1 - (1/729)}{4} = \frac{728}{4 \times 729} = \frac{182}{729}.$
So, the answer given for part (B) is also correct.
Part C
Again, $\displaystyle D = 3^6.$
$\displaystyle N = \sum_{k=1}^3 f(k),~$ as described below:
$f(1) : (1)$ "1", $(1)$ "3", $(4)$ 2's : $\binom{6}{1}\binom{5}{1} = 30.$
$f(2) : (1)$ "1", $(3)$ "3"'s, $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$
$f(3) : (3)$ "1's", $(1)$ "3", $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$
$\displaystyle N = 150.$
$\displaystyle \frac{3^6 - 2^7 + 1}{4 \times 3^6} = \frac{602}{4 \times 729} \neq \frac{150}{729}.$
So, the answer given for part (C) is wrong.
Note that if the answer given for part (C) had been
$\displaystyle \frac{3^6 - 2^7 - 1}{4 \times 3^6}$, then the answer would have been correct.
|
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|
Basic algebraic/arithmetic manipulations I do not know the rules governing the transformation of each member of the following group of equations into the the next one:
$$ 5(1 + 2^{k -1} + 3^{k -1}) - 6(1 + 2^{k -2} + 3^{k -2}) + 2
\\= (5 -6 + 2) + (5 \times 2^{k - 1} - 6 \times 2^{k - 2}) + (5 \times 3^{k -1} - 6 \times 3^{k - 2})
\\= 1 + (5 - 6 \times 2^{-1}) \times 2^{k -1} + (5 - 6 \times 3^{-1}) \times 3^{k -1}
\\= 1 + (5 -3) \times 2^{k-1} + (5 - 2) \times 3^{k -1}
\\= 1 + 2 \times 2^{k -1} + 3 \times 3^{k -1}
\\= 1 + 2^{k} + 3^{k}$$
I need to learn rules such as these from scratch. Where can I learn them?
Could someone label the rule used to form each member of the group of equalities from the previous member?
|
Here's an explanation of some of the lines:
*This line is obtained from line 1 by bringing the coefficients $5$, $6$ into their respective parentheses using the distributive law and then rearranging and regrouping terms using the commutative and associative laws. I assume there are no issues here, but if so, please ask.
*Factors $2^{k-1}$ and $3^{k-1}$ are pulled out of the 2nd and 3rd parentheses, again using the distributive law, but rules of exponents need to be used first. In the 2nd parentheses, the term $-6\cdot2^{k-2}$ has been rewritten as $-6\cdot2^{-1}\cdot2^{k-1}$ using the rule $a^{x+y}=a^xa^y$. Here $a=2$, $x=-1$ and $y=k-1$. In the 3rd parentheses, something similar was done: $-6\cdot3^{k-2}=-6\cdot3^{-1}\cdot3^{k-1}$.
*Arithmetic: $6\cdot 2^{-1}=6\cdot\frac{1}{2}=3$. This uses $a^{-x}=\frac{1}{a^x}$ with $a=2$ and $x=1$.
*Basic arithmetic.
*The law $a^xa^y=a^{x+y}$ is used again here: $2\cdot2^{k-1}=2^1\cdot2^{k-1}=2^{1+k-1}=2^k$ and similarly for the term with base $3$.
|
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|
Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ I want to prove that the below equation can be held.
$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$
Firstly I tried to check the equation with small values of $n$
$$ \text{As } n=2 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \right) + \cos\left(\frac{ 2 \pi \cdot 2 }{ 2 } \right) $$
$$ = \cos\left(\pi\right) + \cos\left(2 \pi\right) $$
$$ = -1+ 1 =0 ~~ \leftarrow~~ \text{Obvious} $$
But
$$ \text{As}~~ n=3 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 3 } \right) +\cos\left(\frac{ 2 \pi \cdot 2 }{ 3 } \right) + \cos\left(\frac{ 2 \pi \cdot 3 }{ 3 } \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + \cos\left( 2\pi \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + 1 =?$$
What formula(s) or property(s) can be used to prove the equation?
|
The cosines are the $x$-coordinates of points on the unit circle equally spaced around the origin. Since the average $x$ value is zero, it follows that the sum of the $x$ values is also zero.
|
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|
If $3\sin x +5\cos x=5$, then prove that $5\sin x-3\cos x=3$
If $3\sin x +5\cos x=5$ then prove that $5\sin x-3\cos x=3$
What my teacher did in solution was as follows
$$3\sin x +5\cos x=5 \tag1$$
$$3\sin x =5(1-\cos x) \tag2$$
$$3=\frac{5(1-\cos x)}{\sin x} \tag3$$
$$3=\frac{5\sin x}{(1+\cos x)} \tag4$$
$$5\sin x-3\cos x=3 \tag5$$
However this should not be true when $\sin x=0$, as division by zero is not defined; and also, if $\sin x=0$, then the expression we have to prove evaluates to $-3$. In other words question is incomplete but my teacher denied it.
Am I correct in my reasoning?
|
If $c=\cos x$ and $s=\sin x$, then you know that$$\left\{\begin{array}{l}3s+5c=5\\c^2+s^2=1.\end{array}\right.$$This system is easy to solve. One of the solutions is $(c,s)=(1,0)$ and the other one is $(c,s)=\left(\frac 8{17},\frac{15}{17}\right)$. In the first case (which is the case that you get when $x=0$), $5s-3c=-3$; in the second case, $5s-3c=3$. So, you are right; some hypothesis is missing. But this also shows that, other than the case in which $x$ is an integer multiple of $2\pi$, the statement is true.
|
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|
Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.
My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ so $K$ is divisible by $4$. Note that any $n\in\Bbb N$ is one of $\overline{0},\overline{1},\overline{2}$ in $\operatorname{mod}3$. Well from this, I can argue anyway but I want to show $K$ is divisible by $3$ more easier or nicer way. Could you help?
|
To prove divisibility by $4$:
$(a^2-b^2)-(a^2-c^2)+(b^2-c^2)=0=\text{even}$
One of the addends must be even and this is possible with integers only if that is a multiple of $4$.
To prove divisibility by $3$:
Pigeonhole principle: at least two of $a^2,b^2,c^2$ must be multiples 9f $3$ or at least two must be one greater than a multiple of $3$. Either way the difference between the identified pair of squares is a multiple of $3$.
QED.
|
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|
Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(\infty)-f(0)]\ln(a/b).$$ Here, $f(x)=\frac{1-\cos(x)}{x},$ hence $I=0.$
So the question is how to show (1), otherwise?
|
Alternatively, integrate by parts
\begin{align}
&\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx\\
= &\int_{0}^{\infty} \frac{6\sin 2x-6\sin 3x}{x} dx\\
=& \>6 \int_{0}^{\infty} \frac{\sin 2x}{2x} d(2x)
- 6 \int_{0}^{\infty} \frac{\sin 3x}{3x} d(3x) \\
=&\>0
\end{align}
|
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|
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is?
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of
$x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP
then $p + q$ is ?
My solution approach :-
I have assumed $α,β,γ,δ$ to be as follows ;
$α = \frac{a}{r^3}$
$β = \frac{a}{r}$
$γ = ar$
$δ = ar^3$
Here the common ratio is $r^2$.
Now $α+β = 1$ and $γ+δ = 4$ and solving these two equations in terms of $a$ and $r$ gives the value of $r= \sqrt{2} \text{ and } \sqrt{-2}$ which leads to the values of $a$ to be $\frac{2\sqrt{2}}{3}$ and $\frac{-2\sqrt{2}}{3}$ respectively.
In both the above cases, $α,β,γ,δ$ turns out to be the same i.e. $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ which gives the value of $p+q = \frac{34}{9}$.
Now if I assume $α,β,γ,δ$ to be as follows ;
$α = a$
$β = ar$
$γ = ar^2$
$δ = ar^3$
Here the common ratio is just $r$.
But in this case $r$ values turn out to be $\pm2$ and for $r=2$ we get $α,β,γ,δ$ to be $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ whereas for $r=-2$ we get $α,β,γ,δ$ to be $-1,2,-4,8$ and hence $p+q = \frac{34}{9} \text{ or }-34 $.
Now I am getting confused why there is this difference in both the approaches. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!
|
The question can also be approached from the direction of working with the zeroes of the polynomials themselves. For $ \ x^2 - x + p \ = \ 0 \ $ and $ \ x^2 - 4x + q \ = \ 0 \ \ , $ we have
$$ \alpha \ , \ \beta \ \ = \ \ \frac{1 \pm \sqrt{1 - 4p}}{2} \ = \ \frac12 \ \pm \ \frac12 \sqrt{1-4p} \ = \ \frac12 \ \pm \ \frac12\sqrt{D_1} \ \ , $$
$$ \gamma \ , \ \delta \ \ = \ \ \frac{4 \pm \sqrt{16 - 4q}}{2} \ = \ 2 \ \pm \ \sqrt{4-q} \ = \ 2 \ \pm \ \sqrt{D_2} \ \ . $$
We wish for the zeroes to lie in a geometric progression $ \ \alpha \ , \ \beta \ , \ \gamma \ , \ \delta \ \ , $ which for real zeroes requires:
• either a ratio $ \ r > 0 \ $ with all of the zeroes positive and in the indicated sequence, or
• a ratio $ \ r < 0 \ $ with the zeroes alternating in sign and falling on the real-number line in the sequence $ \ \gamma \ , \ \alpha \ , \ (0) \ , \ \beta \ , \ \delta \ \ . $
[Note: we are using $ \ r \ $ as the ratio between zeroes here, rather than $ \ r^2 \ . $ ]
We can see already what is necessary for the first case: if $ \ \alpha \ > 0 \ $ (and likewise for the other zeroes) and the zeroes are all to be distinct, we must have $ \ 0 < 1-4p < 1 \ \Rightarrow \ 0 < p < \frac14 \ \ . $ Thus, if $ \ p \ $ and $ \ q \ $ are to be integers, the first case above has no solution. (We will relax this condition later to see what occurs.) On the other hand, if we permit $ \ \alpha < 0 \ $ with the other zeroes having alternating signs, we simply have $ \ 1 < 1-4p \ \Rightarrow \ p < 0 \ \ . $ (We can also find constraints for $ \ q \ $ here, but we'll shortly see that a quite rigid requirement will be set.)
If we first consider the required ratio for each pair of zeroes, we find
$$ \frac12 · ( \ 1 + \sqrt{D_1} \ ) \ = \ r · \frac12 · ( \ 1 - \sqrt{D_1} \ ) \ \ \Rightarrow \ \ D_1 \ = \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ \ , $$
$$ 2 + \sqrt{D_2} \ = \ r · ( \ 2 - \sqrt{D_2} \ ) \ \ \Rightarrow \ \ D_2 \ = \ \left( \frac{2r \ - \ 2}{r \ + \ 1} \right)^2 \ = \ 4 · \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ = \ 4D_1 \ \ . $$
From this, we obtain $ \ 4 - q \ = \ 4 · (1-4p) \ \Rightarrow \ q \ = \ 16p \ \ , $ and consequently that $ \ p + q \ = \ 17p \ \ . $
We now need to "couple" these pairs of zeroes through the condition $ \ \gamma \ = \ r · \beta \ $ or
$$ 2 - \sqrt{D_2} \ = \ r · \frac12 · ( \ 1 + \sqrt{D_1} \ ) \ \ \Rightarrow \ \ 2·2 \ - \ 2·2\sqrt{D_1} \ = \ r · ( \ 1 + \sqrt{D_1} \ ) $$
$$ \Rightarrow \ \ D_1 \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right)^2 \ \ . $$
Equating the two ratio expressions for $ \ D_1 \ $ yields either
$$ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right) \ \ \ \text{or} \ \ \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \ \left( \frac{r \ - \ 4}{r \ + \ 4} \right) \ \ ; $$
the first of these leads to $ \ r^2 + 3r - 4 \ = \ -r^2 + 3r + 4 \ \Rightarrow \ 2r^2 \ = \ 8 \ \Rightarrow \ r^2 \ = \ 4 \ \ , $ while the second equation produces $ \ r^2 + 3r - 4 \ = \ r^2 - 3r - 4 \ \Rightarrow \ 6r \ = \ 0 \ , $ which is a "trivial" result.
The first of the cases (for which we already know $ \ p \ $ and $ \ q \ $ cannot be integral) gives us
$ \mathbf{ r = 2 \ \ : } \quad D_1 \ = \ \left( \frac{2 \ - \ 1}{2 \ + \ 1} \right)^2 \ = \ \frac19 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ \frac29 \ \ \Rightarrow \ \ p+q \ = \ \frac{34}{9} \ \ $
[we do have a rational-number solution] with the zeroes being $ \ \alpha \ , \ \beta \ = \ \frac12 \ \pm \ \frac12\sqrt{\frac19} \ = \ \frac13 \ , \ \frac23 \ $ and $ \ \gamma \ , \ \delta \ = \ 2 \ \pm \ \sqrt{4·\frac19} \ = \ \frac43 \ , \ \frac83 \ \ . $
The alternating-sign case is
$ \mathbf{ r = -2 \ \ : } \quad D_1 \ = \ \left( \frac{[-2] \ - \ 1}{[-2] \ + \ 1} \right)^2 \ = \ 9 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ -2 \ \ \Rightarrow \ \ p+q \ = \ -34 \ \ $
with the zeroes being $ \ \alpha \ , \ \beta \ = \ \frac12 \ \pm \ \frac12·\sqrt{9} \ = \ -1 \ , \ 2 \ $ and $ \ \gamma \ , \ \delta \ = \ 2 \ \pm \ \sqrt{4·9} \ = \ -4 \ , \ 8 \ \ , $ arranged on the $ \ x-$axis in the order $ \ \gamma \ , \ \alpha \ , \ \beta \ , \ \delta \ \ . $
Graphs for the two cases are presented below.
$ \ \ $
We won't need to be concerned about the situation for complex zeroes. We would then have
$$ \alpha \ , \ \beta = \overline{\alpha} \ \ = \ \ \frac12 \ \pm \ i · \frac12 \sqrt{4p-1} \ \ , \ \ \gamma \ , \ \delta = \overline{\gamma} \ \ = \ \ 2 \ \pm \ i · \sqrt{q-4} \ \ , \ \ p > \frac14 \ , \ q > 4 \ \ . $$ Since these are conjugate-pair zeroes which may be expressed as $ \ \zeta · e^{\pm i \phi} \ $ and $ \ \xi · e^{\pm i \psi} \ \ , $ respectively, there is no single complex ratio $ \ r \ = \ \rho · e^{i \theta} \ \ , $ with $ \ \rho \ , \ \zeta \ $ and $ \ \xi \ $ being real numbers, that will put all four zeroes into a geometric progression.
|
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|
Solve : $(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$ I have to find the solution of following differential equation:
$$(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$$
It can be re-written as
$$(2(x - 2y) + 5) \frac{dy}{dx} + (x - 2y) + 3 = 0$$
let $u = (x-2y)$
so, on differentiating both sides with respect to $x$ we get
$$\frac{du}{dx} = 1 - 2\cdot\frac{dy}{dx}$$
$$\therefore \frac{dy}{dx} = \frac{1 - \frac{du}{dx}}{2}$$
using this value in initial equation we get,
$$(2u + 5) (\frac{1-\frac{du}{dx}}{2}) + u + 3 = 0$$
On further solving we get
$$\frac{2u + 5}{4u + 11}.\frac{du}{dx} = 1$$
Integrating both sides with respect to $x$, we get
$$\int \frac{2u + 5}{4u + 11}.du = \int dx$$
$$\implies \int (\frac{1}{2} - \frac{1}{8u + 22})\cdot du = x + C$$
$$\implies \frac{1}{2}\cdot u - \frac{1}{8} \ln(8u + 22) = x + C$$
putting the value of $u$ in the above equation we get,
$$\implies \frac{1}{2} (x-2y) - \frac{1}{8} ln(8(x - 2y)+ 22) = 2 + C$$
we have, when $x = 2 , y = 2.5.$
When I put these value of $x$ and $y$ in the above equation to obtain the value of $C$
$ln(-negative Value)$ is obtained, what is wrong here?
The answer given in the book is :
$$4x + 8y + ln(4x - 8y + 11) = 28$$
|
There's a mistake in following step and it is rectified in next step
$ \frac{1}{2}\cdot u - \frac{1}{8} \ln(8u + 22) = x + C$
$\implies \frac{1}{2}\cdot u - \frac{1}{8} \ln(|8u + 22|) = x + C$
$\implies \frac{1}{2}\cdot (x-2y) - \frac{1}{8} \ln(|8x-16y + 22|) = x + C$
When $x=2$ $y=2.5$, therefore
$\implies \frac{1}{2}\cdot (2-5) - \frac{1}{8} \ln(|8*2-16*2.5 + 22|) = 2 + C$
$\implies \frac{1}{2}\cdot (-3) - \frac{1}{8} \ln(|16-16*2.5 + 22|) = 2 + C$
$\implies - \frac{1}{8} \ln(2) = 3.5 + C$
$\implies - \frac{1}{8} \ln(2)-3.5= C$
Therefore our solution becomes
$\implies \frac{1}{2}\cdot (x-2y) - \frac{1}{8} \ln(|8x-16y + 22|) = x - \frac{1}{8} \ln(2)-3.5$
$\implies 4x-8y - \ln(|8x-16y + 22|) = 8x - \ln(2)-28$
$\implies 4x+8y + \ln(|8x-16y + 22|)-\ln2=28$
$\implies 4x+8y + {\ln(|4x-8y + 11|)}=28$
|
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|
Given $f(r) = r\tan^{-1} \frac{a}{r+b}$ where $a>0$ and $b>0$, find an expression for $r$ in terms of $f(r)$. Is there an analytical solution to the following problem:
Given $f(r) = r\tan^{-1} \frac{a}{r+b}$ where $a>0$ and $b>0$ find an expression for $r$ in terms of $f(r)$.
To me, it seems like there is an analytical solution to this problem. When I plot this out, $f(r)$ resembles a square root function.
I can also represent this equation as follows:
$tan(\frac{f(r)}{r}) = \frac{a}{r+b}$
Is there any way to determine if this has an analytical solution, and if it does, what is the solution? This represents a real geometric problem I just encountered which I've already solved by brute force. But, this equation interests me.
|
Around $r=0$, we can write the infinite sum for$$f=r \tan ^{-1}\left(\frac{a}{r+b}\right)$$ This gives
$$f=r \tan ^{-1}\left(\frac{a}{b}\right)+\sum_{n=1}^\infty \frac{(a-i b) (-b-i a)^{-n}+(a+i b) (-b+i a)^{-n}}{2 (n-1)}\, r^n$$ Truncate it to some order and, for more legibility, let $a=k b$
$$f=r \tan ^{-1}(k)-\frac{k r^2}{b (k^2+1)}+\frac{k r^3}{b^2 \left(k^2+1\right)^2}+\frac{k
\left(k^2-3\right) r^4}{3 b^3 \left(k^2+1\right)^3}+\frac{k\left(1-k^2\right)
r^5}{b^4 \left(k^2+1\right)^4}+O\left(r^6\right)\tag 1$$
Now use series reversion to obtain
$$r=\frac{f}{\tan ^{-1}(k)}+\frac{ k}{b \left(k^2+1\right) \tan ^{-1}(k)^3}f^2+\frac{ k \left(2 k- \tan ^{-1}(k)\right)}{b^2 \left(k^2+1\right)^2 \tan^{-1}(k)^5}f^3+O\left(f^4\right)\tag 2$$
Let us try using $a=2$, $b=3$. For $r=2$, this gives $f=2 \tan ^{-1}\left(\frac{2}{5}\right)\sim 0.761013$.
Using $(2)$, this leads to $r=1.89841$ which does not seem too bad. For sure, this could be significantly improve at the price of more terms. Using for example all the terms present in $(1)$ (that is to say an expansion up to $O\left(f^6\right)$, it would give $r=1.9853$.
Edit
If you want more terms, we could write
$$r=\sum_{n=1}^p \frac {d_n} {\big[b\left(k^2+1\right)\big]^{n-1} x^{2 n-1} }\,f^n +O(f^{(p+1)})\qquad \text{with}\qquad x=\tan ^{-1}(k)$$ and the first $d_n$'s are
$$\left(
\begin{array}{cc}
n & d_n \\
1 & 1 \\
2 & k \\
3 & k (2 k-x) \\
4 & k \left(\left(5-\frac{x^2}{3}\right) k^2-5 x k+x^2\right) \\
5 & -k \left(2 \left(x^2-7\right) k^3-x \left(x^2-21\right) k^2-9 x^2 k+x^3\right)
\\
6 & \left(\frac{x^4}{5}-\frac{28 x^2}{3}+42\right) k^5+\frac{28}{3} x
\left(x^2-9\right) k^4-2 x^2 \left(x^2-28\right) k^3-14 x^3 k^2+x^4 k
\end{array}
\right)$$ The next ones are too long to be typed here.
For the worked example, the above table would give $r=1.99441$
|
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|
If $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$ is divided by $(x^2 +1)$, then find the remainder
If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$
is divided by $(x^
2
+1)$, then the remainder is:
How Do I solve this question without the tedious long division?
Using remainder theorem , we can take $x^3$ common and put $x^2 =-1$ although $x$ is not a real number. By this method, I got the right answer as $-x$. Is it the right way? Because $x$ comes out to be $i$ which is not real.
Also , can I apply remainder theorem to quadratic divisor polynomials in this way?
|
Following may help
$P(x)=x^{19} +x^{17} +x^{13} +x^{11} +x^{7} +x^{5} +x^{3}$
$P(x)=x^{17}(x^2+1)+x^{11}(x^2+1)+x^{5}(x^2+1)+x^3$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5})+x^3$
That $x^3$ looks sad alone , Let's also include it
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5})+x^3+x-x$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5}+x)-x$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5}+x)-x$
Now this is in the form -
$P(x)=Q(x)*d(x)+R(x)$
Hence you get everything now
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|
Mathematicals inequalities For $$x,y,z>0 $$
Prove that
$$(2xyz)^2 \ge (x^3+y^3+z^3+xyz)(x+y-z)(y+z-x)(z+x-y)$$
I have tried a famous inequalities: $$(x+y-z)(y+z-x)(z+x-y) \le xyz$$
So the problem is:
$$3xyz \ge x^3+y^3+z^3$$
But in fact, this isn’t true
Help me plz
|
Suppose $z=\min\{x,y,z\}.$ We can write this inequality as
$$(x^3+y^3+z^3+xyz)\left [xyz-\prod (x+y-z) \right ] \geqslant xyz(x^3+y^3+z^3-3xyz),$$
But
$$xyz-\prod (x+y-z) = (x+y-z)(x-y)^2+z(x-z)(y-z),$$
$$x^3+y^3+z^3-3xyz = (x+y+z)[(x-y)^2+(x-z)(y-z)].$$
Thefore the inequality equivalent to
$$M(x-y)^2+N(x-z)(y-z) \geqslant 0,$$
for
$$\begin{aligned}M &= (x^3+y^3+z^3+xyz)(x+y-z)-xyz(x+y+z) \\& \geqslant (x^3+y^3+z^3+xyz)x-xyz(x+y+z) \\& = x[x^3+(y+z)(y-z)^2] \geqslant 0.\end{aligned}$$
$$N = x^3+y^3+z^3-xy(x+y) = (x+y)(x-y)^2+z^3 \geqslant 0.$$
The proof is completed.
|
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|
How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$? How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$?
\begin{align*}
(x-y+2)(3x+y-4) &= 0\tag{1}\\
(x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}
In equation $1$, it is not possible for both $x-y+2$ and $3x+y-4$ to be nonzero. Like-wise in equation $2$, it is not possible for both $x+y-2$ and $2x-5y+7$ to be nonzero.
Therefore the LHS of either equation must involve a product of $0$ and some number.
This means there are infinitely many solutions to both equations.
However this doesn't shead any light on which points are common between the two equations. I'm confused. It seems I've made an error in my judgement. What should I reconsider?
|
Compare equations indivually, and work case-by-case.
$$x-y+2=0\implies y=x+2, x+y-2=0\implies y=2-x$$
$$x+2=2-x\implies x=0\to y=2$$
So $(0,2)$ is a solution.
$$3x+y-4=0\implies y=4-3x,\\ x+2=4-3x\implies x=\frac12\to y=\frac52$$
So $(\frac12, \frac52)$ is a solution.
You can do the other two.
|
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|
Calculus Infinite Series Homework Problem. I have the problem
"Let
$f(x) = x + \frac{2}{3} x^3 + \frac{2 \cdot 4}{3 \cdot 5} x^5 + \dots + \frac{2 \cdot 4 \dotsm 2n}{3 \cdot 5 \dotsm (2n + 1)} x^{2n + 1} + \dotsb$
on the interval $(-1,1)$ of convergence of the defining series.
(a) Prove that $(1 - x^2) f'(x) = 1 + xf(x).$
(b) Prove that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.$"
I don't exactly know how to prove it without doing part b first, then part a. Would prefer if someone could prove it using the infinite series. If anyone can help me with proving part a, then part b, that'd be greatly appreciated.
|
If you are given a power series $f(x) = \sum_{n = 0}^\infty a_n (x - x_0)^n$ with radius of convergence $R$, then $f$ is differentiable on $(x_0 - R, x_0 + R)$ and the derivative is $f'(x) = \sum_{n = 1}^\infty n a_n (x - x_0)^{n - 1}$.
In other words, you can differentiate term-by-term.
I shall leave it to you to verify that your series does converge on $(-1, 1)$. (For example, use an appropriate version of the root test.)
Recall the notion of a double factorial: We have $$(2n)!! = 2 \cdot 4 \cdot \cdots \cdot (2n)$$ and $$(2n + 1)!! = 1 \cdot 3 \cdot \cdots \cdot (2n + 1).$$
By convention, we have $(-1)!! = 0!! = 1$.
In particular, for $n \ge 1$, we have the identity $\frac{n!!}{n} = (n - 2)!!$.
Now, in your case, we have
$$f(x) = \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1}.$$
and thus,
\begin{align}
f'(x) &= \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!} \cdot (2n + 1) \cdot x^{2n} \\
&= 1 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n}.
\end{align}
Thus, we get
\begin{align}
x^2f'(x) &= x^2 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n + 2} \\
&= x^2 + \sum_{n = 2}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n} \\
&=\sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n}.
\end{align}
Subtracting from the earlier expression, we get
\begin{align}
(1 - x^2)f'(x) &= 1 + \sum_{n = 1}^{\infty}\left[\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!}\right] x^{2n}.
\end{align}
On the other hand, we have the expression
\begin{align}
1 + xf(x) &= 1 + \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 2} \\
&= 1 + \sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 1)!!} x^{2n}.
\end{align}
I leave it to you to check that
$$\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)!!}{(2n - 1)!!}$$
for all $n \ge 1$. Thus, we have shown that
$$(1 - x^2)f'(x) = 1 + xf(x).$$
Now, we wish to conclude that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}$. For this, simply note that $g(x) := \frac{\arcsin x}{\sqrt{1 - x^2}}$ also satisfies the first order ODE
$$y' - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$$
on $(-1, 1)$.
The above is a linear first-order ODE with continuous coefficients on $(-1, 1)$. Thus, it has a unique solution which satisfies $y(0) = 0$.
Since $f$ and $g$ both satisfy the ODE with same initial condition, we see that $f = g$. Thus,
$$\sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1} = \frac{\arcsin x}{\sqrt{1 - x^2}}.$$
|
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|
How to solve $5^{x + 1} = 3^{x + 2}$
Solve $5^{x + 1} = 3^{x + 2}$.
I got this far, but I'm not sure how to continue:
\begin{align}
5^{x + 1} &= 3^{x + 2} \\
(5^x)(5^1) &= (3^x)(3^2) \\
5(5^x) &= 9(3^x)
\end{align}
Where do I go from here?
|
Use natural logarithm:
$$5^{x+1}=3^{x+2}$$
$$\begin{align}&\implies (x+1)\ln 5=(x+2)\ln 3 \\
&\implies x\ln 5+\ln 5=x\ln 3+2\ln 3 \\
&\implies x(\ln 5-\ln 3)=2\ln 3-\ln 5 \\
&\implies x=\frac{2\ln 3-\ln 5}{\ln 5-\ln 3}. \end{align}$$
|
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What is my mistake in finding this pythagorean triplet? Since Project Euler copyright license requires that you attribute the problem to them, I'd like to add that this is about question 9 there.
I am trying to solve this problem on only two brain cells and can't figure out what am I doing wrong. Here is the system for $(a, b, c) \in \mathbb{N}^3$,
\begin{align*}
a^2 +b^2 &= c^2 \\
a+ b + c &= 1000 \\
a &< b < c
\end{align*}
Here is my approach,
\begin{align*}
a + b + c &= 1000 \\
a + b &= 1000 - c &&\text{Subtract } c\\
a^2 + b^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Square both sides}\\
c^2 + 2ab &= 1000^2 + c^2 - 2000c &&\text{Since }a^2 +b^2 = c^2\\
2ab &= 1000^2 - 2000c &&\text{Subtract } c^2\\
\frac{ab}{500} &= 1000 - 2c &&\text{Divide } 1000\\
2c &= 1000 - \frac{ab}{500} &&\text{Rearrange}\\
\end{align*}
Now let $a=5,b=200$,
\begin{align*}
2c &= 1000 - 2\\
2c &= 998 \\
c &= (998 \div 2) = 499 \\
\end{align*}
But certainly these values do not work. I can't see why.
|
I think this comment by @MatthewLeingang explaining @lulu's comment answers the issue with my approach.
What lulu is saying by “not reversible” is that you have shown “If $a, b$, and $c$ are integers such that $a+b+c=1000$ and $a^2+b^2=c^2$, then $2c=1000− (ab/500)$.” That is not the same thing as “If $a$ and $b$ are integers and $2c=1000−(ab/500)$, then $a+b+c=1000$ and $a^2+b^2=c^2$.”
|
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|
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of $\frac{d^2y}{dx^2}$?
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of
$\frac{d^2y}{dx^2}$?
$1)-\frac{x^2+y^2}{y^2}\qquad\qquad2)-\frac{x^2+y^2}{y^3}\qquad\qquad3)\frac{x+y}{x^2+y^2}\qquad\qquad4)\frac{xy}{x^2+y^2}$
Here is my approach:
We have a quadratic equation in $x^2+y^2$. So $x^2+y^2=\frac{3\pm\sqrt5}{2}$ (RHS is a constant). Taking differentiate with respect to $x$,
$$2x+2yy'=0\Rightarrow\quad y'=\frac{-x}{y}$$
$$2+2y'^2+2yy''=0\Rightarrow \quad y''=\frac{-y'^2}{y}\Rightarrow y''=\frac{-x^2}{y^3}$$
Edit:
The second option can be written as $-(\frac{x^2}{y^3}+\frac{1}y)$. It seems it is the correct answer. But why I missed $-\frac1y$ in this approach?
|
Your approach is correct, you just made a small mistake at the last step: from $2+2y'^2+2yy''=0$ we find that
$$y''=-\frac{1+y'^2}{y}=-\frac{1+(-x/y)^2}{y}=-\frac{x^2+y^2}{y^3}.$$
|
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Prove $\sum_{n=2}^{\infty}\frac{1}{n^2+e}<\frac{1}{2}$ Of course, you can use the following formula
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2},$$
but which is too "advanced". We want to find a solution only depending on inequality estimation only.
Maybe, we can obtain
\begin{align*}
\sum_{n=2}^{\infty} \frac{1}{n^2+e}&\le \sum_{n=2}^{100}\frac{1}{n^2+e}+\sum_{101}^{\infty}\frac{1}{n^2}=\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)+\sum_{n=2}^{\infty}\frac{1}{n^2}\\
&=\frac{\pi^2}{6}-1+\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)<\frac{1}{2},
\end{align*}
which is true by checking on machine, but too hard to compute by hand.
|
This is a comment (I haven't enough reputation to comment).
Exact sum result:
$${{i\,\left(\psi_{0}(2-\sqrt{e}\,i)+\gamma\right)}\over{2\,\sqrt{e}
}}-{{i\,\left(\psi_{0}(\sqrt{e}\,i+2)+\gamma\right)}\over{2\,\sqrt{e
}}}$$
or
$$\frac{-3 e-1+(1+e) \pi \sqrt{e} \coth \left(\sqrt{e} \pi \right)}{2 e (1+e)}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4195224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Computation of $\int_0^1 \frac{\arctan^2 x\ln x}{1+x}dx$ I'm searching for a "simple" proof of:
\begin{align}\int_0^1 \frac{\arctan^2 x\ln x}{1+x}dx=-\frac{233}{5760}\pi^4-\frac{5}{48}\pi^2\ln ^2 2+\text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{16}\zeta(3)\ln 2+\frac{1}{24}\ln^4 2+\pi \Im\left(\text{Li}_3\left(1+i\right)\right)-\frac{1}{4}\text{G}\pi\ln 2\end{align}
The context:
I have written a script for Pari GP to search heuristically for certain close-forms.
Testing my script with the above integral i was lucky enough to find something.
The script searchs for integer linear relation between the integral and some constants:
\begin{align}\pi^4,\pi\ln^3 2,\pi^2\ln^2 2,\pi^3\ln 2,\text{Li}_4\left(\frac{1}{2}\right),\zeta(3)\ln 2,\zeta(3)\pi,\ln^4 2,\pi \Im\left(\text{Li}_3\left(1+i\right)\right), \Im\left(\text{Li}_3\left(1+i\right)\right)\ln 2,\text{G}^2,\text{G}\ln^2 2,\text{G}\pi^2,\text{G}\pi\ln 2\end{align}
NB:
Using the algorithm with $\displaystyle\int_0^1 \dfrac{\ln^2(1+x^2)\ln x}{1+x}dx$ gives someting too.
Addendum:
PARI GP script:
beta(n)={intnum(x=0,1,(-log(x))^(n-1)/(1+x^2))};
lindep4(x)={
NAME=["x","Pi^4","Pilog(2)^3","Pi^2log(2)^2","Pi^3log(2)","polylog(4,1/2)","zeta(3)log(2)","zeta(3)Pi","log(2)^4","Piimag(polylog(3,1+I))","log(2)imag(polylog(3,1+I))","Catalan^2","Catalanlog(2)^2","CatalanPi^2","Catalanlog(2)Pi","beta(4)","imag(polylog(4,1+I))"];
VAL=[x,Pi^4,Pilog(2)^3,Pi^2log(2)^2,Pi^3log(2),polylog(4,1/2),zeta(3)log(2),zeta(3)Pi,log(2)^4,Piimag(polylog(3,1+I)),log(2)imag(polylog(3,1+I)),Catalan^2,Catalanlog(2)^2,CatalanPi^2,Catalan*log(2)*Pi,beta(4),imag(polylog(4,1+I))];
L=lindep(VAL);
for(i=2,length(L),if(-L[i]/L[1]>0,print1("+",-L[i]/L[1],NAME[i]));if(-L[i]/L[1]<0,print1(-L[i]/L[1],NAME[i])));
}
for example:
\p 100
lindep4(intnum(x=0,1,atan(x)^2*log(x)/(1+x)))
NB: To improve the script add $\beta(4)$ value and some polygamma values.
PS: the PARI GP script has been updated to take into account more integrals.
Try this one: $\displaystyle \int_0^1\frac{\ln^2(1+x^2)\ln x}{1+x^2}dx$
|
Another framework proposed by Cornel (answer to the second integral, $\displaystyle \int_0^1\frac{\ln^2(1+x^2)\ln x}{1+x^2}\textrm{d}x$)
Observe that $$\int_0^1 \frac{1}{1+x^2}\log^3\left(\frac{2x}{1+x^2}\right)\textrm{d}x$$
$$=\log^3(2)\int_0^1\frac{1}{1+x^2}\textrm{d}x+3\log^2(2)\int_0^1\frac{\log(x)}{1+x^2}\textrm{d}x+3\log(2)\int_0^1\frac{\log^2(x)}{1+x^2}\textrm{d}x$$
$$+\int_0^1\frac{\log^3(x)}{1+x^2}\textrm{d}x-3\log^2(2)\int_0^1\frac{\log(1+x^2)}{1+x^2}\textrm{d}x-6\log(2)\int_0^1\frac{\log(x)\log(1+x^2)}{1+x^2}\textrm{d}x$$
$$-3\int_0^1\frac{\log^2(x)\log(1+x^2)}{1+x^2}\textrm{d}x+3\log(2)\int_0^1\frac{\log^2(1+x^2)}{1+x^2}\textrm{d}x$$$$-\int_0^1\frac{\log^3(1+x^2)}{1+x^2}\textrm{d}x+3\color{blue}{\int_0^1\frac{\log(x)\log^2(1+x^2)}{1+x^2}\textrm{d}x}.$$
*
*Note that the integral in the left-hand side may be beautifully reduced by the variable change $\displaystyle x\mapsto \frac{2x}{1+x^2}$ to $\displaystyle \int_0^1 \frac{1}{1+x^2}\log^3\left(\frac{2x}{1+x^2}\right)\textrm{d}x=\frac{1}{2}\int_0^1 \frac{\log^3(x)}{\sqrt{1-x^2}}\textrm{d}x$, where the last integral is a form involving the derivative of the Beta function.
*Note that all the other resulting integrals in the right-hand side are already known.
*To easily make the connection with the known integrals, for the integrals $\displaystyle \int_0^1\frac{\log^2(1+x^2)}{1+x^2}\textrm{d}x$ and $\displaystyle \int_0^1\frac{\log^3(1+x^2)}{1+x^2}\textrm{d}x$ make the variable change $x \mapsto \tan(x)$ to have a view in terms of trigonometric functions. A relevant link: About the integral $\int_{0}^{\pi/4}\log^4(\cos\theta)\,d\theta$
*Also, you might like to know the following generalized integral, $\displaystyle \int_0^1\frac{\log^{2n}(x)\log(1+x^2)}{1+x^2}\textrm{d}x$, is nicely presented and calculated by Ali Shadhar in his book, An Introduction To The Harmonic Series And Logarithmic Integrals: For High School Students Up To Researchers (see page $149$). The integral easily and naturally reduces to forms involving derivatives of the Beta function.
End of story
|
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"url": "https://math.stackexchange.com/questions/4196102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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|
How many of the first $100$ terms are the same in the arithmetic sequences $2,9,16,\ldots$ and $5,11,17,\ldots$?
If $\{a_n\}$ is an arithmetic sequence with 100 terms where $a_1=2$ and $a_2=9$, and $\{b_n\}$ is an arithmetic sequence with 100 terms where $b_1=5$ and $b_2=11$, how many terms are the same in each sequence?
I think the answer is 17, but how I got it seemed too easy and I just want someone to verify my answer. Here is how I found my solution:
The sequence for
$$a_n= 2, 9, 16, 23, 30, 37, 44, 51, 58, 65, 72,79,86,93,100,107,114,121,\ldots$$
The sequence for
$$b_n= 5, 11,17, 23, 29, 35, 41, 47, 53, 59, 65, 71,77,83,89,95,101,107, \ldots$$
Ignoring the 1st four terms, I noticed that after the 4th term, the same terms appear in every 6th term for $a_n$ and 7th term for $b_n$. Therefore, $100-4=96/6=16$. Then I added $1$ to include $23$.
I hope this makes sense and thank you for your help.
|
The general formulas for the terms are:
\begin{align*}
a_{n} = 2 + (n-1)7 = 7n - 5 \\
b_{n} = 5 + (n-1)6 = 6n - 1
\end{align*}
We are looking for two integers $x$ and $y$ such that:
\begin{align*}
a_{x} &= b_{y} \\
7x - 5 &= 6y - 1 \\
x &= \frac{6y + 4}{7} \\
\end{align*}
In other words we are looking a integer $y$ such that $x$ is a also an integer. Meaning that $6y + 4$ is a multiple of $7$.
\begin{align*}
6y+4 &= 0 \mod 7 \\
6y &= -4 \mod 7 \\
6y &= 3 \mod 7 \\
2y &= 1 \mod 7 \\
2y &= 8 \mod 7 \\
y &= 4 \mod 7 \\
\end{align*}
Therefore $y$ leaves a remainder of $4$ when divided by $7$ and can be written in the form $y=7k+4$. For example, when $k = 0$, we get the pair $y = 4$ and $x = 4$ ($a_{4} = b_{4}$). When $k = 1$, we get the pair $y = 11$ and $x = 10$ ($a_{10} = b_{11}$). The largest $k$ we can plug in before $y$ is above $100$ is $13$, meaning that in total with $k = 0$ there are $14$ possible values of $k$ and $14$ terms in both sequences.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4197691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Solve $\sqrt{\frac{(2 - x)(2 - y)}{(2 + x)(2 + y)} + \frac{(2 - y)(2 - z)}{(2 + y)(2 + z)} + \frac{(2 - z)(2 - x)}{(2 + z)(2 + x)}}$
Find the value of the expression $S$, knowing that $x^2 + y^2 + z^2 + xyz = 4$ and $x, y, z \in (0, \infty)$:
$$
S = \sqrt{\frac{(2 - x)(2 - y)}{(2 + x)(2 + y)} + \frac{(2 - y)(2 - z)}{(2 + y)(2 + z)} + \frac{(2 - z)(2 - x)}{(2 + z)(2 + x)}}
$$
My attempt:
As $x^2 + y^2 + z^2 + xyz = 4$, we can choose $A, B, C \in \left[0, \frac{\pi}{2}\right)$ for which $x = 2 \cos A$, $y = 2 \cos B$ și $z = 2 \cos C$, a well-known substitution identity.
Now, let's prove that:
$$\tan^2 A = \frac{2 - x}{2 + x}$$
It is true that:
$$\cos A = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}$$
Therefore:
$$\tan^2 A = \frac{1 - \cos A}{1 + \cos A}$$
Equivalent to the identity we wanted to prove true.
Let's substitute in the expression:
$$S = \sqrt{\tan^2 \frac{A}{2} \tan^2 \frac{B}{2} + \tan^2 \frac{B}{2} \tan^2 \frac{C}{2} + \tan^2 \frac{C}{2} \tan^2 \frac{A}{2}}$$
Which doesn't seem to be constant. I suspect the problem is actually wrong, because if, instead of one square root, we'd have three square roots, the experession would have evaluated to:
$$S'= \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1$$
Which is calculable by a very well known identity. Could the problem be wrong? Or it's something I'm missing, a further step to solve the initial problem? Thanks in advance!
|
One cannot find the value of $S$, only simplified expression.
Let's denote $x+y+z=a$, $xy+xz+yz=b$, $xyz=c$.
$x^2+y^2+z^2=a^2-2b \Rightarrow a^2-2b+c=4 \Rightarrow 2b=a^2+c-4$.
$S^2=A/B$, $A=(2-x)(2-y)(2+z)+(2-x)(2-z)(2+y)+(2-y)(2-z)(2+x)=$
$3xyz-2(xy+yz+xz)-4(x+y+z)+24=3c-2b-4a+24$,
$B=(2+x)(2+y)(2+z)=xyz+2(xy+yz+xz)+4(x+y+z)+8=c+2b+4a+8$.
Using $2b=a^2+c-4$ one can obtain $A=3c-a^2-c+4-4a+24=2c+32-(a+2)^2$,
$B=c+a^2+c-4+4a+8=2c+(a+2)^2$.
So, the answer is $S=\sqrt{\frac{2c+32-d}{2c+d}}$, where $c=xyz$,
$d=(x+y+z+2)^2$.
The value of $S$ is not constant. $x=2,y=z\approx 0 \Rightarrow S\approx 1$; $x=y=z=1
\Rightarrow S=\sqrt{\frac{1}{3}}$.
Maybe, the word "minimum" or "maximum" is lost before "value".
$S^2=\frac{2c+32-d}{2c+d}$. At constant $c$ $S^2$ is decreasing with
growth of $d$, so the maximum $S$ is corresponding with minimum $d$ and
vice versa.
Minimum $S^2 \to 23-16\sqrt{2}$ is the limit at $x \to 0$ and $y=z
=\sqrt{\frac{4}{x+2}}$. Maximum $S^2\to 1$ is the limit at $x \to 0$ and
$y=x$, $z$ is positive root of $z^2+x^2 z+2 x^2=4$.
Maybe, the original problem should be "to prove that $S<1$". Using
$S^2=\frac{2c+32-d}{2c+d}$, it is equivalent to $d=(x+y+z+2)^2>16
\Leftrightarrow x+y+z>2$.
Using that "geometric mean is not greater than quadratic mean" one can
write $\sqrt[3]{c}=\sqrt[3]{xyz}\leq\sqrt{\frac{x^2+y^2+z^2}{3}}=
\sqrt{\frac{4-xyz}{3}}=\sqrt{\frac{4-c}{3}} \Rightarrow 27c^2\leq
(4-c)^3 \Rightarrow c \leq 1$.
Using that "garmonic mean is not greater than geometic mean" one can
write $\frac{3c}{b}=\frac{1}{\frac{1}{3}\cdot
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} \leq \sqrt[3]{xyz}=
\sqrt[3]{c} \Rightarrow$ $\frac{27c^3}{b^3}\leq c \Rightarrow$
$b^3\geq{27c^2} \Rightarrow b\geq 3 \sqrt[3]{c^2} \geq 3c$ for $c\leq 1$. Using it in $a^2-2b+c=4 \Rightarrow$ $a^2=4+2b-c \geq 4+6c-c > 4$ for $c > 0$. So, $a^2 > 4 \Rightarrow a>2
\Rightarrow d>16 \Rightarrow S^2 < 1 \Rightarrow S < 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4205951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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|
If $\phi(r) =1+\frac 12 +\frac 13… \frac 1r$ and $\sum_{r=1}^{n} (2r+1)\phi (r) =P(n)\phi(n+1)-Q(n)$. Find $P$ and $Q$. I tried making a double sum $$\sum_{r=1}^{n} \sum _{k=1}^r \frac{2r+1}{k}$$
But since the final limits aren’t same the changing of orders cannot be used. Can I get a hint on how to solve it?
|
EDIT:
Using Python, I detected a small error in the final lines, which I have now fixed.
OP:
The sum
$$\sum_{r=1}^n\sum_{k=1}^r\frac{2r+1}{k}=3(1)+5\left(1+\frac{1}{2}\right)+7\left(1+\frac{1}{2}+\frac{1}{3}\right)+\dots+(2n+1)\left(1+\frac{1}{2}+\dots+\frac{2n+1}{n}\right)$$
Can be regrouped into sub-series as this:
$$(3(1)+5(1)+7(1)+\dots)+\left(\frac{5}{2}+\frac{7}{2}+\dots\right)+\left(\frac{7}{3}+\frac{9}{3}+\dots\right)+\dots$$
Each of these forms an arithmetic series sum. The formula for an arithmetic sum is $$\frac{2a+(m-1)r}{2}\cdot m=a+(a+r)+(a+2r)+\dots+(a+(m-1)r)$$
In the expansion of your sum, each sub-series (as I will call the groupings I made) begins with the number $2k+1$ divided by the $k$, where $k$ is the index of the sub-series, so $a=\frac{2k+1}{k}$. Each successive term is an odd number divided by $k$, so the term-term difference is $\frac{2}{k}=r$. Finally, since each sub-series is shorter in length, it turns out that there are $m=n-k+1$ numbers in the sub-series. So, the sum of each sub-series, such as $3+5+7+\dots$ ($k=1$) or $\frac{5}{2}+\frac{7}{2}+\dots$ ($k=2$) is:
$$\begin{align}S(k,n)&=\frac{2\cdot\frac{2k+1}{k}+(n-k+1-1)\cdot\frac{2}{k}}{2}\cdot(n-k+1)\\&=\left(\frac{2k+1}{k}+\frac{n-k}{k}\right)\cdot(n-k+1)\\&=\frac{n+k+1}{k}\cdot(n-k+1)\\&=\frac{n^2+2n+1-k^2}{k}\\&=\frac{(n+1)^2}{k}-k\end{align}$$
Now the original series can be written as the sum of the sums of all these sub-series:
$$\sum_{r=1}^n\sum_{k=1}^r\frac{2r+1}{k}=\sum_{k=1}^nS(k,n)=\sum_{k=1}^n\frac{(n+1)^2}{k}-k$$
Consider $S(k=n+1,n)$: that term would equal $(n+1)^2/(n+1)-(n+1)=0$. Therefore we can increment the number of terms on the above sum safely, as adding zero won't change it.
We now want to evaluate
$$\sum_{k=1}^{n+1}\frac{(n+1)^2}{k}-k=(n+1)^2\left(\sum_{k=1}^{n+1}\frac{1}{k}\right)-k=(n+1)^2\phi(n+1)-\sum_{k=1}^{n+1}k$$
The last sum is a simple triangle formula sum, and so finally we get:
$$(n+1)^2\phi(n+1)-\sum_{k=1}^{n+1}k=(n+1)^2\phi(n+1)-\frac{(n+1)(n+2)}{2}=P(n)\phi(n+1)-Q(n)$$
Where $$P(n)=(n+1)^2\text{ and }Q(n)=\frac{(n+1)(n+2)}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4206092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
|
2 times complete induction only $n\rightarrow n+1$
$2^n+5^n+56=9\cdot m\Rightarrow$
$2^{n+2}+5^{n+2}+56=4\cdot 2^n+25\cdot 5^n+56=3\cdot 2^n+24\cdot 5^n+2^n+5^n+56=3\cdot 2^n+24\cdot 5^n+9\cdot m$
$3\cdot 2^n+24\cdot 5^n+9\cdot m=3(2^n+8\cdot 5^n)+9\cdot m$
$2^n+8\cdot 5^n=3\cdot x$, because $2^{n+1}+8\cdot 5^{n+1}=2^n+8\cdot 5^n+2^n+8\cdot 5^n+3\cdot 8\cdot 5^n=2\cdot 3x+3y $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
}
|
Any trick for evaluating $\left(\frac{\sqrt{3}}{2}\cos(\theta) + \frac{i}{2}\sin(\theta)\right)^7$? Expressions of the form $(a\cos(\theta) + bi\sin(\theta))^n$ come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with $a = b$. Is there some trick to deal with the case of $a \neq b$, for example when the expression is $\left(\frac{\sqrt{3}}{2}\cos(\theta) + \frac{i}{2}\sin(\theta)\right)^7$?
|
How about this:
$$ \frac{\sqrt{3}}{{2}} \cos \theta = \cos 30 \cos \theta= \frac{1}{2} \left[ \cos(30 - \theta)+\cos \left(30+ \theta \right) \right] =\frac12 (b+a)$$
$$ \frac12 \sin \theta= \sin(30) \sin \theta= \frac12 \left[\cos \left( 30 - \theta\right)- \cos (30+ \theta )\right]=\frac12 (b-a)$$
We have:
$$\frac{1}{2^7}\left[ (a+b) + i(b-a)\right]^7$$
Call $a+b=u=\sqrt{3} \cos \theta$, $b-a=v= \sin \theta$, then for the complicated part:
$$ \left[ u+iv\right]^7 = (u^2+v^2)^{\frac72} \left[\frac{u+iv}{(u^2 + v^2)^{\frac12}} \right]^7=(u^2 +v^2)^{\frac72} e^{i7( \tan^{-1} \frac{v}{u}) }$$
Now back substitute and simplify.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4206791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding $a$ such that $f(x)=a\,|x-b|$ is satisfied by $(2,1)$ and $(10,3)$
I was given the following equation: $$f(x)=a\,|x-b|$$ and the information that both $(2,1)$ and $(10,3)$ are solutions for the equation. The question asked to solve for $a$.
This is what I did: $$a=\frac{1}{2-b}\\[10pt]a=\frac{-1}{2-b}\\[10pt]a=\frac{-3}{10-b}\\[10pt]a=\frac{3}{10-b}$$ all of these seem to be possible solutions for the equation, I just need to find the one that will work with both solution sets. I solved and got multiple values for a and b. The ones that worked for both ordered pairs were the solutions $a=\frac12 \text{ and } b=4$ and $a=\frac14\text{ and } b=-2$.
I feel pretty confident about my answer (perhaps wrongly so :) ) but is there a faster, more efficient way to do this?
|
I see two cases. $b\in(2,10)$ and $b\notin(2,10)$
$f(b) = 0$
In the case that $b$ is not in $(2,10)$ we construct the line through the two points. $b$ is the $x$-intercept of this line. $|a|$ is the slope.
$|a| = \frac {3-1}{10-2} = \frac 14$
Since this slope is postive, $a > 0, b < 2$
$0 = \frac 14 (x-2) + 1\\
0 = \frac 14 (x +2)\\
b = -2\\
f(x) = \frac 14 |x+2|$
In the case that $b\in(2,10)$ The line needs to drop one unit in the interval and then rise 3 units.
$a = \frac {3+1}{10-2} = \frac 12$
And we have a negative slope to the left of $b$ and a postive slope to the right.
$0 = \frac 12 (x-10) + 3\\
0 = \frac 12 (x-4)
b = 4$
or
$0 = \frac 12 (x-2) + 1\\
0 = \frac 12 (x-4)
b = 4$
$f(x) = \frac 12 |x-4|$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4208529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Evaluate $\iiint_{V} (x^2+y^2+z^2)\,dx\,dy\,dz$ in the common part of $2az > x^2+y^2$ and $x^2+y^2+z^2 < 3a^2$ Evaluate $$\iiint_{V} x^2+y^2+z^2 \,dx\,dy\,dz$$
Where $V$ (the integration region) is the common part of the paraboloid $x^2 + y^2 \leq 2az$ and the sphere $x^2+y^2+z^2 \leq a^2$.
I first found the intercept of the paraboloid and the sphere: $x^2+y^2 = 2a^2$. In the $z$ direction, the area is bounded by the sphere on top and the paraboloid on the bottom. Hence I setup the integration:
$$4\int_{0}^{\sqrt{2a^2-x^2}} dy \int_{0}^{\sqrt{2}a} dx\int_{(x^2+y^2)/2}^{\sqrt{3a^2-x^2-y^2}} (x+y+z)^2\,dz$$
However, I have been told that the $z$ direction integral should be from $0$ to $\sqrt{3a^2-x^2-y^2}$. I do not understand why this is the case if we are interested in the common part of the circle and paraboloid. Could someone please explain this?
|
The lower bound of $z$ is correct. It should not be zero. However the order of $x$ and $y$ should be corrected - seems a typo.
But it is easier to do this in either cylindrical coordinates or spherical coordinates.
In cylindrical coordinates,
Sphere is $r^2+z^2 = 3 a^2$
Paraboloid is $2az = r^2$
At intersection, $2az + z^2 = 3a^2 \implies z = a, -3a$. If $a \gt 0$, the intersection is above xy-plane.
So at intersection, $r^2 = 2az = 2a^2 \implies r = \sqrt2 a$ as you already found.
So integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{\sqrt2 a} \int_{r^2/2a}^{\sqrt{3a^2 - r^2}} (r^2 + z^2) \ r \ dz \ dr \ d\theta $
|
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|
A JEE Exam problem on determinants and matrices I am first stating the question:
Let $A=\{a_{ij}\}$ be a $3\times 3$ matrix, where
$$a_{ij}=\begin{cases}
(-1)^{j-i}&\text{if $i<j$,}\\
2&\text{if $i=j$,}\\
(-1)^{i-j}&\text{if $i>j$,}
\end{cases}$$
then $\det(3\,\text{adj}(2A^{-1}))$ is equal to __________
I solved this in the following manner:
$$
A=\left[\begin{array}{lcc}
2 & (-1)^{2-1} & (-1)^{3-1} \\
(-1)^{2+1} & 2 & (-1)^{3-2} \\
(-1)^{3+1} & (-1)^{3 + 2} & 2
\end{array}\right]=\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]
$$
$$\begin{aligned}|A| &=2(4-1)+(-2+1)+(1-2) \\ &=6-1-1=4 \end{aligned}$$
$$
\begin{aligned}
& \operatorname{det}\left(3 \operatorname{adj}\left(2 A^{-1}\right)\right) \\
=& 3^{3}\left|\operatorname{adj}\left(2 A^{-1}\right)\right| \\
=& 3^{3}\left|2^{3} \operatorname{adj}\left(A^{-1}\right)\right| \\
=&(3 \times 2)^{3} \times\left(\left|A^{-1}\right|\right)^{2}\\=&6^3\times\Big(\frac14\Big)\\=&13.5
\end{aligned}
$$
Original image
Is my solution correct?
Note: The problem came in the JEE Main Exam of India, on the 20th of July. The answer given for this question in the Answer Key is 108.
|
No, your solution is not correct, but you are almost done.
Looking through the properties of adjugate matrix, we note that if $A$ is a $n\times n$ matrix then $\text{adj}(cA)=c^{n-1}\text{adj}(A)$ (not $c^n$ as you did) and $\text{adj}(A^{-1})=\det(A^{-1})A$. Therefore
$$\begin{align}\det(3\,\text{adj}(2A^{-1}))&=\det(3\cdot 2^{n-1}\det(A^{-1})A)\\
&=(3\cdot 2^{n-1}\det(A^{-1}))^{n}\det(A)
=
\frac{3^n2^{n(n-1)}}{\det(A)^{n-1}}.
\end{align}$$
Hence when $n=3$ and $\det(A)=4$, we find
$$\det(3\,\text{adj}(2A^{-1})=\frac{3^32^{6}}{4^{2}}=27\cdot 4=108$$
which is precisely the given answer.
|
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|
$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displaystyle\frac{b\cdot\frac{a}{b}+a\cdot\frac{b}{a}}{a+b}\ge \biggl[\left(\frac{a}{b}\right)^b\cdot \left(\frac{b}{a} \right)^a\biggl]^\frac{1}{a+b}$ implies $a^a\cdot{b^b}\ge a^b\cdot{b^a}$
Please help me to solve this problem. Thanks
|
For the first inequality:
Let $f(x) = x\ln x$, then $f''(x) = 1/x > 0$ and hence convex. By the definition of a convex function, we have
$$f\left(\frac{a+b}{2}\right) \le \frac{f(a)+f(b)}{2}$$
which, in our case, becomes
$$\frac{a+b}{2} \ln \left(\frac{a+b}{2}\right) \le \frac{a\ln a + b\ln b}{2}$$
or equivalently
$$\left(\frac{a+b}{2}\right)^{a+b} \le a^ab^b$$
|
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|
The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$).
In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square.
Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$.
This results in the solutions $(3,\pm11)$.
If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as
$$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$
where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares.
Thus,
$$x=10a^2+1,$$
$$x^4+x^3+x^2+x+1=5b^2.$$
Unfortunately, this is where I get stuck.
I can substitute the first equation into the second, giving
$$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$
$$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$
but this doesn't seem to be making progress, even with modular arithmetic considerations.
|
Here is an "elementary" proof. The given diophantine equation $x^5 = 1+2y^2$ admits the obvious solution $x=1, y=0$. Exclude this trivial solution and consider $a=x^5$ as an integral parameter which one wants to represent as the value of the quadratic form $t^2+2y^2$, with unknown integers $(t,y)$. Geometrically, the problem is equivalent to find the points of the sublattice $\mathbf Z^2$ of $\mathbf R^2$ which belong to the ellipse with equation $a= t^2+2y^2$. Since the lattice is discrete and the ellipse is compact, the set $S$ of wanted points is finite. If $S$ is not empty and $t=1$, symmetry w.r.t. the $t$-axis imposes that card $S=2$. If one follows this elementary approach, the only reason for the hypothesis $a=x^5$ seems to be the quick growth of the 5-th power, which allows to determine $S$ without too many trials.
NB : As usual, "elementary" methods often mask the power - and slickness - of more elaborate methods. The general process at work here is, as suggested by the answer given by @leoli1, the representation of a positive integer by a binary quadratic positive definite form.
|
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|
Can the formula for the cubes $a^3 + b^3 + c^3 - 3abc$ be generalized for powers other than 3? I recently learnt out about this formula:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - ac - bc)$$
Is there a way of generalizing for powers other than $3$, i.e.
$$a^n + b^n + c^n + \mathop{???} = \mathop{???} $$
where the RHS is in terms of $(a^{n-1} + b^{n-1} + c^{n-1})$, $(a^{n-2} + b^{n-2} + c^{n-2})$, ... , $(a^{1} + b^{1} + c^{1})$, just like the formula for $n=3$.
|
Too long for a comment:
One possible generalization is the circulant determinant, which tells that
$$ \det \begin{pmatrix}
a_1 & a_2 & a_3 & \cdots & a_n \\
a_n & a_1 & a_2 & \cdots & a_{n-1} \\
a_{n-1} & a_n & a_1 & \cdots & a_{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_2 & a_3 & a_4 & \cdots & a_1
\end{pmatrix}
= \prod_{\omega \mathop{:} \omega^n = 1} (a_1 + a_2 \omega + a_3 \omega^2 + \cdot + a_n \omega^{n-1}). $$
When $n = 2$, this gives
$$ a^2 - b^2 = \det\begin{pmatrix} a & b \\ b & a \end{pmatrix} = (a + b)(a - b). $$
When $n = 3$, then with any $\omega$ satisfying $\omega^2 + \omega + 1 = 0$,
\begin{align*}
a^3 + b^3 + c^3 - 3abc
&= \det\begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix} \\
&= (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \\
&= (a + b + c)(a^2 + b^2 + c^2 + (\omega + \omega^2)(ab + bc + ca)) \\
&= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).
\end{align*}
Unfortunately, the formula quickly goes out of control even starting at $n = 4$:
\begin{gather*}
a^4 - b^4 + c^4 - d^4 - 2a^2c^2 + 2b^2d^2 - 4da^2b + 4ab^2c - 4bc^2d + 4cd^2a\\
= (a+b+c+d)(a-b+c-d)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd)
\end{gather*}
|
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|
To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}$ = $(\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8})+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\frac{1}{8}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{15}{56}}{\frac{55}{56}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{3}{11} +\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\frac{1}{18}}$ = $\tan^{-1}\frac{\frac{65}{198}}{\frac{195}{198}}$ = $\tan^{-1}\frac{1}{3}$ = $\cot^{-1}3$
Second Method: we know that $\tan^{-1}x+\cot^{-1}x = \frac{π}{2}$ for $x \in \Bbb R$. So $\cot^{-1}x = \frac{π}{2} - \tan^{-1}x$.
Now $$\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \frac{π}{2} - \tan^{-1}7+ \frac{π}{2} - \tan^{-1}8+ \frac{π}{2} - \tan^{-1}18 \implies \\ \frac{3π}{2} - (\tan^{-1}7+\tan^{-1}8+\tan^{-1}18) = \frac{3π}{2} - \tan^{-1}\frac{7+8+18-7×8×18}{1-7×8-8×18-7×18} \\ = \frac{3π}{2} - \tan^{-1}\frac{-975}{-325} = \frac{3π}{2} - \tan^{-1}3 = π + (\frac{π}{2} - \tan^{-1}3) = π + \cot^{-1}3 .$$ Also we know that $\tan x$ and $\cot x$ are periodic function with period $π$. Please help me in Second Method. Is $π + \cot^{-1}3 = \cot^{-1}3$ ?. If yes, then elaborate it.
|
${tan^{-1}(1/7)+tan^{-1}(1/8)+tan^{-1}(1/18)}$
*
*${tan^{-1} \left[ \frac{ (1/7)+(1/8) } { \left( 1 \right) - (1/7)(1/8) } \right] + tan^{-1}(1/18) }$
*= ${tan^{-1} \frac {3}{11} + tan^{-1} \frac{1}{18}}$
*=${tan^{-1} \left[ \frac {(3/11)+(1/18)} {\left (1 \right) - (3/11)(1/18)} \right] }$
*= ${tan^{-1} \frac{1}{3} }$ or ${cot^{-1} {3}}$
|
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|
Minimum value of algebraic expression This is a high school algebra problem,
If the minimum value of the expression $f(a,b) = \sqrt{a^2 + b^2 - 10a - 10b +50} +\sqrt{b^2 -4b +20} + \sqrt{a^2 - 14a +74}$ is $k$.
Which occurs at $a = \alpha$, $b = \beta$.
Find the value of $k + 4{\alpha} + 3\beta$
It can be seen that the expression can be written as $f(a,b) = \sqrt{(a-5)^2 + (b-5)^2} + \sqrt{(b-2)^2 + 4^2} + \sqrt{(a-7)^2 + 5^2}$
How do I proceed further from here and could there be a geometric solution?
|
Replace $a\to n+5$ and $b\to m+5$
Now the expression simplifies to $ \sqrt{n^2 + m^2} + \sqrt{ 4^2+(m+3)^2} + \sqrt{(n-2)^2 + 5^2}$
This is the sum of distance between the following pairs of points
*
*$(n,0)$ and $(2,5)$
*$(0,m)$ and $(n,0)$
*$(0,m)$ and $(-4,-3)$
If $m\ge0$,
we get $EA>EF$ and $FG+GD>AD$ (refer the following graph). Therefore the path through $A(0,0)$ gives the minimum sum.
If $m\le0$,
construct $AC$ the angle bisector of reflex $∠EAD$. It is easy to see that $∠EAC$ and $∠CAD$ are obtuse therefore we get $EF+FC>EC>EA$ and $CG+GD>CD>AD$ (refer the following graph). Therefore the path through $A(0,0)$ gives the minimum sum.
The minimum is obtained when $(n,m)=(0,0)$ i.e. when $a=b=5$.
|
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|
Proof by induction: induction hypothesis question In this question I found online:
*
*Show that $$ S(n):0^2 + 1^2 + 2^2 + · · · + n^2 =
\frac{n(n + 1)(2n + 1)}{6}$$
I don't understand why for S(k+1) they wrote:
$$S(k+1):1^2+2^2+3^2+⋯+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
instead of:
$$S(k+1):1^2+2^2+3^2+⋯+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
Why is the $k^2$ included in the $S(k+1)$ step I don't get it surely you just substitute $k+1$ for $n$ so I don't know why $k^2$ is needed there because in other proof by induction questions I've done for example
for this proof: $$n<2^n$$
for the $k+1$ step the answer was not $$k + k + 1 < 2^k+1$$
it was:
$$k+1 < 2^k+1$$
EDIT
People have mentioned that my version is correct and they just wrote it in a different way but why could I not prove both sides were equal? This is what I did:
$$S(k+1):1^2+2^2+3^2+⋯+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
I expanded out $$(k+1)^2$$ which gave me $$k^2+2k+1$$
so going back to the k+1 statement we have:
$$S(k+1):1^2+2^2+3^2+⋯+k^2+2k+1=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
I subbed in $$\frac{k(k+1)(2k+1)}{6}$$ for $$1^2+2^2+3^2+⋯+k^2$$ (the induction hypothesis) in the k+1 statement and got:
$$\frac{k(k+1)(2k+1)}{6} + 2k+1=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
long story short my proof failed and I couldn't prove both sides were equal but why was this if my version of S(k+1) was not incorrect?
|
Your mistake is in the last equation. It should read:
$$\frac{k(k+1)(2k+1)}{6} + (k+1)^2= \frac{(k+1)(k(2k+1)+6(k+1))}{6}= \frac{(k+1)(2k^2+7k+6)}{6} \\=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
|
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|
Proof strategy for showing $a^2 + b^2 \neq c^2$ for $a,b$ odd, $c$ even. I am trying to understand a proof that $a^2 + b^2 = c^2$ cannot hold for $a,b$ odd and $c$ even.
The proof I am reading, and I believe I understand its steps, proceeds as follows. If $a,b$ are odd, then $a,b$ are congruent to either $1$ or $3$, mod $4$. (I don't know how to prove this other than by writing out cases, and it seems clear; if anyone has a more formal explanation, I would definitely be interested.) In either event, their square is equal to $1$ mod $4$, so the sum of the squares is equal to $2$, mod $4$. If $c$ is even, then its equivalent to $0$ or $2$ mod $4$, so its square its equivalent to $0$ mod $4$. As $2 \neq 0$ mod $4$, $a^2 + b^2 \neq c^2$.
I believe the argument is that IF $a^2 + b^2 = c^2$, then they would have to be equivalent to each other in any modulus. Because they don't agree mod $4$, that can't be the case.
Is that correct? Is their another way to understand the above statement about congruences mod $4$?
|
Yes your reasoning is right.
For the first doubt note that for $a$ odd we have $a=4k+1$ or $a=4k+3$ and in any case since $a=2h+1$ we have $a^2=4h^2+4h+1\equiv 1 \pmod 4$.
For $c$ even we have that $c^2=4h^2\equiv 0 \pmod 4$.
Therefore we have
$$a^2+b^2=c^2 \implies a^2+b^2 \equiv c^2 \pmod 4$$
but as you noticed the latter is false and then $a^2+b^2=c^2$ is impossible for $a,b$ odd and $c$ even.
|
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|
Troubleshooting a trigonometry/geometry question - spot the (silly) mistake! I'm back again!
Again, another error - not sure if it's them or me this time...
Here is the question
In order to calculate the shaded region, I added some lines to the diagram. I realise that there are other ways to solve this (and indeed the solution give uses a different breakdown and there's seems correct).
I then calculated the area of the sector ADB - the area of the triangle ADB and added this to the area of the sector BOD. I then multiplied that value by 2 to get the area of the shaded region.
Here is my working (Let a = alpha and t = theta)
$Area = 2[\frac12(3r)^2(\frac{a}2)-\frac12(3r)^2\sin(\frac{a}2)+\frac12(2r)^2(\frac{t}2)]$
Giving me a final answer of $r^2[(\frac{9a}2)-9\sin(\frac{a}2)+2t]$
Can you see where I went wrong...?
EDIT:
Here is the solution they provided:
|
The two answers are equivalent.
You have: $r^2[(\frac{9 \alpha}2)-9\sin(\frac{\alpha}2)+2 \theta]$
They have: $ [\frac 9 2(\alpha -\sin \alpha) + 2(\theta - \sin \theta)]r^2$
The triangle $ABD$ is isosceles, so $\frac \alpha 2 + 2 \frac \theta 2=\pi$
$\frac \alpha 2 = \pi-\theta \Rightarrow \sin \frac \alpha 2=\sin(\pi-\theta)=\sin \theta$
That changes your answer to $r^2[9(\pi - \theta)-9\sin \theta+2 \theta]=r^2[9\pi - 9 \sin \theta -7 \theta]$
$\alpha +2\theta = 2\pi \Rightarrow \alpha = 2\pi - 2\theta$
$\sin \alpha = \sin (-2\theta)=-\sin (2\theta)=-2\sin \theta \cos \theta$
That changes their answer to $ [\frac 9 2(2\pi - 2\theta +2\sin \theta \cos \theta) + 2(\theta - \sin \theta)]r^2 = [9\pi - 9\theta +9\sin \theta \cos \theta + 2\theta - 2\sin \theta]r^2 $
$=r^2[9\pi +(9\cos \theta - 2)\sin \theta - 7\theta]$
These look more alike now they are both in terms of $\theta$ but there is still a difference.
We can apply the cosine rule to the triangle $ADB$ to get $\cos (\frac \theta 2)=\frac{2^2+3^2-3^2}{2.2.3}$
So $\cos (\frac \theta 2)=\frac 1 3$
$\cos \theta = \cos (2 \frac \theta 2)=2 \cos^2 (\frac \theta 2) -1 = 2(\frac 1 3)^2-1=-\frac 7 9$
Then $9\cos \theta - 2= -7-2=-9$
which turns their answer into yours.
|
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|
How was the closed form of this alternating sum of squares calculated? I am reading through this answer at socratic.org.
The question is to find the closed form of the sum
$$1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+\ldots.$$
I understand that, if the terms were added, the sum would be
$$
\sum_{n=1}^{N} n^{2}=1^{2}+2^{2}+\ldots+N^{2}.
$$
The person goes on to say, if the series were not alternating, the sum would be
$$
S=\frac{N(N+1)}{2}
$$
But is that correct? I thought the sum of the first $N$ squares would be
$$
\frac{N(N+1)(2N+1)}{6}.
$$
Lastly, I understand moving the $-1$ constant out of the summation as such
$$
=-\sum_{n=1}^{N}(-1)^{n} n^{2}
$$
But I am completely missing how the final closed form was calculated
$$
S_{N}=-\frac{(-1)^{N} N(N+1)}{2}
$$
|
(This answer is an expansion of my comment. It shows an alternative method for computing the sum.)
If $n$ is even, then an easy way to compute $1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2$ is to consider the difference between each pair of terms:
\begin{align}
&1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2 \\[4pt]
=&(1-2)(1+2)+(3-4)(3+4)+\dots +((n-1)-n)((n-1)+n) \\[4pt]
=&-1(1+2)-1(3+4)-\dots-1((n-1)+n) \\[4pt]
=&-1(1+2+3+4+\dots+(n-1)+n) \\[4pt]
=&-\frac{n(n+1)}{2}
\end{align}
If $n$ is odd, then $n-1$ is even, and so we can use the above formula:
$$
1^2-2^2+3^2-4^2+\dots+(n-2)^2+(n-1)^2=-\frac{(n-1)n}{2} \, .
$$
Then, adding $n^2$ to both sides, we find that
$$
1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2=\frac{n(n+1)}{2} \, .
$$
Therefore, regardless of the parity of $n$,
$$
1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2=(-1)^{n+1}\cdot\frac{n(n+1)}{2} \, .
$$
|
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|
How to solve a limit that can be factored but doesn't help? I saw examples that can be factored, eliminating the part that causes the indetermination, none of this type. The other option is by rationalize but dont know how to apply it here.
$$\lim_{x \to 4} \frac{2x^2+7x+5}{x^2-16}$$
I tried by factoring, doesn't help
$$\lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x-4)(x+4)} \\$$
UPDATE: I make a mistake, the numerator is ${2x^2+7x+5}$ not ${x^2+7x+5}$, really sorry.
|
HINT
Your limit is in the form
$$\lim_{x \rightarrow 4} \frac{2x^2+7x+5}{(x-4)(x+4)} = \lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x+4)}\cdot \lim_{x \rightarrow 4} \frac{1}{x-4}$$
with
$$\lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x+4)} =\frac{65}{8}$$
then all boils down in that one
$$\lim_{x \rightarrow 4} \frac{1}{x-4} $$
|
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|
Determine the smallest $k$ such that $\log (1 +e^x) < k + x ?$ Determine the smallest $k$ such that $\log (1 +e^x) < k + x $ where $k$ is constant and $x \in (0,\infty)$
My attempt :$$\log(1+e^x)= e^x -\frac{e^{x}}{2} +\frac{e^{2x}}{3}+.....$$
$$e^x = 1 +x+x^2/2^2+...$$
$$\log(1+e^x)= (1+x+x^2/2^2+....)-\frac{1}{2}(1+x+x^2/2^2+...)\\+\frac{1}{3}(1+2x+(2x)^2/2^2+...)-\frac{1}{4}(1+3x+(3x)^2/2^2+...$$
$$=(1-\frac{1}{2} +\frac{1}{3} -\frac{1}{4}+...) + x(1-1/2+2/3-3/4+....) + \text{higher degree} $$
$$\implies\log(1+e^x) < \log 2 + x \implies k=\log 2$$
|
$f(x)=e^x (e^k -1)>1$ is strictly increasing.
$\\ $ $f(x\to +0)>1 \implies k>\ln 2 $ or $\inf k =\ln 2 $.
|
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|
Finding the inverse of a block $2\times2$ square matrix $ \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ where $A$ is a square invertible matrix. The hint I got was to rewrite the original matrix as a product of 3 matrices and use the property for inverse of product of matrices $(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}$
$\begin{align} \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} &= \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix} \begin{bmatrix} I & A \\ 0 & I \end{bmatrix}
\end{align}\\ $
Then
$\begin{align}
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1}= \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} ^{-1} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1} \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}^{-1}
\end{align}$
The first 2 matrices on the RHS are invertible but the third is not, so I might have gone down the wrong path with the product of 3 matrices I think.
The actual inverse for $\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ is $\begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}$ which can be found by using the formula at the end of section 3 in Lu and Shiou. I would like to know how to solve the problem without resorting to using a formula. Thanks for any tips.
|
Ben Grossmann already pointed out that you have a flaw in the product. You can't factor invertible matrix into a product of matrices some of which are not invertible. This is an easy consequence of Binet-Cauchy formula: $\det(AB) = \det A\cdot\det B$.
There is an approach that doesn't require anything smart to find the inverse. All you have to remember is how you invert a $2\times 2$ matrix:
$$
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix}
d & -b\\
-c & a
\end{pmatrix}.
$$
Now, this formula uses that multiplication in a field is commutative, but let us inspect how far can this idea get us in your case:
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & -A\\
-A^T & I
\end{pmatrix} = \begin{pmatrix}
-AA^T & 0\\
0 & -A^TA
\end{pmatrix}.
$$
Multiply everything on the right by \begin{pmatrix}
-(AA^T)^{-1} & 0\\
0 & -(A^TA)^{-1}
\end{pmatrix} to get
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & -A\\
-A^T & I
\end{pmatrix}\begin{pmatrix}
-A^{-T}A^{-1} & 0\\
0 & -A^{-1}A^{-T}
\end{pmatrix} = \begin{pmatrix}
I & 0\\
0 & I
\end{pmatrix},
$$
that is
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & A^{-T}\\
A^{-1} & -A^{-1}A^{-T}
\end{pmatrix} = \begin{pmatrix}
I & 0\\
0 & I
\end{pmatrix}.
$$
|
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|
Find the limit $\lim_{x \to 0}{\left(\frac{\tan x}{x}\right)}^{1/{x^2}}$ using l'Hôpital's Rule. Here is the limit I'm trying to find.
$$\lim_{x \to 0}{\left(\frac{\tan x}{x}\right)}^{1/{x^2}}$$
Now, since it takes an indeterminate form at $x=0$, I'm taking natural logarithm and trying to find the following limit.
$$
\begin{align}
L
& = \lim_{x \to 0}\ln{\bigg(\frac{\tan{x}}{x}\bigg)}^{1/{x^2}} \\
& = \lim_{x \to 0}\frac{\ln{\left(\frac{\tan{x}}{x}\right)}}{x^2}
\end{align}
$$
Now, l'Hôpital's Rule cannot be applied here since the numerator
$ f(x)
= \ln\left({\frac{\tan{x}}{x}}\right)
$
is undefined at $x = 0$. What should be done here?
|
For this type of problem, it is simpler to get the limit using the Maclaurin series of the tangent function and the logarithm function. Letting $z = 1/x^2$ we can write:
$$\begin{align}
\frac{1}{x^2} \log \bigg( \frac{\tan x}{x} \bigg)
&= \frac{1}{x^2} \Bigg[ \log \tan (x) - \log(x) \Bigg] \\[6pt]
&= \frac{1}{x^2} \Bigg[ \bigg( x + \frac{x^3}{3} + \frac{2 x^5}{15} + \frac{17 x^7}{315} + \cdots \bigg) - \log(x) \Bigg] \\[6pt]
&= \frac{1}{x^2} \log \bigg( 1 + \frac{x^2}{3} + \frac{2 x^4}{15} + \frac{17 x^6}{315} + \cdots \bigg) \\[6pt]
&= z \log \bigg( 1 + \frac{1}{3z} + \frac{2}{15 z^2} + \frac{17}{315 z^3} + \cdots \bigg) \\[6pt]
&= z \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \bigg( \frac{1}{3z} + \frac{2}{15 z^2} + \frac{17}{315 z^3} + \cdots \bigg)^n \\[6pt]
&= z \Bigg[ \frac{1}{3z} + O(z^{-2}) \Bigg] \\[6pt]
&= \frac{1}{3} + O(z^{-1}). \\[6pt]
\end{align}$$
Taking $x \rightarrow 0$ gives $z \rightarrow \infty$ which gives:
$$\frac{1}{x^2} \cdot \log \bigg( \frac{\tan x}{x} \bigg)
= \frac{1}{3} + O(z^{-1}) \rightarrow \frac{1}{3},$$
so we have:
$$\bigg( \frac{\tan x}{x} \bigg)^{1/x^2} \rightarrow \exp(1/3) = 1.395612.$$
|
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|
Intuition behind getting two straight lines as result Question:
Find the equation of the straight line that passes through $(6,7)$ and makes an angle $45^{\circ}$ with the straight line $3x+4y=11$.
My solution (if you want, you can skip to the bottom):
Manipulating the given equation to get it to the slope-intercept form,
$$3x+4y=11...(i)$$
$$\implies 4y=-3x+11$$
$$\implies y=\frac{-3}{4}x+\frac{11}{4}$$
Let, the slope of (i) is $m_1=\frac{-3}{4}$, and the slope of our desired equation is $m_2$. Now, according to the question,
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}...(1)$$
$$\implies 1=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2}$$
$$\implies \pm \frac{3}{4}+m_2=1-\frac{3}{4}m_2...(ii)$$
Picking positive value from (ii),
$$\frac{3}{4}+m_2=1-\frac{3}{4}m_2$$
$$\implies m_2(1+\frac{3}{4})=1-\frac{3}{4}$$
$$\implies m_2=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}$$
$$\implies m_2=\frac{1}{7}$$
Picking negative value from (ii),
$$-\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -\frac{3}{4}-m_2=1-\frac{3}{4}m_2$$
$$\implies -m_2(1-\frac{3}{4})=1+\frac{3}{4}$$
$$\implies m_2=-\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$$
$$\implies m_2=-7$$
Picking $m_2=\frac{1}{7}$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=\frac{1}{7}$$
$$\implies 7y-49=x-6$$
$$\implies -x+7y-43=0$$
$$\implies x-7y+43=0...(iii)$$
Picking $m_2=-7$, the equation of the straight line that passes through $(6,7)$,
$$\frac{y-7}{x-6}=-7$$
$$\implies -7x+42=y-7$$
$$\implies 7x+y-49=0...(iv)$$
The general form of equation (1) is,
$$\tan\theta=\pm \frac{m_1-m_2}{1+m_1m_2}$$
Here, $\pm$ has been included to include both the acute and the obtuse angles that are formed when two lines with slopes $m_1$ and $m_2$ intersect each other.
Now, I used this equation to find the straight line that makes $45^{\circ}$ with (i). Why am I getting 2 values of $m_2$ when there is only one value of $m_2$ in the general form of the equation? How can I reconcile between my getting of 2 values of $m_2$ with the $\pm$ sign arising due to the acute and obtuse angles?
|
You have a right triangle with the right angle at (6, 7) and the line 3x+ 4y= 11 as hypotenuse. The "two lines" are the two legs of the triangle.
|
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|
A square root of −4 modulo the ideal generated by $(x^2 + 1)^2$ Let $S= \mathbb R[x]$ denote the polynomial ring in one variable over the field $\mathbb R$ of real numbers. Find a
monic polynomial of least degree in $S$ that is a square root of $−4$ modulo the ideal $I$ generated by
$(x^2 + 1)^2$.
Let $p(x)$ be a monic polynomial that is a square root of $−4$ modulo the ideal $I$.
Then ${(p(x))}^2 = -4 + \langle(x^2 + 1)^2\rangle$.
$p(x)$ is of the form $a_0 + a_1x + a_2x^2 + x^3$. Thus, $(a_0 +a_1x + a_2x^2 + x^3)^2 = -4 + \langle (x^2 + 1)^2\rangle$.
Solving this rigorously (reducing and equating) may get such $p(x)$.
But Is there any other simple way?
|
The basis $1,x,x^2,x^3$ has no relation to $(x^2+1)^2$, which is the polynomial which you are trying to go modulo. Instead, what is usually done is traditionally called "lifting" as suggested in the comments by Jyrki. (So it wasn't misread, it was a hint).
The idea of lifting is simple : if an equation is true modulo $p(x)$, then it's also true modulo $q(x)$ where $q$ is a factor of $p$. Therefore, it should be possible to solve the equation modulo $q(x)$ and then "lift" it to a solution modulo $p(x)$, using the solution obtained modulo $q(x)$.
In this scenario, $x^2+1$ is a factor of $(x^2+1)^2$, so you want to first solve this equation modulo $x^2+1$, and then modulo $(x^2+1)^2$, by lifting.
Let's first solve $q(x)^2 = -4 \pmod{x^2+1}$. Now, here the degree of the polynomial $q$ can be chosen to be at most $1$ i.e. $q(x) = ax+b$ for some $a,b$. Then, note that $(ax+b)^2 = a^2x^2+2abx+b^2$ , which modulo $x^2+1$ becomes $2abx + (b^2-a^2)$. Thus, $2abx + (b^2-a^2) = 4 \pmod{x^2+1}$, hence $2ab=0$ and $b^2-a^2 = 4$. We easily get $b=0,a=\pm 2$ from here. Sticking with $a=2$ gives us one of the solutions, $q(x) = 2x$. (And the set of all solutions would be $2x$ or $-2x$ modulo $x^2+1$).
Now, suppose that $p(x)$ satisfies $p(x)^2 = -4 \pmod{(x^2+1)^2}$. Then, $p$ also satisfies $p(x)^2 = -4 \pmod{x^2+1}$, but then we've already seen that one solution to this is $q(x)$! Therefore, there could be a solution $p(x) \equiv q(x) \equiv 2x\pmod{x^2+1}$. In fact, there is one. Similarly, there will also be a solution for $p(x) \equiv -2x \pmod{x^2+1}$, as I urge you to calculate (later on, you can see the similarity between the solution you get from here, and the solution I'm about to derive : they will be $-1$ times each other).
Thus, suppose that $p(x) = (ax+b)(x^2+1) - 2x$ for some $a,b$ (following the realization that $p$ has degree at most $3$). We need this to satisfy $p(x)^2 = -4 \pmod{(x^2+1)^2}$. While squaring $p(x)$, the entire $(ax+b)(x^2+1)$ term squares to something that is $0$ modulo $(x^2+1)^2$. Neglecting this gives $$
2(ax+b)(x^2+1)(-2x) + (4x^2) = -4 \pmod{(x^2+1)^2}
$$
which ... we don't need to expand and make life complicated! Why? Because taking the $-4$ to the other side :
$$
2(ax+b)(x^2+1)(-2x) + (4x^2 +4) = 0 \pmod{(x^2+1)^2} \\
\implies 2(ax+b)(-2x)(x^2+1)+4(x^2+1)=0 \pmod{(x^2+1)^2} \\
\implies (x^2+1) \left[(2(ax+b)(-2x) + 4)\right] = 0 \pmod{(x^2+1)^2}
$$
which happens if and only if
$$
(2(ax+b)(-2x) + 4) = 0 \pmod{x^2+1}
$$
and expanding this out gives us $-4ax^2-4bx + 4 = 0 \pmod{x^2+1}$, which then simplifies to $-4bx + 4a+4 =0 \pmod{x^2+1}$. As the LHS is linear, this can occur only if $b = 0$ and $a = -1$. Thus, the final $p(x)$ is $-x(x^2+1) - 2x = -x^3-3x$.
To confirm this, one can check that $(-x^3-3x)^2 =-4 \pmod{(x^2+1)^2}$.
The other solution, will in fact be $x^3+3x$, which you can note is a different element modulo $(x^2+1)^2$. Our analysis confirms that these are the only two solutions.
|
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|
Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If we can prove that $\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$ is always greater than $0$ we can find the minima of the function. How can we prove that the term is always positive?
And how can we prove that $\displaystyle \int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$
|
If the function you are integrating is always positive, then the integral must be positive as well. Take a look at the function being integrated. We have
$$f(x) = \frac{x^{n+2}}{(x+1)^2(n+2)}$$ with $0\leq x \leq 1$. Given that $x\in [0,1]$, is it true that $f(x)$ is always positive?
I assume also that you mean to find the maxima of the $f(x)$ as above. To do this, we can rely on the fact that $f$ is a differentiable function. Hence we can use the first derivative test to find the local maxima, then compare the values of the local maxima to the endpoints $x=0$ and $x=1$.
|
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|
Independent chances of $3$ events ${a\over{a+x}}$, ${b\over{b+x}}$, ${c\over{c+x}}$ Here's a problem from my probability textbook:
Of three independent events the chance that the first only should happen is a; the chance of the second only is $b$; the chance of the third only is $c$. Show that the independent chances of the three events are respectively$${a\over{a+x}}, \quad {b\over{b+x}}, \quad {c\over{c+x}},$$where $x$ is a root of the equation$$(a+x)(b+x)(c+x) = x^2.$$
Here's what I did. We have$$a = p_1(1 - p_2)(1 - p_3), \quad b = (1-p_1)p_2(1-p_3), \quad c = (1-p_1)(1-p_2)p_3.$$Without loss of generality let's consider $a$. We have$$p_1 = {a\over{(1 - p_2)(1 - p_3)}}.$$If we assume the result we want to show, then this equals$$p_1 = {a\over{\left(1 - {b\over{b+x}}\right)\left(1 - {c\over{c+x}}\right)}} = {a\over{{{x^2}\over{(b+x)(c+x)}}}} = {a\over{a+x}}.$$However, we assumed in part what we wanted to show, which possibly makes this circular.
Another observation I noticed is that$$p_1p_2p_3 + p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 + (1-p_1)(1-p_2)(1-p_3) + a + b + c = 1.$$However, despite what I've tried, I'm stuck and do not know how to proceed further. Could anybody help me? Is there a way to turn my circular approach into a noncircular one?
|
To get corner cases out of the way, if $p_1=1$ then $b=c=0$, and $x=1-a$ will satisfy all the wanted properties. If $p_1=0$, then $a=0$ and the problem reduces to a similar one with only two events. The same goes for $p_2$ and $p_3$, so from here on assume $0<p_1<1$, $0<p_2<1$, and $0<p_3<1$.
From your observation
$$a = p_1(1 - p_2)(1 - p_3), \quad b = (1-p_1)p_2(1-p_3), \quad c = (1-p_1)(1-p_2)p_3$$
we can get
$$ (1-p_1)(1-p_2)(1-p_3) = \frac{1-p_1}{p_1} a = \frac{1-p_2}{p_2} b = \frac{1-p_3}{p_3} c $$
If we call this value $x = (1-p_1)(1-p_2)(1-p_3)$, then $x>0$ and
$$ x = \frac{1-p_1}{p_1} a $$
$$ \frac{x}{a} = \frac{1}{p_1} - 1 $$
$$ \frac{1}{p_1} = \frac{a+x}{a} $$
$$ p_1 = \frac{a}{a+x} $$
Solving for $p_2$ and $p_3$ works just the same.
To show that $(a+x)(b+x)(c+x) = x^2$, it will be easier to look at the left side divided by $x^3$:
$$ \begin{align*} \frac{(a+x)(b+x)(c+x)}{x^3} &= \left(1+\frac{a}{x}\right) \left(1+\frac{b}{x}\right) \left(1+\frac{c}{x}\right) \\
&= \left(1+\frac{p_1}{1-p_1}\right) \left(1+\frac{p_2}{1-p_2}\right) \left(1+\frac{p_3}{1-p_3}\right) \\
&= \frac{1}{(1-p_1)(1-p_2)(1-p_3)} = \frac{1}{x}
\end{align*} $$
Now just multiply both sides by $x^3$ to get the polynomial equation in the question.
|
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|
What is the value of the $\measuredangle IEP$ in the figure below? In a right triangle ABC straight at B, $\measuredangle C = 37^\circ. $If $E$ is an excenter in relation to BC, I is an incenter and $P$ is the point of tangency of the circle inscribed with AC.
calculate $\measuredangle IEP$
My progres::
I made the drawing and marked the relationships found
Let $KE \perp CB$
AE is angle bissector = $\frac{53}{2} \implies \measuredangle AEJ = \frac{137}{2}\\
JEK =90^\circ\implies \measuredangle KEI = \frac{53}{2}$
but I still haven't found the link to find $\angle IEP$
|
HINT:
$\small{\triangle ABC}$ is a $\small{3,4,5}$ triangle and $\small{\triangle AJE}$ is a right triangle with perpendicular sides in the ratio $\small{1:2}$. Therefore say $\small{AB=3, BC=4, AC=5}$ and $\small{BJ=JE=3}$. This leads $\small{AE=3\sqrt 5}$.
Let $\small{AB}$ touches incircle at $\small Q$, then define $\small{AP=x, BQ=y, CP=z}$. Then $$\small{\begin{align}x+y &=3 \\ y+z &=4 \\ x+z&=5 \\ \end{align}}$$ By solving equations, we get $\small{x=2, y=1, z=3}$.
So we know $\small{AP=2, AE=3\sqrt 5}$ and $\small{\angle PAE=26.5^\circ}$. Thus by applying cosine rule and sine rule to $\small{\triangle APE}$ we can evaluate $\small{\angle IEP}$.
(Reminder: $\small{\tan 26.5^\circ\approx\frac 12}$) By cosine rule, $$\small{\cos 26.5^\circ=\frac{2}{\sqrt 5}=\frac{2^2+(3\sqrt 5)^2-(PE)^2}{2\cdot 2\cdot 3\sqrt 5}}$$ $$\small{\implies PE=5}$$ By sine rule, $$\small{\frac {5}{\sin 26.5^\circ}=\frac{2}{\sin \theta}}$$ $$\small{\implies \theta =\arcsin \left(\frac {2\sqrt 5}{25}\right)\approx 10.30^\circ=\angle IEP}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4251567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
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Fractions in Questions and Answers
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