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Question Sequences and Series Prove that the sequence $(a_n)_{n \geq 1}$ defined by $$ a_n= \sum_{k=1}^{n} \frac{1}{k} - \ln(n+1)$$
is increasing and bounded. It is on the study guide for my final exam, which is tomorrow so I am trying to figure it out. Thanks
Can someone please help?
| The part that $a_n$ is increasing is already done. We focus on proving $a_n$ is bounded above. We have:
$a_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} - \ln(n+1) = 1 + \dfrac{1}{2} + \cdots + \dfrac{1}{n} - \ln n - \ln\left(1+\frac{1}{n}\right) = 1 + \dfrac{1}{2} + \cdots + \dfrac{1}{n-1} - \ln n+ \left(\dfrac{1}{n} - \ln\left(1+\frac{1}{n}\right)\right)$. Apply the MVT for the function $f(x) = x - \ln(1+x)$ on $(0,x)$ we have: $x - \ln(1+x) = f'(c)\left(x-0\right) = \dfrac{cx}{1+c} < cx, 0 < c < x$. Let $x = \dfrac{1}{n} \Rightarrow \dfrac{1}{n} - \ln(1+\frac{1}{n}) < c\dfrac{1}{n} < \dfrac{1}{n^2}$. By repeating this pattern back to $\ln 2$ we have that $a_n < 1 + \dfrac{1}{2^2} + \cdots + \dfrac{1}{n^2} < \dfrac{\pi^2}{6}$, proving boundedness of $a_n$.
| {
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Which (convergent) series can one find the sum of? I know about geometric series and how one can find the sum when they are convergent.
I also have heard that one can prove that the $p$-series
$$\sum_{n=1}^{\infty} \frac{1}{n^2}$$
has sum $\pi^2/6$ (but I don't know how this is actually proved).
This might be an unanswerable question, but are there any tools or general rules that will tell you when you can find the exact sum of a given series? Are there, for example, rules/tools for how to find the exact value of any $p$-series? (like with geometric series).
| Thomas Andrews is right when he says that there are no general rules. Often this type of problem, we must use a more intuitive method.
I know the p-series for p = 2. The resolution method in this series comes from Euler himself.
(1) $\sin(x) = x - x^3/3! + x^5/5! -x^7/7! + ...$ (Taylor series)
(2) $\sin(x)/x = 1 - x^2/3! + x^4/5! -x^6/7! + ...$ (divided by x)
(3) Now, the roots (intersections with the x-axis) of sin(x)/x occur precisely at x = $n\pi$ where n = ±1,±2,±3,... Let us assume we can express this infinite series as a (normalized) product of linear factors given by its roots, just as we do for finite polynomials
\begin{align} \frac{\sin(x)}{x} & {} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ & {} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align}
(4) If we formally multiply out this product and collect all the $x^2$ terms (we are allowed to do so because of Newton's identities), we see that the $x^2$ coefficient of sin(x)/x is
$$
1/\pi^2 + 1/(4\pi^2) + 1/(9\pi^2) + ... = (1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}
$$
(5) But from the original infinite series expansion of sin(x)/x, the coefficient of $x^2$ is −1/(3!) = −1/6. These two coefficients must be equal...
$$
-1/6 = -(1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}
$$
(6) Multiplying through both sides of this equation by $-\pi^2$ gives the sum of the reciprocals of the positive square integers.
$$
\sum_{n=1}^\infty\frac{1}{n^2} = \pi^2/6
$$
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Basic Logs, simplifying So basically the question goes: $\log_{14} 2 = a$, $\log_{14} 3 = b$, solve for $\log_7 24$. I have attached my work so far and the answer is ${3a+b}\over{1-a}$... I just got stuck at the end. Thanks
| In your approach you tackle the problem using the definition: $\log_x y = z \iff y=x^z.$ You just need to solve $\log_7 2$, and you'd be done. Jump to the second-to-last line for that.
We know that $\log_{14} 2 = a$ means $2=14^a$, and $\log_{14} 3 = b$ means $3=14^b$.
We'd like to solve for $\log_{7} 24 = x.$ In other words, $24=7^x$.
Now notice that $2\cdot 2\cdot 2 \cdot 3= 24$.
Substituting what we know from our first line above, we have:
$$14^a 14^a 14^a 14^b = 14^{3a+b} = (2\cdot 7)^{3a+b} = \color{green}{2^{3a+b} 7^{3a+b} = 7^x}.$$
Taking $\log_7$ of both sides of the equation in green, we have
$$(3a+b)\log_7 2 +(3a+b) = x.$$
So, $$(3a+b)(\log_7 2 + 1) = x.$$
But what is $\log_7 2$? Well, notice that $2=14^a=2^a7^a$, which is equivalent to $2^{1-a}=7^a$. So $(1-a)\log_7 2 = a$. Therefore $\log_7 2 = \dfrac{a}{1-a}$.
So, $$(3a+b)\left(\frac{a}{1-a} + 1\right) = x.$$
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Iterated limits of $\frac{x-y}{x^3-y}$ Why it the following limits look like this: $$\lim_{x\rightarrow -1} \frac{x-y}{x^3-y}=1$$ but suprisingly $$\lim_{y\rightarrow -1} \frac{x-y}{x^3-y}=\frac{1}{1-x+x^2}$$I thought that after substituting the value of $y=-1$ the limit will be equal $\frac{1}{x^2}$. I do not know what am i doing wrong.
| $$\lim_{x\to -1} \frac{x-y}{x^3-y}=\frac{-1-y}{-1-y}=1$$
Whereas for your second limit,
$$\lim_{y\to -1} \frac{x-y}{x^3-y}=\frac{x+1}{x^3+1}$$
Using the identity $a^3+b^3=(a+b)(a^2+b^2-ab)$,
$$\frac{x+1}{x^3+1}=\frac{1}{x^2-x+1}$$
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Show that $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$ Let $n(\neq 0,1)$ be a square-free integer. Suppose that $|a^2 - nb^2|$ is a prime integer for $a,b \in \mathbb{Z}$. Show that $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$. Then, show that $p = |a^2-nb^2|$ is not irreducible in $\mathbb{Z}[\sqrt{n}]$.
I can't prove the first part. I think the question is a bit un clear too. Are all $a + b\sqrt{n}$ is an irreducible element of $\mathbb{Z}[\sqrt{n}]$? Or just those with norm being a prime? For the second, I find it weird because i thought all primes in an integral domain is irreducible. Is $\mathbb{Z}[\sqrt{n}]$ not an integral domain? If so, can someone give me an example of a zero divisor in $\mathbb{Z}[\sqrt{n}]$ or a proof of $\mathbb{Z}[\sqrt{n}]$ not being an integral domain?
Thanks.
My attempt:
Claim 1
$N((a+b\sqrt{n})(c+d\sqrt{n}) = N(a+b \sqrt{n})N(c+d\sqrt{n})$
Proof
\begin{align*}
& N((a+b\sqrt{n})(c+d\sqrt{n}) \\
&= N((ac + bdn) + (ac + bd)\sqrt{n}) \\
&= |(ac+bdn)^2 - n(ac+bd)^2| \\
&= |(a^2c^2 + 2abcdn + b^2d^2n^2) - n(a^2c^2 + 2abcd + b^2d^2)| \\
&= |a^2c^2 + 2abcdn + b^2d^2n^2 - a^2c^2n - 2abcdn - b^2d^2n| \\
&= |a^2c^2 + b^2d^2n^2 - a^2c^2n - b^2d^2n| \\
&= |(a^2 - nb^2)(c^2 - nd^2)| \\
&= |a^2 - nb^2||c^2 - nd^2)| \\
&= N(a+b\sqrt{n})N(c+ d \sqrt{n})
\end{align*}
Claim 2.
$a+b\sqrt{n}$ is a unit in $\mathbb{Z}[\sqrt{n}]$ if and only if $N(a+b\sqrt{n}) = 1$.
Proof
Since $a+b\sqrt{n}$ is a unit in $\mathbb{Z}[\sqrt{n}]$, we can find $(c+d \sqrt{n}) \in \mathbb{Z}[\sqrt{n}]$ such that $(a+b\sqrt{n})(c+d \sqrt{n}) = 1$. This gives us $N(a+b\sqrt{n})N(c+ d \sqrt{n}) = N((a+b\sqrt{n})(c+d \sqrt{n})) = N(1) = 1$. From the way the norm is defined, $N(a+b\sqrt{n}) \geq 0$ and $N(c+ d \sqrt{n}) \geq 0$. Therefore, we must have $N(a+b\sqrt{n}) = 1$ and $N(c+ d \sqrt{n}) = 1$.
I am having trouble proving the other direction
Proof for the question
First part
Suppose $a + b\sqrt{n}$ is reducible in $\mathbb{Z}[\sqrt{n}]$, then $a + b\sqrt{n} = (\alpha_1 + \beta_1 \sqrt{n})(\alpha_2 + \beta_2 \sqrt{n})$ for some non-units $\alpha_1 + \beta_1 \sqrt{n}, \alpha_2 + \beta_2 \sqrt{n} \in \mathbb{Z}[\sqrt{n}]$. This gives us $ |a^2 - nb^2| = N(a + b\sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})(\alpha_2 + \beta_2 \sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})N(\alpha_2 + \beta_2 \sqrt{n})$. By claim 2, $N(\alpha_1 + \beta_1 \sqrt{n}) \neq 1$ and $N(\alpha_2 + \beta_2 \sqrt{n} \neq 1$ as $\alpha_1 + \beta_1 \sqrt{n}, \alpha_2 + \beta_2 \sqrt{n}$ are non-units. This implies that $N(\alpha_1 + \beta_1 \sqrt{n}) = p$ and $N(\alpha_2 + \beta_2 \sqrt{n}) = p$ as $p$ is prime. This presents us with a contradiction as $p = N(a + b\sqrt{n}) = N(\alpha_1 + \beta_1 \sqrt{n})N(\alpha_2 + \beta_2 \sqrt{n}) = p^2$. Hence, $a + b\sqrt{n}$ is irreducible in $\mathbb{Z}[\sqrt{n}]$.
Second part
If $p$ is irreducible in $\mathbb{Z}[\sqrt{n}]$, then if $p = ab$ for some $a,b \in \mathbb{Z}[\sqrt{n}]$, $a$ is a unit or $b$ is a unit . This gives us $p^2 = N(p) = N(a)N(b)$. If $a$ is a unit, then $N(b) = p^2$.
I don't know how to continue
If $|ab| \neq |a||b|$, then my proof probably fail, but i think this is true.
| Note that $(a^2-nb^2)(c^2-nd^2)=(ac+bdn)^2-n(ad+bc)^2$, which is the multiplicativity of norms, related to $(a-b\sqrt n)(c-d\sqrt n)=(ac+bdn)-(ad+bc)\sqrt n$.
Suppose you have a factorisation $a-b\sqrt n=(c-d\sqrt n)(e-f \sqrt n)$ then you also have $$\pm p=a^2-nb^2=(c^2-nd^2)(e^2-nf^2)$$ and this is a factorisation in $\mathbb Z$, so one of the factors, say the first, must be $\pm 1$.
This then gives $(c-d\sqrt n)(c+d\sqrt n)=c^2-nd^2=\pm 1$ so that $c-d\sqrt n$ is a unit (we have found an explicit multiplicative inverse).
For the second part $\pm p=(a+b \sqrt n)(a-b\sqrt n)$
Suppose $1=(a\pm b\sqrt n)(c\pm d\sqrt n)=(ac+bdn)\pm (ad+bc)\sqrt n$ (taking the same sign both times) so that one factor is a unit. We know that $n$ is square free, so $\sqrt n$ is irrational. But this equation expresses $\sqrt n$ as a rational number unless $ac+bdn=1$ and $ad+bc=0$. Let $q=c^2-nd^2$
Now we have $\pm pq=(a^2-nb^2)(c^2-nd^2)=(ac+bdn)^2-n(ad+bc)^2=1$ which is a contradiction in $\mathbb Z$ because $p$ is not a unit (and we cannot, from this equation, have $q=0$).
The language of norms codifies such calculations. The fact that norms are multiplicative allows us to take questions about factorisation from a quadratic ring (or ring of integers in any number field) into $\mathbb Z$. It is important at this stage to get an accurate understanding of what is happening here. It is particularly useful to note that the formula for the norm gives an explicit inverse for an element of norm $\pm 1$ and therefore a direct proof that an element of norm $\pm 1$ is a unit.
Note that it is not necessarily true that a prime in $\mathbb Z$ is a prime in the quadratic field. If we set $n=-1$, for example, we find $2=i(1-i)^2$ where $i$ is a unit, and $5=(1+2i)(1-2i)$. The prime $5$ splits into two distinct factors, while the double factor for $2$ is an example of "ramification". It turns out that this is a significant difference.
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Intriguing Indefinite Integral: $\int ( \frac{x^2-3x+1/3 }{x^3-x+1})^2 \mathrm{d}x$
Evaluate
$$\int \left( \frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 \mathrm{d}x$$
I tried using partial fractions but the denominator doesn't factor out nicely. I also substituted $x=\dfrac{1}{t}$ to get
$$\frac{-1}{9} \int \left(\frac{t^2-9t+3}{t^3-t^2+1}\right)^2 \, \mathrm{d}t $$
But I don't know how to solve this either.
Please Help.
Thanks in advance.
| The goal is to represent
$$\dfrac{dt}{dx} = \dfrac19 \cdot \dfrac{(3x^2-9x+1)^2}{(x^3-x+1)^2} = \dfrac{d(u/v)}{dx} = \dfrac{vdu/dx - u dv/dx}{v^2}$$
Hence, it is tempting to choose $v=(x^3-x+1)$. This means we need $u$ such that
$$(x^3-x+1)u'(x) - (3x^2-1)u = (x^2-3x+1/3)^2$$
Hence, $u(x)$ must be a quadratic as $u(x) = ax^2+bx+c$. This gives us
$$(x^3-x+1)(2ax+b) - (3x^2-1)(ax^2+bx+c) = (x^2-3x+1/3)^2$$
Comparing coefficient of $x^4$, we obtain
$$2a - 3a = 1 \implies a = -1$$
Comparing coefficient of $x^2$, we obtain
$$-2a + a -3c = 9 + 2/3 \implies 3c = -a-29/3= 1-29/3 = -26/3 \implies c = -26/9$$
Comparing the constant term, we obtain
$$b +c = 1/9 \implies b = 3$$
Hence, we have
$$\dfrac{d}{dx}\left(\dfrac{-x^2+3x-26/9}{x^3-x+1}\right) = \dfrac19 \cdot \dfrac{(3x^2-9x+1)^2}{(x^3-x+1)^2}$$
| {
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Find ${\rm d}y/{\rm d}x$ and simplify as much as possible, $y=x/(2x+5)^3$ My answer is $-6x(2x+5)^2+(2x+5)^{-3}$. I am wondering if my answer is right and whether I am able to simplify more?
| You can use the quotient rule directly: $$\frac{{\rm d}y}{{\rm d}x} = \frac{(2x+5)^3-6x(2x+5)^2}{(2x+5)^6} = \frac{2x+5-6x}{(2x+5)^4} = \frac{-4x+5}{(2x+5)^4}.$$
This in fact indicates that you've got a little mistake: it's $(2x+5)^{-4}$ instead of $(2x+5)^2$ in what you have written there. I, particularly, prefer to leave stuff in a single fraction.
Another approach: writing $y = x(2x+5)^{-3}$ and using the product rule, we have: $$\frac{{\rm d}y}{{\rm d}x} = (2x+5)^{-3} - x(-3(2x+5)^{-4}2) = (2x+5)^{-3}-6x(2x+5)^{-4}.$$
Your mistake probably was thinking that $(({\rm stuff})^{-3})' = -3({\rm stuff})^{-2}$, because $2 < 3$, and also a sign mistake. You decrease the exponent, so $-3 \to -4$.
| {
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Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$
$$2.~~y^{a-b}=z^{c-a}$$
$$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.
Thank you.
| First let's assume that there are no indeterminations:
$x > 0 \land y > 0 \land z > 0 \land a \neq b \neq c$
And we have to prove that $k = 1$ in:
$x^{b+c−a}⋅y^{c+a−b}⋅z^{a+b−c} = k$
Using the asker equations from cross multiplication:
$x^{b+c−a}⋅x^{a-c}⋅y^{a}⋅z^{a}⋅x^{a-b} = x^a⋅y^a⋅z^a = (xyz)^a = k$
From this method, we can also obtain:
$(xyz)^b = k$
and
$(xyz)^c = k$
But then:
$k^{\frac{1}a} = k^{\frac{1}b} = k^{\frac{1}c} \implies k = 1$
What's more:
$xyz = 1$
| {
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Prove these two functions have the same coefficients We have $\displaystyle p(x) = \frac{1}{1-x}\cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \cdot \ ...$ and $\displaystyle q(x) = (1+x)\cdot(1+x^2) \cdot (1+x^3)\cdot \ ...$.
Let's say that these two functions have coefficients $a_n$ and $b_n$ respectively.
Now I have to prove that $a_n = b_n$ for all $n$.
I know that $\displaystyle \frac{1}{1-x} = \sum_{n=0} ^\infty x^n$ and $\displaystyle \frac{1}{1-x^3} = \sum_{n=0} ^\infty x^{3n}$ etc.
But how about the terms of $q(x)$? And how can I get one power series for $p(x)$ and $q(x)$?
| The coefficient of $x^n$ in the product
$$ (1+x)(1+x^2)(1+x^3)\cdot\ldots $$
gives the number of ways of writing $n$ as a sum of distinct natural numbers, while the coefficient of $x^n$ in the product:
$$ \frac{1}{1-x}\cdot\frac{1}{1-x^3}\cdot\ldots = (1+x+x^2+\ldots)(1+x^3+x^6+\ldots)\cdot\ldots $$
gives the number of ways to write $n$ as a sum of different natural numbers, where every number occurs an odd number of times. Now have a look at the paragraph "Odd parts and Distinct parts" in the following Wikipedia page. We have:
$$\prod_{n\geq 1}(1+x^n) = \prod_{n\geq 1}\frac{1-x^{2n}}{1-x^n}=\frac{\prod_{n\geq 1}(1-x^{2n})}{\prod_{n\geq 1}(1-x^{2n})\prod_{n\geq 1}(1-x^{2n-1})}=\prod_{n\geq 1}\frac{1}{1-x^{2n-1}}.$$
This is just a special case of Glaisher's theorem.
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Volume of a parallelepiped when not given values for three vectors There is a parallelepiped determined by three dimensional vectors x, y, and z. The volume of this parallelepiped is $11$. What is the volume of the parallelepiped determined by the three dimension vectors x+2y, y+2z, z+2x?
I tried substituting constant values for x, y, and z, but it became very messy quickly. How should I approach this problem?
| Hint:
Volume of original parallelepiped = $|\det(A)|$, where the columns of $A$ are $\mathbf x$, $\mathbf y$, $\mathbf z$.
Volume of new parallelepiped = $|\det(B)|$, where the columns of $B$ are $\mathbf x +2\mathbf y$, $\mathbf y+2\mathbf z$, $\mathbf z+2\mathbf x$.
Alternative Hint:
Volume of original parallelepiped = $|\mathbf x \cdot (\mathbf y \times \mathbf z)|$.
Volume of new parallelepiped = $|(\mathbf x +2\mathbf y) \cdot [(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)]|$.
EDIT: It's been long enough that I'd like to just write the answer down.
Determinant Approach
If we consider a determinant not as a function on square matrices, but as a function on the columns (or rows) of square matrices, then we can see that $\det(\mathbf x, \mathbf y, \mathbf z) = 11$ from the problem and the above hint.
The $\det$ function is multilinear and antisymmetric. Thus $$\begin{align}\det(\mathbf x +2\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x) &= \det(\mathbf x,\mathbf y+2\mathbf z, \mathbf z+2\mathbf x) + 2\det(\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x) \\ &= [\det(\mathbf x,\mathbf y, \mathbf z+2\mathbf x) + 2\det(\mathbf x,\mathbf z, \mathbf z+2\mathbf x)] \\ &\ \ \ \ + [2\det(\mathbf y, \mathbf y, \mathbf z+2\mathbf x) + 4\det(\mathbf y, \mathbf z, \mathbf z+2\mathbf x)] \\ &= \det(\mathbf x,\mathbf y, \mathbf z) + 2\det(\mathbf x,\mathbf y, \mathbf x) + 2\det(\mathbf x,\mathbf z, \mathbf z) \\ &\ \ \ \ + 4\det(\mathbf x,\mathbf z, \mathbf x) + 2\det(\mathbf y, \mathbf y, \mathbf z) + 4\det(\mathbf y, \mathbf y, \mathbf x) \\ &\ \ \ \ + 4\det(\mathbf y, \mathbf z, \mathbf z) + 8\det(\mathbf y, \mathbf z, \mathbf x) \\ &= 11 + 2(0) + 2(0) + 4(0) + 2(0) + 4(0) + 4(0) + 8(11) \\ &= 99\end{align}$$
Note that this looks like a lot of work, but after you've practiced it a few times you'll be able to immediately look at an expression like the one above and mentally throw out all of the zero terms. The way you do this is basically just look for which combinations don't give you more than $1$ of each column vector. Starting with $(\mathbf x +2\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x)$, the only non-null combos are going to be $(\mathbf x, \mathbf y, \mathbf z)$ and $(2\mathbf y, 2\mathbf z, 2\mathbf x)$ as those are the only choices without a repeated $\mathbf x$, $\mathbf y$, or $\mathbf z$.
Triple Scalar Product Approach
We know that $|\mathbf x \cdot (\mathbf y \times \mathbf z)| = 11$, so then we just have to reduce the following expression to something in terms of this one:
$$\begin{align}(\mathbf x +2\mathbf y) \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} &= \mathbf x \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} + 2\mathbf y \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot\color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x) +2\mathbf z \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &\ \ \ \ + 2\mathbf y \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x) +2\mathbf z \times (\mathbf z + 2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x)\color{red}{]} +2\mathbf x \cdot \color{red}{[}\mathbf z \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &\ \ \ \ + 2\mathbf y \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x)\color{red}{]} +4\mathbf y \cdot\color{red}{[}\mathbf z \times (\mathbf z + 2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot (\mathbf y \times \mathbf z) + 2\mathbf x \cdot (\mathbf y \times \mathbf x) + 2\mathbf x \cdot (\mathbf z \times \mathbf z) + 4\mathbf x \cdot (\mathbf z \times \mathbf x) \\ &\ \ \ \ + 2\mathbf y \cdot (\mathbf y \times \mathbf z) + 4\mathbf y \cdot (\mathbf y \times \mathbf x) \\ &\ \ \ \ +4\mathbf y \cdot (\mathbf z \times \mathbf z) + 8\mathbf y \cdot (\mathbf z \times \mathbf x) \\ &= \mathbf x \cdot (\mathbf y \times \mathbf z) + 0 + 0 + 0 +0 + 0 + 0 +8 \mathbf x \cdot (\mathbf y \times \mathbf z) \\ &= 99\end{align}$$
Here we've made use of such properties as the distributivity of the cross and dot products, the fact that scalar multiplication can basically be done in whichever order (for example, in the case of the dot product: $\alpha (\mathbf a \cdot \mathbf b) = (\alpha \mathbf a) \cdot \mathbf b = \mathbf a \cdot (\alpha \mathbf b)$), and the circular shift property of the triple scalar product.
And again, some of these steps could have been done mentally by someone proficient in the application of these products.
| {
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How to prove this definition of the Riemann Zeta Function? I have to prove that
$\sum_{n=1}^{\infty} \frac{1}{n^s}$ = $\frac{1}{1-2^{1-s}} \sum_{n=1}^{\infty} \frac{-1^{n+1}}{n^s} $
I know I am supposed to show my work, but I have no clue on how to proceed.
Thanks!
| $$
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^s}&=
\sum_{n=1}^{\infty} \frac{1}{(2n)^s}+\sum_{n=1}^{\infty} \frac{1}{(2n-1)^s}\\
&=\frac1{2^s}\sum_{n=1}^{\infty} \frac{1}{n^s}+\sum_{n=1}^{\infty} \frac{1}{(2n-1)^s}
\end{align}
$$
Hence
$$
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^s}=(1-\frac{1}{2^s})\sum_{n=1}^{\infty} \frac{1}{n^s}
$$
Now we have
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}&=\sum_{n=1}^{\infty} \frac{1}{(2n-1)^s}-\sum_{n=1}^{\infty} \frac{1}{(2n)^s}\\
&=(1-\frac{1}{2^s})\sum_{n=1}^{\infty} \frac{1}{n^s}-\frac1{2^s}\sum_{n=1}^{\infty} \frac{1}{n^s}\\
&=(1-\frac1{2^{s-1}})\sum_{n=1}^{\infty} \frac{1}{n^s}
\end{align}
$$
| {
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Find the range of the given function $f$: $f(x) = \frac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) + \frac{x+1}{x^2 + 2x + 5} $
Find the range of the following function.
$$\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) + \cfrac{x+1}{x^2 + 2x + 5} $$
where $\sin^{-1}x \ $ and $ \ \tan^{-1}x $ are inverse trigonometric functions.
This is what I've tried yet.
Let : $\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) = g(x)$ and $ \cfrac{x+1}{x^2 + 2x + 5} = h(x)$ .
Therefore,
$$f(x) = g(x) + h(x)$$
I can write $h(x)$ as : $$h(x) = \cfrac{x+1}{(x+1)^2 + 4} = \cfrac{1}{(x+1) + \cfrac{4}{x+1}}$$
So, the function $f(x)$ becomes:
$$f(x) = \cfrac{1}{\pi} \left(\sin^{-1} x +\tan^{-1} x\right) + \cfrac{1}{(x+1) + \cfrac{4}{x+1}}$$
Domain of $g(x)$ is $[-1,1] \cap \mathbb{R} = [-1,1]$
Domain of $h(x)$ is $\mathbb{R} - \{-1\}$
Therefore, $dom(f(x)) = (-1,1]$
Now, finding maximum value of $h(x)$
$$\begin{align}
\cfrac{dh(x)}{dx} =& \cfrac{d}{dx} \left( (x+1) + \cfrac{4}{x+1} \right) \\
= & 1 - \cfrac{4}{(x+1)^2}\end{align}$$
Putting $h'(x)$ equal to zero to find critical points, I got:
$$\begin{align} 1 - \cfrac{4}{(x+1)^2} =& 0 \\
(x+1) =& \pm 2 \\
\boxed{x = 1 \ \text{OR} \ x = -3} \rightarrow \textsf{CRITICAL POINTS} \end{align}$$
How to go further from here? Is this method the perfect one for solving this type of questions? Or is there any other trick to go with it?
| Note that the domain of $g(x)=\frac{1}{\pi}(\sin^{-1}x+\tan^{-1}x)$ is $[-1,1]$ and the domain of $h(x)=\frac{x+1}{x^2+2x+5}$ is $(-\infty,\infty)$. Thus the domain of $f(x)=g(x)+h(x)$ is $[-1,1]$. Note that also
$$ g'(x)=\frac{1}{\pi}\left(\frac{1}{\sqrt{1-x^2}}+\frac{1}{x^2+1}\right)>0$$
and
$$ h'(x)=\frac{4-(x+1)^2}{(x+1)^2+4}>0 \text{ for }x\in[-1,1]$$
and hence $f(x)=g(x)+h(x)$ is increasing in $[-1,1]$. So the range of $f(x)$ is
$$[f(-1),f(1)]=[-\frac{3}{4},1].$$
| {
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14 pencils handed out to 6 people. Each person has at least 1 pencil. Person 6 no more than 3 pencils. We have 14 indistinguishable pencils and we want to hand out all of the pencils to 6 people and we want everyone to get at least one pencil. However, we do not want person 6 to get more than 3 pencils. How many different ways could this be done?
Is the answer $\binom{10}{6}$ correct?
Reserve 1 pencil per person and reserve 3 pencils to person 6. So we now have 6 pencils left that we need to pass out to 5 people (since person 6 can no longer receive a pencil). So the problem is the same as asking how many ways can pass out 6 pencils to 5 people? Or am I looking at this wrong?
| Here is an alternative method using the Inclusion-Exclusion Principle.
If we hand out $14$ pencils to six people so that each person gets at least $1$ in such a way that person $6$ gets at most $3$ pencils, then we need to find the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 14 \tag{1}$$
in the positive integers subject to the restriction that $x_6 \leq 3$.
As you observed, we can first give each person one pencil to reduce the problem to distributing eight pencils to six people so that person $6$ gets no more than $2$ additional pencils. Let $y_k = x_k - 1$ for $1 \leq k \leq 6$. Then
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 & = 14\\
y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 + y_5 + 1 + y_6 + 1 & = 14\\
y_1 + y_2 + y_3 + y_4 + y_5 + y_6 & = 8 \tag{2}
\end{align*}
where equation 2 is an equation in the nonnegative integers subject to the restriction that $y_6 \leq 2$. Moreover, equation 2 has the same number of solutions has equation 1.
If we do not restrict the value of $y_6$, then the number of solutions to equation 2 is the number of ways we can insert five addition signs in a row of eight ones, which is
$$\binom{8 + 5}{5} = \binom{13}{5}$$
However, we must eliminate those solutions in which $y_6 \geq 3$. Suppose $y_6 \geq 3$. Then $z_6 = y_6 - 3 \geq 0$. Then the number of solutions we must eliminate is the number of solutions in the nonnegative integers of the equation
\begin{align*}
y_1 + y_2 + y_3 + y_4 + y_5 + z_6 + 3 & = 8\\
y_1 + y_2 + y_3 + y_4 + y_5 + z_6 & = 5 \tag{3}
\end{align*}
Since equation 3 has
$$\binom{5 + 5}{5} = \binom{10}{5}$$
solutions in the nonnegative integers, the number of ways $14$ pencils can be distributed to six people so that each person receives at least one pencil and person $6$ receives no more than three pencils is
$$\binom{13}{5} - \binom{10}{5}$$
As you can check, this is equivalent to the answer that Gamamal obtained by considering the cases $y_6 = 0$, $y_6 = 1$, and $y_6 = 2$ separately, that is,
$$\binom{13}{5} - \binom{10}{5} = \binom{12}{8} + \binom{11}{7} + \binom{10}{6} = 1035$$
| {
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Prove $\sum\limits_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$
Prove $$\sum_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$$
| By the functional identities for the $\Gamma$ function we have that:
$$\prod_{k=1}^{n}(3k+1)=3^n\cdot\frac{\Gamma\left(\frac{4}{3}+n\right)}{\Gamma\left(\frac{4}{3}\right)}\tag{1}$$
hence:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{n!}{3^n\prod_{k=1}^{n}(3k+1)}&=&\Gamma\left(\frac{4}{3}\right)\sum_{n\geq 1}\frac{\Gamma(n+1)}{9^n\cdot \Gamma\left(n+\frac{4}{3}\right)}=\frac{1}{3}\sum_{n\geq 1}\frac{B\left(n+1,\frac{1}{3}\right)}{9^n}\\&=&\frac{1}{3}\int_{0}^{1}\sum_{n\geq 1}\frac{(1-x)^n x^{-2/3}}{9^n}\,dx=\frac{1}{3}\int_{0}^{1}\frac{1-x}{8+x}x^{-2/3}\,dx\end{eqnarray*}$$
and the last integral can be solved by using the substitution $x=y^3$, leading to:
$$ \sum_{n\geq 1}\frac{n!}{3^n\prod_{k=1}^{n}(3k+1)}=\int_{0}^{1}\frac{1-y^3}{8+y^3}\,dy = -1+\frac{\sqrt{3}}{8}\,\pi+\frac{3}{8}\,\log 3.\tag{2} $$
If now we multiply both sides of $(2)$ by $4$ we prove tha stated identity.
| {
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Solving the recurrence relation obtained from the power series method Assuming the solution to my differential equation is of the form $y=\sum_{n=0}^\infty a_nx^n$, I was able to get to the recurrence relation.
The recurrence relation is $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$
From this I get
$$a_2 = \frac {-3}{2}a_0$$ $$a_3 = 0 \implies a_{\text{odd}} = 0 \tag{except maybe $a_1$} \\ a_4 = \frac 14a_2 = \frac {-3}8a_0 \\ a_6 = \frac 3{10}a_4 = \frac {-9}{80}a_0 \\ \vdots$$
Thus my solution should be of the form $$y= a_1(x) + a_0 (1-\frac 32x^2 -\frac 38x^4 -\frac 9{80}x^6 + \cdots)$$
How do I simplify this -- that is write the infinite series for the even terms? I've been trying for a while, but I'm not sure what the standard technique is for this.
|
$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$
$$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$
$$a_{2n+1} = 0$$
$$\begin{align}
%
a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)} \\
& \times \dots \\
& \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0}
\end{align}$$
And canceling out terms:
$$\begin{align}
%
a_{n} &= a_0~3^{n/2}~\frac {1}{(n-1)(n-0)} \times \frac {1}{(n-2)} \times \frac {1}{(n-4)} \\
& \times \dots \\
& \times \frac {1}{6} \times \frac{1}{4}\times \frac{(-1)}{2}
\end{align}$$
$$a_{n} = a_0~3^{n/2}~\frac{-1}{n-1}~\frac{1}{(n-0)(n-2)(n-4) \dots (4)(2)}$$
And rewriting it in various ways:
$$a_{n} = a_0~3^{n/2}~\frac{-1}{2(n/2)-1}~\frac{2^{-n/2}}{(n/2-0)(n/2-1)(n/2-2) \dots (2)(1)}$$
$$f_n(x) = a_{2n}x^{2n} = a_0~(3/2)^{n}~\frac{-1}{2n - 1}~\frac{x^{2n}}{(n-0)(n-1)(n-2) \dots (2)(1)}$$
$$f_n(x) = \frac{-a_0}{2n - 1}~\frac{\left(3x^2/2\right)^n}{n!}$$
$$-a_0~g_n\left( \frac{3x^2}{2} \right) = f_n(x) \iff \color{darkred}{g_n(x) = \frac{1}{2n - 1}~\frac{x^n}{n!}}$$
This $g$ doesn't look like any common taylor series, so checking wolfram online:
$$\sum_{n=0}^{\infty} \frac{1}{2n - 1}~\frac{x^n}{n!} = \sqrt{\pi x}~~ {\rm erfi}(\sqrt{x}) - e^{x}$$
where erfi is the imaginary error function. So while your recursion has the above closed form, your series does not have an elementary closed form.
| {
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Reccurence equation $f(n)=6f(n-1)-9f(n-2)+(n^2+1)3^n$ $f(n)=6f(n-1)-9f(n-2)+(n^2+1)3^n$
The root for the above relation is 3 two times. So its general term will be:
$f(n) = c_{1}3^n + c_{2}n3^n + something$
According to my notes $something: n^2(n^2p_{2,0}+np_{1,0}+p_{0,0})3^n = (n^2+1)3^n$
That should have been correct if it was: $+n^23^n$But it is: $+(n^2+1)3^n = 3^nn^2+3^n$
so it should be: $n^2(n^2p_{2,0}+np_{1,0}+p_{0,0})3^n + n^2(p_{0,1})3^n=(n^2+1)3^n$
Is my reasoning wrong?
| i will try a change of variable $$u_n = f_n3^{-n}, \quad f_n = 3^nu_n.$$ the recurrence relation for $u_n$ is $$3^nu_n = 6\, 3^{n-1}u_{n-1} - 9\, 3^{n-2}u_{n-2}+(n^2+1)\, 3^n $$ dividing out by $3^n,$ gives you $$Lu_n = u_n- 2u_{n-1}+u_{n-2}= n^2+1 \tag 1$$
we see that
$$\begin{align}L(n+1)^4 &= (n+1)^4 -2n^4 +(n-1)^4=12n^2+2,\\
L(n+1)^2 &=(n+1)^2-2n^2+(n-1)^2= 2\end{align}$$
therefore a particular solution is $$u_n = \frac1{12}\left((n+1)^4 -1\right), \quad f(n) = \frac1{12}\left((n+1)^4 -1\right)3^n.$$
| {
"language": "en",
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How many numbers smaller than one million, their sum of digits is at least 20?
How many numbers smaller than one million, their sum of digits is at least 20?
My attempt:
Since I don't know how to handle the "at least" part, I'll be using a complement:
The general case is all the digit combination up to one million: $9\cdot10^5$
Complement: the sum is at most 20.
Now we need to find all the solutions for $x_1+...+x_6=n : n\in[0,20]:x_i\in[0,9]$
So I found the generating sums function: $\frac{(1-x^{10})^6}{(1-x)^7}$
And the coefficients I got from that for $x^{20}$ are: $\binom{20+6}{6}-6\binom{10+6}6+\binom 6 2$
So the final answer is: $9\cdot10^5-\left(\binom{20+6}{6}-6\binom{10+6}6+\binom 6 2\right)=717803$
But I'm not sure I got the general case right, is it really all the 6 digit combinations?
| To simplify the calculations, we will work with nonnegative integers.
There are $1,000,000$ nonnegative integers less than $1,000,000$ since $0$ is nonnegative.
If the sum of the digits of a nonnegative integer is not at least $20$, then the sum of its digits is at most $19$.
A number less than $1,000,000$ has at most six digits. We can append leading zeros to a number with fewer than six digits to represent it as a six-digit number. For instance, we regard $473$ as $000473$.
Thus, we wish to exclude nonnegative integers whose digits satisfy the inequality
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \leq 19$$
subject to the restrictions that $x_k \leq 9$ for $1 \leq k \leq 6$.
To handle the inequality, let $d = 19 - (x_1 + x_2 + x_3 + x_4 + x_5 + x_6)$.
Then $d$ is a nonnegative integer and
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d = 19$$
If there were no restrictions on the size of $x_k$ for $1 \leq k \leq 6$, then the number of solutions of this equation in the nonnegative integers would be equal to the number of ways we can place six addition signs in a row of $19$ ones, which is
$$\binom{19 + 6}{6} = \binom{25}{6}$$
However, we have counted solutions in which one of the $x_k$'s is greater than $9$. Note that at most one $x_k$ can exceed $9$ since $10 + 10 = 20 > 19$.
Suppose that $x_1 \geq 10$. Let $y_1 = x_1 - 10$. Then $y_1$ is a nonnegative integer and
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 19\\
y_1 + 10 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 19\\
y_1 + x_2 + x_3 + x_4 + x_5 + x_6 + d & = 9
\end{align*}
This equation has
$$\binom{9 + 6}{6} = \binom{15}{6}$$
solutions in the nonnegative integers. Since any of the six $x_k$'s could exceed $9$, the number of nonnegative integers less than $1,000,000$ such that the sum of the digits is less than $20$ is
$$\binom{25}{6} - \binom{6}{1}\binom{15}{6}$$
a total that includes the number $0$.
Thus, the number of nonnegative integers less than $1,000,000$ such that the sum of the digits is at least $20$ is
$$1,000,000 - \left[\binom{25}{6} - \binom{6}{1}\binom{15}{6}\right]$$
| {
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Solve the integral $\int \frac{dx}{\:\sqrt[4]{\left(x+2\right)^5\cdot \left(x-1\right)^3}}$ Here is a indefinite integral must be solved. Help, who knows. Although it would be like casual.
$$\int \frac{dx}{\:\sqrt[4]{\left(x+2\right)^5\cdot \left(x-1\right)^3}}$$
| I think I may see something.
$$u=x+2,du=dx$$
$$\int\frac{dx}{\sqrt[4]{(x+2)^5(x-1)^3}}=\int\frac{du}{\sqrt[4]{u^5(u-3)^3}}=$$
$$\int\frac{du}{u^2\sqrt[4]{(\frac{u-3}{u}})^3}=\int\frac{(1-\frac3u)^{-3/4}du}{u^2}$$
$$t=1-\frac3u,dt=\frac{3du}{u^2}$$
$$\int\frac{(1-\frac3u)^{-3/4}du}{u^2}=\frac13\int t^{-3/4}dt=\frac43\sqrt[4]t=\frac43\sqrt[4]{1-\frac3{x+2}}+C$$
| {
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IVP to $y'=\frac{2xy^2+2x}{x^2+1}$ Find a solution of $y'=\frac{2xy^2+2x}{x^2+1}$ with given IVP $y(0)=\sqrt{3}$.
My solution:
$\int \frac{1}{y^2+1}dy=\int \frac{2x}{x^2+1}dx$ $\Rightarrow \tan^{-1}(y)=\log(x^2+1)+c, c\in \mathbb{R}$ $\Rightarrow y= \tan (\log(x^2+1)+c),c\in \mathbb{R} $
$y(o)=\sqrt{3}=\tan(c)$ $ \Rightarrow c=\frac{4 \pi}{3}$
$\Rightarrow y= \tan (\log(x^2+1)+\frac{4 \pi}{3})$
Determining the domain of defeniton:
$\frac{\pi}{2}= (\log(x^2+1)+\frac{4 \pi}{3})$ $ \Rightarrow x= \pm \sqrt{ e^{ \frac{-5\pi}{6}}-1}$
$-\frac{\pi}{2}=(\log(x^2+1)+\frac{4 \pi}{3})$ $\Rightarrow x=\pm \sqrt{ e^{ \frac{-11\pi}{6}}-1}$
So domain of definition would be $D=]-\sqrt{ e^{ \frac{-11\pi}{6}}-1}, \sqrt{ e^{ \frac{-5\pi}{6}}-1}[$
Is this correct?
| The solution and the steps leading up to them are all correct. You can always check by plugging the expression you obtained back into the differential equation. Using the chain rule, the derivatives of elementary functions, and trigonometric identities, you should find that both sides are equal to
$
\dfrac{2x \sec^2\left(\log(x^2+1)+\frac{4\pi}{3}\right)}{x^2 + 1}.
$
The domain of definition, on the other hand, isn't correct. Notice that the ends you gave for the maximal interval of existence are square roots of negative numbers, since $e^x < 1$ for all $x < 0$. The core issue is that $\log(x^2 + 1) + \frac{4\pi}{3} \geq \frac{4\pi}{3} > \pm \frac{\pi}{2}$, and so the two equations you were trying to solve in fact have no real solutions. In other words, the singularities of $\tan$ at $\pm\frac{\pi}{2}$ do not lead to singularities of your expression for $y$.
But $\pm\frac{\pi}{2}$ aren't the only forbidden inputs for $\tan$. Your expression for $y(x)$ blows up to $\pm\infty$ at $x$ wherever $\tan$ has a singularity at $\log(x^2+1) + \frac{4\pi}{3}$. What you need to do is to find the closest such $x$'s to the left and to the right of $0$.
| {
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Find $\int\limits_{0}^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$ I have to find $$\int_0^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$$
I tried to do it by substitution $t = \tan(\frac{x}{2})$
Then we have that $$\cos(x) = \frac{1-t^2}{1+t^2} \quad dx = \frac{2\,dt}{1+t^2}$$ but then also limits of integration are changing so we have
$$\int\limits_{t(0)}^{t(2\pi)} \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int\limits_0^0 \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = 0$$
I figured out that it is not correct because $\tan(\frac{\pi}{2})$ is not defined and $t(\pi) = \tan(\frac{\pi}{2})$ and $\pi \in [0, 2\pi]$. How can I "repair" that and do it right?
| Hint:
Substitute u=tan(x/2). Then transform the integrand using sinx = 2u/(u^2+1), cosx = (1-u^2)/(u^2+1)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding quadratic factors Show that $(x-√3)$ and $(x+√3)$ are factors of $x^4+x^3-x^2-3x-6$. Hence write down one quadratic factor of $x^4+x^3-x^2-3x-6$, and find a second quadratic factor of this polynomial.
My attempt:
$f(x)=x^4+x^3-x^2-3x-6$
$f(3^\frac12)=(3^\frac12)^4+(3^\frac12)^3-(3^\frac12)^2-3(3^\frac12)-6=0$
$f(-3^\frac12)=(-3^\frac12)^4+(-3^\frac12)^3-(-3^\frac12)^2-3(-3^\frac12)-6=0$
I know the first quadratic factor is $x^2-3$ but I don't know how to arrive to the second quadratic factor.
| You can solve all the questions in one formula: since you correctly identified $x^2-3$ as the quadratic factor, try polynomial long division by $x^2-3$. This will
a. prove that $\pm \sqrt{3}$ are roots of the polynomials (since the remainder is zero),
b. give the other quadratic factor (as the quotient).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An inverse hyperbolic cosine approximation Given $-1<a<b<0<c<d<1$, $\gamma=\frac{a+b}{a-b}$, how do we approximate $\left\lceil \frac{\cosh^{-1}\gamma}{\log\gamma}\right\rceil$ as
$$
\left\lceil \sqrt{\frac{a}{b}}\left(1+\frac{1}{3}\frac{b}{a}+O(b^2/a^2)\right)\right\rceil?
$$
In general how to approximate $\left\lceil \frac{\cosh^{-1}\delta}{\log\gamma}\right\rceil$ where $\delta=\frac{c+d}{c-d}$?
| Presumably you're taking $b/a \to 0$ here. If $t = b/a$, then $\gamma = (1+t)/(1-t) = 1 + 2 t + 2 t^2 + O(t^3)$. Now if $\gamma = 1 + u$, so $u = 2 t + 2 t^2 + O(t^3)$,
$$\eqalign{\cosh^{-1}(\gamma) &= \log\left(\gamma + \sqrt{\gamma^2-1}\right) = \log\left(1+u + \sqrt{2u + u^2}\right)\cr
&= \log\left(1 + \sqrt{2} u^{1/2} + u + \dfrac{\sqrt{2}}{4} u^{3/2} + O(u^{5/2})\right)\cr
&= \sqrt{2} u^{1/2} - \dfrac{\sqrt{2}}{12} u^{3/2} + O(u^{5/2}) \cr
\log(\gamma) &= \log(1+u) = u - \dfrac{u^2}{2} + O(u^3)\cr
\dfrac{\cosh^{-1}(\gamma)}{\log(\gamma)} & = \sqrt{2} u^{-1/2} + \dfrac{5 \sqrt{2}}{12} u^{1/2} + O(u^{3/2})\cr
&= t^{-1/2} + \dfrac{1}{3} t^{1/2} + O(t^{3/2})\cr
}$$
as desired.
| {
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"source": "stackexchange",
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Cubic root of unity Is there anyway to solve this without substituting with the values?
Prove that: $$\frac{1+10w^2}{1-2w} + \frac{2+17w}{2+3w} = 6$$. (Where $w$ & $w^2$ are the cubic roots of unity)
| Using $w^2=-w-1$ gives you
$$\begin{align}\frac{1+10w^2}{1-2w}+\frac{2+17w}{2+3w}&=\frac{(1+10(-w-1))(2+3w)+(2+17w)(1-2w)}{(1-2w)(2+3w)}\\&=\frac{-64w^2-34w-16}{2-w-6w^2}\\&=\frac{-64(-w-1)-34w-16}{2-w-6(-w-1)}\\&=\frac{6(5w+8)}{5w+8}\\&=6\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How does $x^2+4xy-6x+4y^2-12y+9=0$ represent a straight line. I need to show $x^2+4xy-6x+4y^2-12y+9=0$ is a straight line. But I only know of a straight line in the form $y=mx+c$. Any help?
| Note that
\begin{align}
0 & = x^2+4y^2+9 + 4xy-6x-12y\\
& = x^2 + (2y)^2 + (-3)^2 + 2\cdot x \cdot (2y) + 2 \cdot x \cdot (-3) + 2 \cdot (2y) \cdot (-3)\\
& = (x+2y-3)^2
\end{align}
This gives us
$$x+2y-3 = 0$$
which indeed is a straight line.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to use an ML estimate to show the solution to an integral I have a question I needed to show that
$$\lim_{R\to\infty} \int_{C_R} \frac {z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$
For $C_R$ the circle with radius R, center z=0 and positively oriented.
Which I have done using an ML estimate.
The next part of the question is to show
$$\int_{C} \frac {z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$
where C is the circle radius 5, center z=2, positively oriented in the complex plane.
I'm just not too sure how to use the ML estimate to show the second part of the question, because he hasn't done any examples in class and I haven't been able to find any similar questions online.
Thank you!
| There are four poles of $\frac{z^2+4z+7}{(z^2+4)(z^2+2z+2)}$ inside $|z-2|<5$ are at $z=\pm 2i$ and $z=-1\pm i$. The residues are
$$\begin{align}
&-\frac{7}{20}-i\frac{13}{40} \cdots \text{at z= 2i}\\
&-\frac{7}{20}+i\frac{13}{40} \cdots \text{at z= -2i}\\
\end{align}$$
and
$$\begin{align}
&\frac{7}{20}-i\frac{1}{5} \cdots \text{at z= -1+i}\\
&\frac{7}{20}+i\frac{1}{5} \cdots \text{at z= -1-i}
\end{align}$$
Inasmuch as these add to zero, the residue theorem guarantees that
$$\oint_C \frac{z^2+4z+7}{(z^2+4)(z^2+2z+2)} dz=0$$
where $C$ is defined by $|z-2|=5$.
| {
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Representing a Decimal as a Fraction - 2 Methods So I am trying to represent the number 0.71717171 · · · as a fraction and have managed to do it using algebra. I was told I was supposed to solve it using a geometric sum.
Could someone guide me through the steps of solving such a question with geometric sums?
My method:
Let $x = 0.\overline{71}$
Then $100x = 71.\overline{71}$
Consequently, $$100x - x = 71.\overline{71} - 0.\overline{71}$$
$$99x = 71$$
$$x=\frac{71}{99}$$
Any help would be greatly appreciated. Thank you.
| Using geometric sums:
$$x=\dfrac{7}{10}+\dfrac{1}{10^2}+\dfrac{7}{10^3}+\dfrac{1}{10^4}+\dfrac{7}{10^5}+\ldots$$
$$x= \left( \dfrac{7}{10}+\dfrac{7}{10^3}+\dfrac{7}{10^5}+\ldots \right)+\left( \dfrac{1}{10^2}+\dfrac{1}{10^4}+\dfrac{1}{10^6}+\ldots \right) $$
Here you have a decreasing geometric with ratio $r<1$, so:
$$x=\left( \dfrac{\dfrac{7}{10}}{1-\dfrac{1}{10^2}} \right)+ \left( \dfrac{\dfrac{1}{10^2}}{1-\dfrac{1}{10^2}} \right)$$
$$x=\frac{71}{99}$$
| {
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Global maximum and minimum of $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ with Lagrange multipliers? The global maximum and the global minimum of the function $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ can be found using Lagrange multipliers.
$\nabla f = \lambda \nabla g$
$g(x,y,z)=x^2+2y^2+3z^2-6=0$
This results into:
$$yz = \lambda 2x$$
$$xz = \lambda 4y$$
$$xy=\lambda6z $$
When looking at the equation $x^2+2y^2+3z^2-6=0$, it can be seen that there are six solutions $(x,y,z)$ with two coordinates equal to zero: $(0,0,\pm \sqrt 2 ), (0, \pm \sqrt 3, 0 ) , (\pm \sqrt 6, 0, 0 ) $
But I got stuck here. Could someone help me with this?
| I think you may have confused yourself in computing and interpreting your result.
First, in solving the set of "Lagrange equations" you found, you can follow the suggestions of Mann or KittyL, or try it this way. We may go ahead and solve each equation for the multiplier $ \ \lambda \ $ thus,
$$ \lambda \ = \ \frac{yz}{2x} \ = \ \frac{xz}{4y} \ = \ \frac{xy}{6z} \ \ . $$
This can be read as three equations, each involving a pair of ratios (one of the equations is redundant). Since we may neglect the possibility of any of the variables having the value zero, as this would automatically produce zero for the value of $ \ f \ $ , we may solve any two of these equations to find, say,
$$ \frac{yz}{2x} \ = \ \frac{xz}{4y} \ \ \Rightarrow \ \ \frac{y}{2x} \ = \ \frac{x}{4y} \ \ \Rightarrow \ \ 2 x^2 \ = \ 4 y^2 \ \ \text{or} \ \ 2 y^2 \ = \ x^2 \ \ , $$
$$ \frac{yz}{2x} \ = \ \frac{xy}{6z} \ \ \Rightarrow \ \ \frac{z}{2x} \ = \ \frac{x}{6z} \ \ \Rightarrow \ \ 2 x^2 \ = \ 6 z^2 \ \ \text{or} \ \ 3 z^2 \ = \ x^2 \ \ , $$
with the remaining ratio equation telling us nothing new. [We encounter similar systems of non-linear equations for many "rectangular box" volume optimization problems, for which we may also discard zero variable values.]
Inserting these into the constraint equation produces
$$ x^2 \ + \ 2y^2 \ + \ 3z^2 \ = \ 6 \ \ \Rightarrow \ \ 3 \ x^2 \ = \ 6 \ \ \Rightarrow \ \ x \ = \ \pm \sqrt{2} \ \ , $$
$$ y^2 \ = \ \frac{1}{2} \ x^2 \ = \ 1 \ \Rightarrow \ \ y \ = \ \pm 1 \ \ , \ \ z^2 \ = \ \frac{1}{3} \ x^2 \ = \ \frac{2}{3} \ \Rightarrow \ \ z \ = \ \pm \frac{\sqrt{6}}{3} \ \ . $$
The solution points lie on the surface of a triaxial ellipsoid centered on the origin. Because the function we are extremizing, $ \ f(x, \ y, \ z) \ = \ x \ y \ z \ $ , has an anti-symmetry about the origin and the constraint surface is symmetric about the origin, we may expect that the extrema will be arranged with such symmetry. There are in fact eight extremal points, one in each octant of $ \ \mathbb{R}^3 \ $ ; these form four pairs on lines through the origin . The absolute value of the function is the same for all eight points and the "partners" will have values with opposite signs.
We can then group the solution points as
maximal value : $$ \ \frac{2 \sqrt{3}}{3} \ \ \text{at} \ \ ( \ \sqrt{2} \ , \ 1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ , \ \ ( \ \sqrt{2} \ , \ -1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ,( \ -\sqrt{2} \ , \ -1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ ( \ -\sqrt{2} \ , \ 1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ; $$
minimal value : $$ \ -\frac{2 \sqrt{3}}{3} \ \ \text{at} \ \ ( \ \sqrt{2} \ , \ -1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ , \ \ ( \ \sqrt{2} \ , \ 1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ ,( \ -\sqrt{2} \ , \ 1 \ , \ \frac{\sqrt{6}}{3} \ ) \ \ ( \ -\sqrt{2} \ , \ -1 \ , \ -\frac{\sqrt{6}}{3} \ ) \ \ . $$
These are confirmed by user64494's Maple computation, although we may discard the results with a function value of zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\rightarrow0}\frac{\cos^{2}\left(\frac{1}{x}\right)\left(1-\cos x\right)}{x}$ $$\lim_{x\rightarrow0}\frac{\cos^{2}\left(\frac{1}{x}\right)\left(1-\cos x\right)}{x}=\lim_{x\rightarrow0}\frac{\left(1-O\left(\frac{1}{x^{2}}\right)\right)^{2}\left(1-1+O\left(x^{2}\right)\right)}{x}=\lim_{x\rightarrow0}\frac{\left(1-2O\left(\frac{1}{x^{2}}\right)+\left(O\left(\frac{1}{x^{2}}\right)\right)^{2}\right)O\left(x^{2}\right)}{x}=\lim_{x\rightarrow0}\frac{O\left(x^{2}\right)-2O\left(x^{2}\right)\cdot O\left(\frac{1}{x^{2}}\right)+O\left(x^{2}\right)\cdot\left(O\left(\frac{1}{x^{4}}\right)\right)}{x}=\lim_{x\rightarrow0}\frac{O\left(x^{2}\right)-2\cdot O\left(1\right)+\left(O\left(\frac{1}{x^{2}}\right)\right)}{x}$$
$$=\lim_{x\rightarrow0}O\left(x\right)-2\cdot O\left(\frac{1}{x}\right)+\left(O\left(\frac{1}{x^{3}}\right)\right) = O\left(\frac{1}{x^{3}}\right)$$
Therefore, the limit doesn't exist.
Did I evaluate it correctly?
Update: the answer is supposed to be 0.
| Much simpler (and no asymptotic analysis required): it it is enough to observe that
$0\le \cos^2\dfrac1x\le 1\,$ and $\,\,\lim\limits_{x\to0}\dfrac{\cos x-1}x=\cos'0 =0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the remainder $ax+b$ when a cubic polynomial $P(x)$ is divided by $x^2-1$? If a cubic polynomial $P(x)$ with real coefficients has remainder 3 when divided by $x-1$ and remainder -7 when divided by $x+1$,
What is the remainder $ax+b$ when divided by $x^2-1$?
I see that since $$P(n) = ax^3 + bx^2 + cx + d$$ then $$P(1) = a + b + c + d = 3 $$ while $$P(-1) = -a + b - c + d = -7$$
I know $x^2-1$ is $(x-1)(x+1)$
I'm stuck here - how can I solve this? Could you please help?
| We have
\begin{align}
p(x) & = (x-1)q(x) +3\\
p(x) & = (x+1)r(x) - 7\\
p(x) & = (x^2-1)h(x) + (ax+b)
\end{align}
This gives us $p(1) = 3$, $p(-1) = -7$. Hence, we have
$$a+b = 3 \text{ and }-a+b = -7$$
Trust you can finish now.
| {
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Intuitive understanding of factoring quadratic equations When factoring a second degree equation $ax^2 + bx + c$ you find the roots then take $a(x - \text{root})(x - \text{root})$. I am wondering why this works. Sorry if poorly phrased question.
| We can show the solutions to $ax^2+bx+c=0$ is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ by completing the square and solving for $x$. So that means the factored form of $ax^2+bx+c$ can be written as $a(x-\frac{-b + \sqrt{b^2-4ac}}{2a})(x-\frac{-b - \sqrt{b^2-4ac}}{2a})$. Expanding this gives: $a(x+\frac{b - \sqrt{b^2-4ac}}{2a})(x+\frac{b + \sqrt{b^2-4ac}}{2a})=a(x^2+(\frac{b- \sqrt{b^2-4ac}}{2a}+\frac{b+\sqrt{b^2-4ac}}{2a})x+\frac{b^2-(b^2-4ac)}{4a^2})=(x^2+\frac{b}{a}x+\frac{4ac}{4a^2})=ax^2+bx+c$
| {
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"timestamp": "2023-03-29T00:00:00",
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solving second order non homogeneous differential equation I am given a non homogeneous differential equation
$$
y''+4y = 3 \csc 2x.
$$
When I try to find value of the Wronskian $W(y_{1},y_{2}),$ the result is zero. I can't solve it the differential equation because of division by zero.
| Start by writing down the charactersitic equation of the associated homogeneous equation and solving for the roots.
$$m^2 + 4 = 0$$
$$m^2 = -4 = 4i^2$$
$m_1= 2i$ and $m_2=-2i$
Thus $y_h=c_1\cos2x +c_2\sin2x$
Now calculate the Wronskian and determine the particular solution.
$w= \left[
\begin{array}{ c c }
\cos(2x) & \sin(2x) \\
-2\sin(2x) & 2\cos(2x)
\end{array} \right] = 2\cos^2(2x)+2\sin^2(2x) = 2(\cos^2(2x)+ \sin^2(2x)) = 2 \times 1 = 2
$
$w_1= \left[
\begin{array}{ c c }
0 & \sin(2x) \\
3\csc(2x) & 2\cos(2x)
\end{array} \right] = -3\csc(2x)\sin(2x) = -3\times\frac{\sin(2x)}{\sin(2x)} = -3
$
$w_2= \left[
\begin{array}{ c c }
\cos(2x) & 0 \\
-2\sin(2x) & 3\csc(2x)
\end{array} \right] = 3\cos(2x)\csc(2x) = 3\times\frac{\cos(2x)}{\sin(2x)} = 3\times{\cot(2x)}
$
Thus
$u_1'=\frac{w_1}{w} = \frac{-3}{2} \implies u_1= \frac{-3}{2}x $
$u_2'= \frac{w_2}{w} = \frac{3}{2}\cot(2x) \implies u_2= \frac{3}{4}\ln|\sin(2x)|$
Then $$y = y_h + y_p = y_h + uy_1 + uy_2$$
$$y= c_1\cos(2x) +c_2sin(2x) - \frac{3}{2}x\cos(2x) +\frac{3}{4}\sin(2x)\ln|\sin(2x)|$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$ |f''(x)+2xf'(x)+(x^2+1)f(x)|\leq1 $ for all $x$. Prove $ \lim_{x\rightarrow\infty }f(x)=0$ Let $f(x):(0, \infty)\rightarrow \mathbb{R} $ be a twice continuously differentiable function
such that
| $ f''(x)+2xf'(x)+(x^2+1)f(x) |\leq1 $ for all $x$. Prove $ \lim_{x\rightarrow\infty }f(x)=0$
I think this can be solved by applying L'Hospital rule twice on $\frac{e^{x^2}f(x)}{e^{x^2}}$. My problem is, Is it possible to apply the rule without knowing about numerator . Also I like to see different types of proofs
| We have that
$$\lvert f''(x)+2xf'(x)+(x^{2}+1)f(x)\rvert\le1$$
Consider the function $g(x)=e^{\frac{x^2}{2}}f(x)$ then
$$\begin{align}
g'(x)&=xe^{\frac{x^2}{2}}f(x)+e^{\frac{x^2}{2}}f'(x)\\
g''(x)&=e^{\frac{x^2}{2}}f(x)+x^2e^{\frac{x^2}{2}}f(x)+xe^{\frac{x^2}{2}}f'(x)+xe^{\frac{x^2}{2}}f'(x)+e^{\frac{x^2}{2}}f''(x)\\
&=e^{\frac{x^2}{2}}(f''(x)+2xf'(x)+(x^2+1)f(x))
\end{align}$$
So we have that:
$$\left\lvert\frac{d^{2}}{dx^{2}}\left(e^{\frac{x^2}{2}}f(x)\right)\right\rvert\le e^{\frac{x^2}{2}}=\left((x^2+1)e^{\frac{x^2}{2}}\right)\frac{1}{x^2+1}=\frac{1}{x^2+1}\cdot\frac{d^2}{dx^2}\left(e^{\frac{x^2}{2}}\right)$$
So
$$\left\lvert\frac{\frac{d^2}{dx^2}\left(e^{\frac{x^2}{2}}f(x)\right)}{\frac{d^2}{dx^2}\left(e^{\frac{x^2}{2}}\right)}\right\rvert\le\frac{1}{x^2+1}\to0 \text{ as }x\to \infty$$
Now use the general form of L'hopital's rule (proved using the Stolz-Cesaro Theorem) to conclude that
$$\lim_{x\to\infty}\frac{\frac{d}{dx}\left(e^{\frac{x^2}{2}}f(x)\right)}{\frac{d}{dx}\left(e^{\frac{x^2}{2}}\right)}=\lim_{x\to\infty}\frac{\frac{d^2}{dx^2}\left(e^{\frac{x^2}{2}}f(x)\right)}{\frac{d^2}{dx^2}\left(e^{\frac{x^2}{2}}\right)}=0$$
Since $\frac{d}{dx}\left(e^{\frac{x^2}{2}}\right)=xe^{\frac{x^2}{2}}\to\infty$ as $x\to\infty$, applying the previous discussion again we conclude that:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^{\frac{x^2}{2}}f(x)}{e^{\frac{x^2}{2}}}=\lim_{x\to\infty}\frac{\frac{d}{dx}\left(e^{\frac{x^2}{2}}f(x)\right)}{\frac{d}{dx}\left(e^{\frac{x^2}{2}}\right)}=0$$
| {
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Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)$
Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)\;,$ Where $a,b\in \mathbb{R}$
$\bf{My\; Try::}$ First we have to calculate $\max(a^2+b,b^2+a) = \left\{\begin{matrix}
a^2+b& \;,a^2+b>b^2+a \\\\
b^2+a& \;,a^2+b\leq b^2+a \\
\end{matrix}\right.$
Now For $\bf{I^{st}}$ case, Here $f(a,b) = a^2+b\;\;,$If $ a^2+b>b^2+a\Rightarrow (a-b)\cdot (a+b-1)>0$
Similarly For $\bf{II^{st}}$ case, $f(a,b) = b^2+a\;\;,$If $ a^2+b\leq b^2+a\Rightarrow (a-b)\cdot (a+b-1)\leq 0$
Now I did not understand how can i solve it, Help me
Thanks
| (Update)
I assume that you want to compute the quantity $$Q:=\min_{(a,b)\in{\Bbb R}^2}\max\{a^2+b,b^2+a\}\ .$$
The difference
$$(b^2+a)-(a^2+b)=(b-a)(b+a-1)$$
vanishes when $u:=b-a=0$, or when $v:=a+b-1=0$. This suggest that we introduce $u$ and $v$ as new variables, which leads to
$$2a=1-u+v,\quad 2b=1+u+v\qquad\bigl((u,v)\in{\Bbb R}^2\bigr)\ .\tag{1}$$
One then computes
$$4(a^2+b)=3+4v+(u-v)^2,\quad 4(b^2+a)=3+4v+(u+v)^2\ .$$
It follows that
$$\max\{a^2+b,b^2+a\}={1\over4}\bigl(3+4v+\max\{(u-v)^2,(u+v)^2\}={1\over4}\bigl(3+4v+\bigl(|u|+|v|\bigr)^2\bigr)\ .$$
Now we have to minimize the right hand side over $(u,v)\in{\Bbb R}^2$. For given $v$ this is minimal when $u=0$, so that we are left with the task to minimize
$$f(v):={1\over4}(3+4v+v^2)={1\over4}\bigl((v+2)^2-1\bigr)\ .$$
The minimum $-{1\over4}$ is taken when $v=-2$. Putting $u=0$, $v=-2$ in $(1)$ leads to $a=b=-{1\over2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Check my general solution to the differential equation? Given differential equation:
$$y' = \frac{(2xy^{3}+4x)}{(x^{2}y^{2}+y^{2})}$$
This is the general solution that I got for the above differential equation:
$$\frac{1}{3} \ln{\lvert y^3+2\rvert}=\ln{\lvert x^2+1\rvert}+\ln C$$
Please check if it is correct?
And for the solution satisfying y(1)=0, this is what I got:
$$(y^3+2)^{\frac{1}{3}}=(x^2+1)+C$$
| Rearranging the differential equation gives
$$ \frac{y^2 y'}{y^3+2} = \frac{2x}{1+x^2} $$
Integrating this does give you the general solution you have,
$$ \frac{1}{3}\log{(2+y^3)} = \log{(1+x^2)}+A, $$
$A$ a constant. Exponentiating both sides gives
$$ y^3+2 = e^{3A}(1+x^2)^3 = B(1+x^2)^3, $$
where I have relabelled the constant as $B$.
To find the solution that satisfies y(1)=0, you have to put $y=0$ and $x=1$ to find the constant $A$ (your solution should not have an undetermined constant if you have one first-order equation and one boundary condition!):
$$ 2 = B \cdot 2^3, $$
so we find that $B=1/4$, or
$$ y^3+2 = \frac{1}{4}(1+x^2)^3. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many ways are there to do this so that no officer picks $3$ students from the same high school? There are $18$ students, three (distinct) students each from $6$ different high schools.
There are $6$ admissions officers, one from each of $6$ colleges. Each of the officers successively picks a subset of
three of the $18$ students to go to their college (once a student is chosen, another college cannot choose him later).
How many ways are there to do this so that no officer picks $3$ students from the same high school?
Answer is:
$\dfrac{18!}{(3!)^6} - 6\dfrac{6 \cdot 15!}{(3!)^5} + {6 \choose 2} \dfrac{6 \cdot 5 \cdot 12!}{(3!)^4} \dots $
Can someone please explain?
| Suppose the colleges are college A, B, C, D, E, F (we can properly order them alphabetically).
By inclusion-exclusion the number of ways no college violates the "don't pick all students from the same high school" rule is:
$$N_0 = N_{\geq 0} - N_{\geq1} + N_{\geq2} - N_{\geq3} + N_{\geq4} - N_{\geq5} + N_{\geq6}$$
Where $N_{\geq k}$ represents the number of ways at least $k$ colleges violate the rule.
$N_{\geq0}$ is simply the number of ways that the colleges can pick the students without regard to the "don't take from a single highschool" rule. By multiplication principle college A picks three students out of the 18 total, college B picks three students out of the remaining 15, etc... for a total number of $\binom{18}{3}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{18!}{(3!)^6}$ number of ways.
For $N_{\geq 1}$, first pick the college that is guaranteed to be in violation of the rule (there are possibly others, hence why we are using inclusion-exclusion): there are six colleges to choose from. Now, take the offending college and have them pick a highschool to take all three students from: there are six highschools to choose from. The remainder of the colleges can take randomly from the remaining students for a total of: $\binom{6}{1}\cdot 6\cdot \binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} = 6\cdot 6\cdot \frac{15!}{(3!)^5}$
For $N_{\geq 2}$, first pick two colleges that are guaranteed to be in violation of the rule: there are $\binom{6}{2}$ choices. From here, let the offending colleges in alphabetical order choose which highschool they want to take from. The first offending college has 6 choices, the second offending college has 5 choices. The remaining colleges each take randomly from the remaining students for a total of $\binom{6}{2}\cdot 6\cdot 5\cdot \frac{12!}{(3!)^4}$.
The remaining cases are similar for a final total of:
$$N_0 = \frac{18!}{(3!)^6} - \binom{6}{1} \cdot 6 \frac{15!}{(3!)^5} + \binom{6}{2}\cdot 6\cdot 5\frac{12!}{(3!)^4} - \binom{6}{3}\cdot 6\cdot 5\cdot 4\frac{9!}{(3!)^3} + \binom{6}{4}\cdot 6\cdot 5\cdot 4\cdot 3\frac{6!}{(3!)^2} - \binom{6}{5}\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \frac{3!}{(3!)^1} + \binom{6}{6}\cdot 6!\cdot \frac{0!}{(3!)^0}$$
(note: it is impossible for 7 or more colleges to violate the rule since there are only six colleges that we are worried about)
In case it wasn't made perfectly clear, in response to your comment "but I don't understand why that should be multiplied by an additional 6, or why the next term is multiplied by 30", these numbers appear because in our multiplication principle setup they represent the number of ways of having the violating colleges choose which highschools they will take students from. Note that since the colleges are distinct, order here matters so we use a permutation instead of a combination.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Range of the function $f(x) = |x-6|+x^2-1$
find the Range of $f(x) = |x-6|+x^2-1$
$$ f(x) = |x-6|+x^2-1 =\left\{
\begin{array}{c}
x^2+x-7,& x>0 .....(b) \\
5,& x=0 .....(a) \\
x^2-x+5,& x<0 ......(c)
\end{array}
\right.
$$
from eq (b) i got $$f(x)= \left(x+\frac12\right)^2-\frac{29}4 \ge-\frac{29}4$$
and from eq (c) i got $$f(x)= \left(x-\frac12\right)^2+\frac{19}4 \ge\frac{19}4$$
and eq(b) tells me that it also passes through 5 and so generalize all this and found its range is $\left[-\frac{29}4 , \infty\right)$
but the graph says its range is $(5, \infty)$
| You should have $x-6<0$, $x-6=0$ and $x-6>0$ respectively. Always look to the entire expression within the absolute value.
Oh, and another thing: While finding such minimum, you need to check whether it is in the domain. For example, you have $x>0$ (should be $x>6$) for (b), but the minimum is given at $x=-\frac{1}{2}$.
| {
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Find the area of the region $ABCD$. In the Figure $\square PQRS$ is a square with side $2\sqrt6$.
By joining the midpoints another square $\square WXYZ$ is formed .
Circles are drawn with $4$ vertices as the center and
radius equal to the side of the square $\square WXYZ$.
Find the area common to all the $4$ circles .
$a.)6\pi\left(\dfrac{\sqrt3-1}{2}\right)\\
b.)4\pi-3\sqrt3\\
c.)\dfrac12\left(\pi -3\sqrt3\right)\\
d.)4\pi-12\left(\sqrt3 -1\right)\quad \LARGE \color{green}{\checkmark}\\$
So i have to find the area of region $ABCD$.
I have found that $WX=WY=YZ=XZ=2\sqrt3.$
Let Region $WBA$=Region $XAD$=Region $ZCD$=Region $YBC=a$
and
Region $WAX$=Region $XDZ$=Region $ZCY$=Region $YBW=b$
and
Region $ABCD=m$
$Area(\square WXYZ)=12\\
m+4a+4b=12$.
Area of sectors.
$Area$(sector $WZY$)=$Area$(sector $XYZ$)=$Area$(sector $WXY$)=$Area$(sector $WZX$)=$m+3a+2b=3\pi$.
Now i m stucked.
I m looking for simple short way.
I have studied maths upto $12th$ grade.
| Each side of smaller square WXZY is calculated as $$WX=\sqrt{(WQ)^2+(QX)^2}=\sqrt{(\sqrt{6})^2+(\sqrt{6})^2}=2\sqrt{3}$$
The area common to four circles each having a radius $a$ & center at the vertice of a cube with side $a$ is given by the general expression (obtained by using integration) $$\bbox[4pt, border:1px solid blue;]{A_{common}=a^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)}$$ Hence, by setting $a=2\sqrt{3}$ in the above formula, we get the requited common area ABCD $$=(2\sqrt{3})^2\left(\frac{\pi-3(\sqrt{3}-1)}{3}\right)=4\left(\pi-3(\sqrt{3}-1)\right)=4\pi-12(\sqrt{3}-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Express $w=f(z)=\frac{1}{(1-z)^2}$ in the form $w=u(x,y)+iv(x,y)$ I start by writing $f(z)$ as $$\frac{1}{(1-(x+iy))^2}$$
and then I expand the bottom to get $$\frac{1}{(1-2x+x^2-y^2) + i(2y-2xy)}$$
The answer says $$w=\frac{(1+x^2-2x-y^2)-(i(2xy-2y)}{(1+x^2-2x-y^2)^2+(2xy-2y)^2}$$
How do I get to this stage?
| Here is another way: $u(z) = \operatorname{re} f(z) = {1 \over 2} (f(z)+ \overline{f(z)})$, $v(z) = \operatorname{im} f(z) = {1 \over 2i} (f(z)- \overline{f(z)})$.
In this case, we have (also using $w \bar{w} = |w|^2$.)
\begin{eqnarray}
u(z) &=& {1 \over 2} ({1 \over (1-z)^2}+{1 \over (1-\bar{z})^2}) \\
&=& {1 \over 2} {(1-\bar{z})^2 + (1-z)^2 \over |1-z|^4 }\\
&=& {1 \over 2} {2 - 2 (z + \bar{z}) + z^2 + \bar{z}^2 \over |1-z|^4 } \\
&=& {1 \over 2} {2 - 2 \cdot 2 \operatorname{re}(z) + 2 \operatorname{re}(z^2) \over |1-z|^4 } \\
&=& { 1 - 2x +x^2-y^2\over ((1-x)^2+y^2)^2}
\end{eqnarray}
Similarly:
\begin{eqnarray}
v(z) &=& {1 \over 2i} ({1 \over (1-z)^2}-{1 \over (1-\bar{z})^2}) \\
&=& {1 \over 2i} {(1-\bar{z})^2 - (1-z)^2 \over |1-z|^4 }\\
&=& {1 \over 2i} {2 (z - \bar{z}) + \bar{z}^2 - z^2 \over |1-z|^4 } \\
&=& {1 \over 2i} {2 \cdot 2i \operatorname{im}(z) - 2 i \operatorname{im}(z^2) \over |1-z|^4 } \\
&=& { 2y - 2xy\over ((1-x)^2+y^2)^2}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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More than 4 at Even throw of Fair Dice A dice is being thrown till we get number greater than 4 at Even Throw. What is the Probability of this Event?
I have two approaches:
Method $1$.
The probability of number greater than $4$ is $\frac{1}{3}$ and Probability of number less than or equal to $4$ is $\frac{2}{3}$
let $T_1$, $T_2$, $T_3$ $\cdots$ be Throw numbers. Then the Required Probability is
$$P(A)=P(T_1T_2)+P(T_1T_2T_3T_4)+P(T_1T_2T_3T_4T_5T_6)+\cdots$$
But $P(T_1T_2)=\frac{2}{3}.\frac{1}{3}$
$P(T_1T_2T_3T_4)=\frac{2}{3}.\frac{2}{3}.\frac{2}{3}.\frac{1}{3}$ and so on finally we get an infinite GP as
$$P(A)=\frac{1}{3}\left(\frac{2}{3}+\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^5+\cdots\right)=\frac{2}{5}$$.
In Method $1$, I considered that at odd throws we get number less than or Equal to $4$ and at Even throw a number greater than $4$.
Method $2$. In This i will consider that at odd throws we can get any number, but still i need Probability of number greater than $4$ at Even throw.
Now here $P(T_1T_2)=1. \frac{1}{3}=\frac{1}{3}$
$P(T_1T_2T_3T_4)=1.\frac{2}{3}.1.\frac{1}{3}=\frac{1}{3}.\frac{2}{3}$
$P(T_1T_2T_3T_4T_5T_6)=1.\frac{2}{3}.1.\frac{2}{3}.1.\frac{1}{3}=\frac{1}{3}.\left(\frac{2}{3}\right)^2$ Hence
$$P(A)=\frac{1}{3}\left(1+\frac{2}{3}+\left(\frac{2}{3}\right)^2+\cdots\right)$$
hence $$P(A)=1$$...
Can any one Resolve this Ambiguity
| If you insist that number greater than 4 comes on an even throw, you can ignore all the odd throws. They are not events, because they cannot determine success or failure. The simple answer is you keep trying with some chance each time. Eventually you will succeed. The probability is $1$. You need to understand that=that is the point of the problem. Your error in method 1 is that all the $T_n$ with $n$ odd are $1$ because that doesn't matter.
| {
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Is it possible to express $x^4-x^3+3x^2-4x+6$ as a product of polynomials of smaller degree with integer coefficients? Is it possible to express $x^4-x^3+3x^2-4x+6$ as a product of polynomials of smaller degree with integer coefficients?
My attempt: By equating the polynomial to $0$, one obtains $x=1\pm i, \pm\frac{1}{2}i(\sqrt{11}\mp i)$. From this, one can write the polynomial given in the question as the
$$(x^2-2x+2)(x^2+x+3)$$
by using Viete formula.
However, I'm not sure whether I can solve the polynomial before hand and use the answers to express it as product of polynomial of smaller degree.
| Your polynomial has no integer root since such a root must be a divisor of $6$. More over it can have no negative root since, if $x>0$, $p(x)> 0$. Thus one has to test the values $\,1,2 ,3,6$ and none is a root.
Thus a factorisation must have the form:
$$x^4-x^3+3x^2-4x+6=(x^2+ax+b)(x^2+a'x+b')$$
which leads to the system:
$$\begin{cases}
a+a'=-1\\aa'+b+b'=3\\ab'+ba'=-4\\bb'=6
\end{cases}$$
Furthermore, $b$ and $b'$ have to be positive: if they were negative, the factors, hence $p(x)$, would have a negative root.
Theere remains two possibilities: $b=2,\enspace b'=3\,$ or $\, b=1,\enspace b'=6$. From the second equation, we deduce that $aa'=-2$ or $aa'=-4$.
Now we have to solve for the classical problem of finding two numbers, knowing their sum, $-1$ and their product, $-2$ or $-4$. They're roots of one of these equations:
$$t^2+t-2=0,\quad t^2+t-4=0.$$
It happens that the first one has integer roots, $1$ and $-2$, while the second has not.
Thsi third equation will let us know which root is $a$ and which is $a'$. It can be written as $\,3a+2a'=-4$, which enforces $a=-2,\ a'=1$. Finally we obtain the factorisation:
$$x^4-x^3+3x^2-4x+6=(x^2-2x+2)(x^2+ x+3). $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Use quadratic formula to find upper and lower limits of an expression
Using quadratic formula show that $\frac{x^2-x+1}{x^2+x+1}$ lies
between $3$ and $\frac{1}{3}$ for all real values of $x$.
Let $\frac{x^2-x+1}{x^2+x+1}=y$, then $\frac{x^2+1}{-x}=\frac{y+1}{y-1}$
$\therefore (y-1)x^2+(y+1)x+(y-1)=0$, if $x$ is real then
$(y+1)^2-4(y-1)(y-1)\geq 0$
$-3y^2+10y-3 \geq 0$
$(y-3)(-3y+1) \geq 0 \Rightarrow y>3 \text{ AND } y <\frac{1}{3}$
What am I doing wrong?
| Use the symmetries of the function. First, note that $\,y(-x)=\dfrac 1{y(x)}$, hence it is enough to obtain bounds for $x>0$.
Let's rewrite the expression:
$$y=\frac{x^2-x+1}{x^2+x+1} =\frac{x+\cfrac 1x-1}{x+\cfrac 1x+1}$$
so, setting $u=x+\dfrac1x$, we have
$$y=\dfrac{u-1}{u+1}=1-\frac2{u+1},$$
which is an increasing function of $u$. Note that, if $x>0$, $u\ge 2$, so, for $x\ge 0$, we have:
$$1-\frac23=\frac13\le y \le 1$$
If $x<0$, we deduce that
$$1\le y(-x)=\frac 1{y(x)}\le 3.$$
Grouping these inequalities, we obtain that in all cases:
$$\frac13\le y\le 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$? if we have $a+b+c=1$ and $ab+bc+ac=\frac{1}{3}$ then what is the value of $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
and
$$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}$$.
from the hypothesis we have also $a^2 +b^2 +c^2 =\frac{1}{3}$ and also we have $a^3 +b^3+c^3=3abc$ but I don't know how should get the result. it will be great if you share your Idea with me,thanks.
| Hint:since
$$a+b+c=1,ab+bc+ac=\dfrac{1}{3}$$ so
$$(a+b+c)^2-3(ab+bc+ac)=0\Longrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\Longrightarrow a=b=c$$
| {
"language": "en",
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Find an orthogonal basis for the space spanned by the columns of the given matrix. Let $$X = \begin{pmatrix}
1 & 1 & 4 \\
1 & 2 & 1 \\
1 & 3 & 0 \\
1 & 4 & 0 \\
1 & 5 & 1 \\
1 & 6 & 4 \end{pmatrix}$$
It is immediately clear to me that the columns of $X$ form a basis for the space spanned by the columns of $X$.
How does one generate an orthogonal basis for the space spanned by the columns of $X$? I know that for each element $x \neq y$ in this basis that $x\cdot y = 0$, but this doesn't help me actually generate the basis.
| To apply Gran-Schmidt, I used
$$u_k = v_k - \sum\limits_{j=1}^{k-1}\operatorname{proj}_{u_j}(v_k)$$
where $$\operatorname{proj}_{u_j}(v_k) = \dfrac{v_k \cdot u_j}{u_j \cdot u_j}u_j\text{.}$$
We have $$u_1 = v_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
1 \\
1 \\
1\end{pmatrix}\text{.}$$
Now $$\operatorname{proj}_{u_1}(v_2) = \dfrac{v_2 \cdot u_1}{u_1 \cdot u_1}u_1 = \dfrac{21}{6}u_1\text{.}$$
Thus, $$u_2 = v_2 - \dfrac{21}{6}u_1 = \begin{pmatrix}
-5/2 \\
-3/2 \\
-1/2 \\
1/2 \\
3/2 \\
5/2\end{pmatrix}\text{.}$$
Also,
\begin{align*}
\operatorname{proj}_{u_1}(v_3) &= \dfrac{v_3 \cdot u_1}{u_1 \cdot u_1}u_1 = \dfrac{10}{6}u_1 \\
\operatorname{proj}_{u_2}(v_3) &= \dfrac{v_3 \cdot u_2}{u_2 \cdot u_2}u_2 = \dfrac{0}{17.5}u_2 = 0\text{.}
\end{align*}
Thus, $$u_3 = v_3 - \dfrac{10}{6}u_1 = \begin{pmatrix}
7/3 \\
-2/3 \\
-5/3 \\
-5/3 \\
-2/3 \\
7/3\end{pmatrix}\text{.}$$
Hence $\{u_1, u_2, u_3\}$ is an orthogonal basis for the space spanned by the columns of $X$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$ $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$$
Assume that $x^2 - 6x + 12 = (x - 3)(x - 3) + 3 = (x - 3)^2 + 3$,
then $t = x - 3 \rightarrow dt = dx$,
since $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx} = \int_{0}^{3} \frac{12}{t^2 + 3}\,dt$$.
However, I am unsure as to how to continue.
| It is $\frac{1}{3} \int{\dfrac{12}{\left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1} \,dt}$, since $\dfrac{1}{t^{2} + 3} = \dfrac{1}{3 \left(\dfrac{t^{2}}{3} + 1 \right)} = \dfrac{1}{3 \left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1}$.
| {
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Question about a trigonometry proof? I just want to ask how can you prove that 2α is twice the value of α in the following figure that depicts a proof of an arctangent identity (and likewise, for β as well).
| let $$a = \tan^{-1}\left(\frac{1-y}x\right), b = \tan^{-1}\left(\frac{1-x}y\right) .$$ then
$$\begin{align}\tan(a+b) &= \frac{\tan a + \tan b}{1- \tan a \tan b} \\
&= \frac{\frac{1-y}x+\frac{1-x}y}{1-\frac{1-y}x \frac{1-x}y} \\
&= \frac{y+x-y^2-x^2}{xy-(1-x-y+xy)} \\
&=\frac{(x+y)(1-x-y)}{-(1-x-y)} \\
&= -x-y\end{align}$$
don't you need $x+y = -1$ for $a+b = \pi/4?$
| {
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How to simplify $\frac{\sqrt{x^3}}{\sqrt[3]{x^4}}$? Please, could someone help simplify and show the steps on how to simplify $$\frac{\sqrt{x^3}}{\sqrt[3]{x^4}}?$$
Thank you.
| $$\large{\sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^{\frac{3}{2}}}$$
Now
$$\large{\sqrt[3]{x^4} = (x^4)^{\frac{1}{3}} = x^{\frac{4}{3}}}$$
Now you divide these two $$\large{\frac{x^{\frac{3}{2}}}{x^{\frac{4}{3}}} = x^{\frac{3}{2}-\frac{4}{3}} = \color{red}{x^{\frac{1}{6}} =\sqrt[6]{x}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
If $\tan x=\sqrt{\frac{a}{b}}$ where a,b are positive real numbers and x is in 1st quadrant then find the value of $\sin x\sec^7x+\cos x\csc^7x$ The answer is $\frac{(a+b)^3(a^4+b^4)}{(ab)^{\frac{7}{2}}}$. I just want to now how to do it.
| We very well know that $$\color {blue}{\sec^2 x-\tan^2 x=1}$$ $$\implies \sec^2 x=1+\tan^2 x=1+\left(\sqrt{\frac{a}{b}}\right)^2=1+\frac{a}{b}=\frac{a+b}{b}$$ $$\implies \sec x=\pm \sqrt{\frac{a+b}{b}}$$ But x lies in the first quadrant i.e. $\color{blue}{0<x<\frac{\pi}{2}}$ hence $\sin x$, $\cos x$, $\csc x$ & $\sec x$ all have positive values in the I-quadrant, thus we have $$\color {blue}{ \sec x= \sqrt{\frac{a+b}{b}}}$$ $$\implies \cos x=\frac{1}{\sec x}=\frac{1}{\sqrt{\frac{a+b}{b}}}=\sqrt{\frac{b}{a+b}}$$ $$\implies \sin x=\frac{\tan x}{\sec x}=\frac{\sqrt{\frac{a}{b}}}{\sqrt{\frac{a+b}{b}}}=\sqrt{\frac{a}{a+b}}$$ $$\implies \csc x=\frac{1}{\sin x}=\sqrt{\frac{a+b}{a}}$$
Now substituting the above values, we have $$\sin x\sec^7 x+\cos x\csc^7 x=\sqrt{\frac{a}{a+b}}\left(\sqrt{\frac{a+b}{b}}\right)^7+\sqrt{\frac{b}{a+b}}\left(\sqrt{\frac{a+b}{a}}\right)^7 $$ $$=\frac{(a+b)^3}{b^3} \sqrt{\frac{a}{a+b}}\sqrt{\frac{a+b}{b}}+\frac{(a+b)^3}{a^3}\sqrt{\frac{b}{a+b}}\sqrt{\frac{a+b}{a}}$$ $$=\frac{(a+b)^3}{b^3}\sqrt{\frac{a}{b}}+\frac{(a+b)^3}{a^3}\sqrt{\frac{b}{a}} $$ $$=(a+b)^3\left[\frac{a^{\frac{1}{2}}}{b^{\frac{7}{2}}}+\frac{b^{\frac{1}{2}}}{a^{\frac{7}{2}}}\right]$$ $$=(a+b)^3\left[\frac{a^4+b^4}{a^{\frac{7}{2}}b^{\frac{7}{2}}}\right]$$$$=\color{purple}{\frac{(a+b)^3(a^4+b^4)}{(ab)^{\frac{7}{2}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Obfuscated proofs I am, just for fun, looking for long and complicated proofs for statements which can be proven rather easily and much faster. The proof itself still has to be correct however.
While the proof should be obfuscated, all parts should have some relevance. So do not prove Fermat‘s last theorem and end with "ah, by the way: 1+1=2, so the statement follows.
It is also boring to obfuscate simple arithmetic; one can prove "1+1=2" in 100 pages only using addition, subtraction, multiplication and division – but that is not fun.
I rather look for some very interesting obfuscation of a proof. Maybe a statement of elementary number theory can be proven in a "nice" complicated way. Or maybe one can use functional analysis to prove basic analysis stuff etc.
Here is one (not that good) example: Theorem: For $a, b \in \mathbb{R}$ it holds that $(a-b)^2 = a^2 - 2ab + b^2$.
Proof: Let $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x^2$. As $f$ is analytical the Taylor expansion of $f$ converges. Therefore
$$
\begin{align*}
x^2
&= f(x) \\
&= Tf(x,b) \\
&= \sum_{n=0}^\infty \frac{f^{(n)}(b)}{n!} (x-b)^n \\
&= \frac{b^2}{0!} + \frac{2b}{1!} (x-b) + \frac{2}{2!} (x-b)^2 + \sum_{n=3}^\infty \frac{0}{n!} (x-b)^n\\
&= b^2 + 2bx - 2b^2 + (x-b)^2 \\
&= -b^2 + 2bx + (x-b)^2,
\end{align*}
$$
ie.
$$
x^2 + b^2 - 2bx = (x-b)^2
$$
and for $x = a$ the theorem follows.
| Proof: $0a = 0$
$$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2$$
$$a^2 = a^2 + 2(0) + 0^2$$ $$0 = 2(0) + 0^2$$ $$2(0) = (1+1)0 = 1(0) + 1(0) = 0 + 0 = 0$$ $$\dfrac{0^2}{2} >= \sqrt{0^2} = 0$$ $$0^2 >= 2(0)$$
We have proven that $$2(0) = 0$$
Therefore, $$(a+0)^2 = a^2 = a^2 + 2(0) + 0^2 = a^2 + 0$$ $$a^2 = a^2 + 0$$ $$0 = 0$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the value of $a^4+b^4+c^4$?
Consider $a,b,c$ such that $a+b+c =1, a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$. Find the value of $a^4+b^4+c^4$, if possible.
Trial: I observe that
\begin{align}
a^4+b^4+c^4 &=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\\&=2^2-2[(ab+bc+ca)^2-2abc(a+b+c)]\\&=4-2[(ab+bc+ca)^2-2abc]
\end{align}
Then
\begin{align}
(a+b+c)^2 &=a^2+b^2+c^2+2(ab+bc+ca)\\\implies 1^2= 2+ 2(ab+bc+ca)\\\implies ab +bc+ca &=-\dfrac12
\end{align}
Then I am stuck. Please help.
| Hint:let $$x_{n}=a^n+b^n+c^n$$
and
$$x_{n+2}=(a+b+c)x_{n+1}-(ab+bc+ac)x_{n}+abc\cdot x_{n-1}$$
so
$$x_{4}=(a+b+c)x_{3}-(ab+bc+ac)x_{2}+abc\cdot x_{1}$$
and we easy to find $ab+bc+ac,abc$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+\zeta$ using trigonometric substitution We know that $\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+\zeta$.
I tried to verify formula using trigonometric substitution and I had some problems.
Here is all my steps:
$\int\frac{1}{x^2-a^2}dx=\frac{1}{a}\int\frac{sec(\theta)\:d\theta}{tg(\theta)}=\frac{1}{a}\int{csc(\theta)}\:d\theta=-\frac{1}{a}\ln\left(\frac{x}{\sqrt{x^2-a^2}}+\frac{a}{x}\right)+\zeta\Rightarrow\theta=\sec^{-1}(\frac{x}{a})$.
How can I continue such that to get $\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$ ?
| $\DeclareMathOperator{\sech}{sech}$If it matters, the step where you integrate $\csc$ looks awry; should be
$$
\frac{1}{a} \int \csc\theta\, d\theta
= -\frac{1}{a} \log\left\lvert\csc\theta + \cot\theta\right\rvert + C
= -\frac{1}{a}\log\left\lvert\frac{x + a}{\sqrt{x^{2} - a^{2}}}\right\rvert + C.
$$
Then factor the radicand as a difference of squares and use properties of logarithms.
Yet another approach (in addition to partial fractions and verifying the formula via differentiation), incidentally, is to use the hyperbolic substitution $x = a\tanh u$, $dx = a\sech^{2} u\, du$ which leads to
$$
\int \frac{dx}{x^{2} - a^{2}}
= \int \frac{a \sech^{2} u\, du}{a^{2} \sech^{2} u}
= \frac{1}{a} u + C
= \frac{1}{a} \tanh^{-1} \frac{x}{a} + C
= \frac{1}{2a} \log \left\lvert \frac{x - a}{x + a} \right\rvert + C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Classifying a singularity $f(z)=\frac{\sin z}{\cos(z^3)-1}$ $$f(z)=\frac{\sin z}{\cos(z^3)-1}$$
I need to classify the singularity at $z=0$ and calculate the residue.
Apparently the method of evaluating this is through series, but that seems strange to me since I have only encountered apparently trivial cases through cauchy's theorem. What is the easiest approach to calculating the residue?
| As you mention, we can understand this through series. (One could also proceed using the formula that Arthur mentions in his comment).
Firstly, we know that $\sin z = z - z^3/6 + \dots$ and $\cos(z^3) - 1 = (1-1) - z^6/2 + \ldots = -z^6/2 + \dots$. So the ratio looks like
$$ \frac{z - z^3/6}{-z^6/2} = \frac{1 - z^2/6}{-z^5/2}.$$
We've omitted higher order terms because they do not contribute here. This tells us that we have a pole of order $5$.
You also ask how to compute the residue, which is the coefficient of the degree $-1$ term. Let's show the let's use series method of getting the residue.
We now know that
$$ \frac{\sin z}{\cos z^3 - 1} = \frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots$$
This also means that
$$ \sin z = (\cos z^3 - 1)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We can now compare series expansions to arrive at the answer. In particular,
$$ z - z^3/6 + z^5/120 + \dots = (-z^6/2 + z^{12}/6 + \dots)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We know that $a_{-5} = -2$ from above. Notice that $a_{-4} = a_{-2} = a_0 = 0$, which we can see by noticing that the products with $-z^6$ have degrees that don't appear on the left or by noticing that the left side is an odd function and $\cos z^3 - 1$ is even, so the Laurent expansion must be an odd function.
So how about $a_{-1}$? Notice that the only way to multiply two terms on the right to get a term like $z^5$ is to multiply $-z^6/2$ and $a_{-1}/z$. So by comparing coefficients, we see that
$$ a_{-1} \cdot \frac{-1}{2} = \frac{1}{120},$$
so that
$$a_{-1} = -\frac{1}{60}.$$
So this is the residue. *As an aside, notice that we didn't even have to compute the $a_{-3}$ term. This sort of thing can happen when we have really sparse series, like $\cos z^3 - 1$. In these cases, we can very quickly compute residues with series, especially when many things can be done quickly in your head, as in this case.*$\diamondsuit$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example:
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way?
| One method that avoids fractions until the very end is to multiply through by $4a$.
\begin{align}
ax^2 + bx + c &= 0 \\
4a^2x^2 + 4abx + 4ac &= 0 \\
4a^2x^2 + 4abx + b^2 &= b^2 - 4ac \\
(2ax + b)^2 = b^2 - 4ac \\
2ax + b &= \pm \sqrt{b^2 - 4ac} \\
x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 4
} |
Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1 = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red 4$ and $46 = 6\times 7 + \color \red 4$.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c \geq x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. $x$ is supposed to be the largest possible value that holds the assertion. Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?
| Your claim is not correct. Let $a=13$, $b=7$, $c=1$, $x=2$.
Clearly, $a=b=c = 1 \mod x$. So your assumptions are met.
Then
$$ a-b = 13-7 = 6$$
$$ b-c = 7-1 = 6 $$
$$ a-c = 13-1 = 12.$$
Obviously, $\mathrm{gcd}(6,6,12) = 6 \neq 2 = x$.
This leads to a contradiction, and hence your statement is not correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 4
} |
Find sum of series I need to find the sum of the following series:
$$\sum_{n=2}^\infty \ln\left(1-\frac 1{n^2}\right)$$
How to proceed with this?
| Another way to compute the product.
$$\sin(z) = z\prod_{n=1}^\infty \left(1-\frac{z^2}{\pi^2n^2}\right)$$
$$\frac{\sin(z)}{z(1-\frac{z^2}{\pi^2})} = \prod_{\color{red}{n=2}}^\infty \left(1-\frac{z^2}{\pi^2 n^2}\right)$$
$$\lim_{z\rightarrow \pi}\frac{\sin(z)}{z(1-\frac{z}{\pi})(1+\frac{z}{\pi})} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi}\frac{\color{blue}{-(z-\pi)}+ \color{green}{(z-\pi)^3/6+\cdots}}{(1-\frac{z}{\pi})} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi}\frac{\color{blue}{-(z-\pi)}+\color{green}{(z-\pi)^3/6+\cdots}}{-\frac{1}{\pi}(z-\pi)} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi} \color{blue}{ \frac{ -(z-\pi) }{-\frac{1}{\pi} (z-\pi)}} + \color{green}{\frac{(z-\pi)^3/6+\cdots }{-\frac{1}{\pi} (z-\pi)}} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi}\lim_{z\rightarrow \pi} \color{blue}{ \pi} +\color{green}{ \pi(z-\pi)^2/6+\cdots } = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi} (\color{blue}{\pi} + \color{green}{0} )= \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2\pi} \color{blue}{\pi}= \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
$$\frac{1}{2} = \prod_{n=2}^\infty \left(1-\frac{1}{ n^2}\right)$$
So $$ \sum_{n=2}^\infty \log(1-\frac{1}{n^2}) = -\log(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim\limits_{x\to\ 0}\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$ without L'Hospital's rule I have no idea how to solve this limit. I tried transforming the expression
$$\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$$
in
$$e^{\ln\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}}$$
but that didn't work.
| we will use maclaurrin series to find an approximation for
$$\begin{align} \cos x \cos^{1/2}(2x)\cos^{1/3}(3x) &=\left(1 - \frac12 x^2 + \cdots\right)\left(1 - \frac12 (2x)^2 + \cdots\right)^{1/2}\left(1 - \frac12 (3x)^2 + \cdots\right)^{1/3} \\
&= \left(1 - \frac12 x^2 + \cdots\right) \left(1 - x^2 + \cdots\right) \left(1 - \frac32 x^2 + \cdots\right)\\
&=1 - 3x^2 + \cdots\end{align} $$ therefore $$\lim_{x \to 0}\frac{1-\cos x (\cos 2x)^{{1}/{2}} (\cos 3x)^{{1}/{3}}}{x^2} = 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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A unusual inequality about function $\ln$ These day,I met a unusual inequality when I solve a difficult problem, and proving the inequality means I have done the work!
Could you show me how to prove it or deny it? By the way, I believe that it's true!
Prove that, for all $t > 0$,
\begin{align*}
&4\ln t\ln (t + 2) - \ln t\ln (t + 1) - 3\ln t\ln (t + 3)\\
+ &4\ln (t + 1)\ln (t + 3) - 3\ln (t + 1)\ln (t + 2) - \ln (t + 2)\ln \left( {t + 3} \right)>0.
\end{align*}
Let
$$f\left( t \right) = 4\ln t\ln \left( {t + 2} \right) - \ln t\ln \left( {t + 1} \right) - 3\ln t\ln \left( {t + 3} \right) + 4\ln \left( {t + 1} \right)\ln \left( {t + 3} \right) - 3\ln \left( {t + 1} \right)\ln \left( {t + 2} \right) - \ln \left( {t + 2} \right)\ln \left( {t + 3} \right),$$
We have
$$f'\left( t \right) = \frac{{2\left[ {{t^2}\ln t - 3{{\left( {t + 1} \right)}^2}\ln \left( {t + 1} \right) + 3{{\left( {t + 2} \right)}^2}\ln \left( {t + 2} \right) - {{\left( {t + 3} \right)}^2}\ln \left( {t + 3} \right)} \right]}}{{t\left( {t + 1} \right)\left( {t + 2} \right)\left( {t + 3} \right)}}.$$
Let
$$g\left( t \right) = {t^2}\ln t - 3{\left( {t + 1} \right)^2}\ln \left( {t + 1} \right) + 3{\left( {t + 2} \right)^2}\ln \left( {t + 2} \right) - {\left( {t + 3} \right)^2}\ln \left( {t + 3} \right),$$
we got
$$g'\left( t \right) = 2\left[ {t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right)} \right].$$
And let
$$h\left( x \right) = t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right),$$
we have
\begin{align*}
h'\left( x \right) &= \ln t - 3\ln \left( {t + 1} \right) + 3\ln \left( {t + 2} \right) - \ln \left( {t + 3} \right)\\
&= \ln \frac{{t{{\left( {t + 2} \right)}^3}}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}} = \ln \left[ {1 - \frac{{2t + 3}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}}} \right] < 0.
\end{align*}
However, it seems that there are no use!
| EDIT: I made a typo, entering the function into the graphing software, leading to the following believable conjecture (based on the incorrect graph). I'll leave the answer up for posterity, but please don't upvote it.
Conjecture:
*
*$f(t) > 0$ for $0 < t < 1$,
*$f(t) < 0$ for $t > 1$,
*$f(1) = 0$, and $t = 1$ is the only root,
*$f'(t) < 0$ for all $t > 0$ ($f$ is always decreasing),
*$f''(t) > 0$ for all $t > 0$ ($f$ is always concave up).
Here's a better graph, which suggests that the inequality in the question is correct.
By the way, you can have a look at the graph and manipulate the zoom here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
} |
Generating Functions with Fibonacci a) Let
\begin{align*}
F_{\text{even}}(x) &= F_0x^0 + F_2x^2 + F_4x^4 + F_6x^6 + F_8x^8 + \cdots \\
&= x^2 + 3x^4 + 8x^6 + 21x^8 + \cdots
\end{align*}
be the generating function whose coefficient of $x^n$ is the $n^{\text{th}}$ Fibonacci number for even $n$, and is zero for odd $n$.
Write $F_{\text{even}}(x)$ as a rational function (that is, as a simplified quotient of polynomials).
b) For what ordered pair of constants $(a,b)$ is it true that $F_{2n}=aF_{2n-2}+bF_{2n-4}$ for all integers $n\ge 2$?
How can I approach this with generating functions?
| By making use of
\begin{align}
F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta}
\end{align}
where $2 \alpha = 1+ \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$, then
\begin{align}
\sum_{n=0}^{\infty} F_{2n} \, x^{2n} &= \frac{1}{\alpha - \beta} \, \left( \frac{1}{1-\alpha^{2} x^{2}} - \frac{1}{1 - \beta^{2} x^{2}} \right) = \frac{F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}}
\end{align}
where $L_{n}$ are the Lucas numbers.
By using the difference equation $F_{2n+4} = a F_{2n+2} + b F_{2n}$ then
\begin{align}
\sum_{n=0}^{\infty} F_{2(n+2)} \, x^{2n} &= a \sum_{n=0}^{\infty} F_{2(n+1)} \, x^{2n} + b \sum_{n=0}^{\infty} F_{2n} \, x^{2n} \\
\sum_{n=2}^{\infty} F_{2n} \, x^{2n-4} &= a \sum_{n=1}^{\infty} F_{2n} \, x^{2n-2} + b \sum_{n=0}^{\infty} F_{2n} \, x^{2n} \\
\frac{1}{x^{4}} \, \left( - F_{2} x^{2} + \frac{F_{2} x^{2}}{1 - L_{2} x^{2} + x^{4}} \right) &= \frac{a}{x^{2}} \, \frac{F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}} + \frac{b \, F_{2} \, x^{2}}{1 - L_{2} x^{2} + x^{4}} \\
\end{align}
which becomes $F_{2} L_{2} x^{4} - F_{2} x^{6} = a F_{2} x^{4} + b F_{2} x^{6}$ and leads to $a = L_{2} = 3$ and $b = -1$. This can be verified in the following way:
\begin{align}
3 F_{2n-2} - F_{2n-4} &= 3 F_{2n-2} - ( F_{2n-2} - F_{2n-3} ) \\
&= F_{2n-1} + F_{2n-2} \\
&= F_{2n}
\end{align}
which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof coutinuity $f:=\left\{\begin{matrix}
\frac{xy}{\sqrt{x}+y^2}, x,y\neq 0\ begin & \\
0, x,y=0 &
\end{matrix}\right.$
Is f continuous in (0,0)?
My idea is:
$\left | f(x,y) \right | = \left | \frac{x*y}{\sqrt{x}+y^2} \right |\leq \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right|$
we know:
$\sqrt{x}+\sqrt{y}>\sqrt{x+y} \implies \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right| \leq\left | \frac{x*y}{\sqrt{x+y}} \right|$
we know:
$\sqrt{x+y} \geq 2*\sqrt{x}\sqrt{y} \implies \left | \frac{x*y}{\sqrt{x+y}} \right| \leq \left | \frac{x*y}{(x*y)^{\frac{1}{4}}} \right|=(x*y)^{\frac{3}{4}}$
with $(x,y) \implies 0$ we get $(x*y)^{\frac{3}{4}}$ =0
Can someone please check my suggestion?
| Try setting $y=kx$ for some real number $k$, obviously non-zero, and consider the behaviour of the function with pairs constrained to such a line.
| {
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"url": "https://math.stackexchange.com/questions/1324302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$x^3-9x-5=0$, then what is $x^4-18x^3-81x^2-12$ If we have $x^3-9x-5=0$, then what $x^4-18x^3-81x^2-12$ equals to?
This is a multiple choices question.
A)$5$ B)$25$ C)$42$ D)$67$ E)$81$.
My attempt,
By long division of polynomials, we have $x^4-18x^3-81x^2-12=(x-18)(x^3-9x-5)+(-72x^2-157x-102)$
Since $x^3-9x-5=0$, whatever $x$ could be, $x^4-18x^3-81x^2-12=-72x^2-157x-102$
How to proceed?
P/S. Conclusion, NONE of the options can be right.
| Look at the decomposition
$$x^4-18x^3-81x^2-12=x^2(x^2-18x-81)-12$$
Observe the roots of the equation $x^3-9x-5=0$ must belong to the following two intervals: $(-3,-2)$ and $(3,4)$.
The axis of symmetry of this function $f=x^2-18x-81$ is $x=9$, so the trend of the function is decreasing for $x < 9$, so observe:
$$\begin{array} {cccc} f(-3)<0&f(-2)<0&f(3)<0&f(4)<0\end{array}$$
So the function $f=x^2-18x-81<0$ in the interval of $(-3,-2)$ to $(3,4)$.
Since $x^2>0$ and $-12<0$, then $x^2(x^2-18x-81)-12$ must be less than zero. Since all the given options are positive, none of them can be right.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How should I find Splitting Field of $x^3-2$ over $\mathbb Q$. How should I find Splitting Field of $x^3-2$ over $\mathbb Q$.
**My try **:
$x^3-2=(x-2^\frac{1}{3})(x^2+2^\frac{1}{3}x+2^\frac{2}{3})$
On solving I am getting the roots as $2^\frac{1}{3},\dfrac{2^\frac{1}{3}[-1+\sqrt 3 i]}{2},\dfrac{2^\frac{1}{3}[-1-\sqrt 3 i]}{2},$
But I dont know the answer is given that roots are $2^\frac{1}{3},2^\frac{1}{3}\omega ,2^\frac{1}{3}\omega ^2$ where $\omega $ is cube root of unity
| The roots you find are correct; just observe that
$$
2^{1/3}\frac{-1+i\sqrt{3}}{2}=2^{1/3}\omega
$$
and that $\omega$ is indeed a cube root of unity.
You can also proceed by steps. It's clear that $\sqrt[3]{2}$ must belong to the splitting field, so we can start adding it. Now, in $\mathbb{Q}(\sqrt[3]{2})$, we can factor the polynomial as
$$
(x-\sqrt[3]{2})(x^2+\sqrt[3]{2}\,x+\sqrt[3]{4})
$$
If $\alpha$ is a root of the second factor, it must belong to the splitting field $F$, but then also $\omega=\alpha/\sqrt[3]{2}\in F$. The other root is $\sqrt[3]{4}/\alpha$.
From
$$
\alpha^2+\sqrt[3]{2}\,\alpha+\sqrt[3]{4}=0
$$
we get
$$
\omega^2+\omega+1=0
$$
so $\omega$ is a (non real) root of $x^3-1=(x-1)(x^2+x+1)$, hence a cubic root of unity. The other non real root is its conjugate $\bar{\omega}=\omega^{-1}=\omega^2$.
Thus
$$
F=\mathbb{Q}(\sqrt[3]{2},\alpha,\sqrt[3]{4}/\alpha)=
\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\,\omega,\sqrt[3]{2}/\omega)=
\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}\,\omega,\sqrt[3]{2}\,\omega^2)=
\mathbb{Q}(\sqrt[3]{2},\omega)
$$
(the last equality can be easily proved by double inclusion).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convert the given set into roster notation: Problems:
Find $A$, where
1) $A = \{x \in \mathbb{R} \mid x^4 - 1 = 0\}$
2) $A = \{x \in \mathbb{C} \mid x^4 - 1 = 0\}$
Attempt:
Solving the equation:
$x^4 - 1 = (x^2)^2 - 1^2 $
$= (x^2 - 1)(x^2 + 1) $
$= (x-1)(x+1)(x^2 + 1) $
$= (x - 1)(x + 1)(x^2 - i^2) $
$= (x-1)(x+1)(x-i)(x+i) $
1) Since $x$ is defined as a real number in this case, $A= \{1,-1\}$.
2) Since $x$ is defined as a complex number in this case, $A= \{1,-1, i, -i\}$.
Is my work correct?
| Yes, your answer and notation are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x} \, dx$? Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$?
The answer is supposed to be ($\arctan \tan^2 x + C$), but I obtained ($-\arctan \cos2x + C$) as follows. Please identify the error.
$$\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$$
$$= \int \frac {\sin 2x}{(\sin^2 x+ \cos^ 2x)^2 - 2\cos^2 x\sin^2 x}dx$$
$$= \int \frac {\sin 2x}{1- \frac{\sin^2 2x}{2}}dx$$
$$= \int \frac {2\sin 2x}{2- \sin^2 2x}dx$$
$$= \int \frac {2\sin 2x}{1 + \cos^2 2x}dx$$
$1 + \cos^2 2x = t ; dt = 2(\cos 2x)(-\sin 2x)(2)dx; 2\sin 2x.dx= \frac{-dt}{2\cos 2x} = \frac{-dt}{2(\sqrt{t-1})};$
$$\int \frac{\frac{-dt}{2\sqrt {t-1}}}{t} = -1/2\int \frac{dt}{t.\sqrt{t-1}}$$
$\sqrt{t-1} = u; du=\frac{dt}{2\sqrt{t-1}}; 2u.du = dt$;
$$= -1\int \frac{du}{u^2+1} = -\arctan{u} = -\arctan \sqrt{t-1} = -\arctan \sqrt{1+\cos^2 2x-1} = -\arctan \cos2x$$
Also, is there an easier method to this problem?
| the answer is supposed to be $\arctan(\tan^2x)+c$, which you can obtain by substituting $t=\tan^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Solve $7^x \equiv 6 \pmod{17}$ given 3 is a primitive root $\bmod 17$ It's easy to show that 3 is a primitive root $\bmod 17$,
but how do I use it prove the congruence?
Is there a general way to solve any congruence of the form $a^x \equiv b \pmod{c}$ if you know a primitive root $\bmod c$ and c is big (without brute force)?
| You can reduce the amount of brute force quite significantly. Using an interval $d$ of approximately $\sqrt{c}$, make an "island" of values $a\cdot g^i$ with $i\in \{0,d{-}1\}$, then calculate all $g^{jd}$ and identify the coinciding value.
$\begin{array}{c|c}
k\ (d=4) &g^k & a\cdot g^k & b\cdot g^k \\ \hline
0 & 1 & 7 & 6 \\
1 & 3 & \color{red}{4} & \color{violet}{1} \\
2 & 9 & 12 & 3 \\
3 & 10 & 2 & 9 \\
4 & 13\\
8 & 16\\
12 & \color{red}{4}\\
16 & \color{violet}{1}\\
\end{array}$
giving, $\bmod 17$, $7\cdot 3\equiv 3^{12}$ and $6\cdot 3\equiv 3^{16}$ , so $7\equiv 3^{11}$ and $6\equiv 3^{15}$.
Then we need to find $11x\equiv 15 \bmod 16 (=\phi(17))$, which means finding the inverse of $11\bmod 16$, which here is $3$, giving $x\equiv 15\cdot 3 \equiv 13\bmod 16$.
Obviously for this exercise in small numbers this is more work thn simply calculating the powers of $7 \bmod 17$, but for larger numbers it avoids calculating all powers.
Because $3$ is a primitive root $\bmod 17$, we are guaranteed to find suitable exponents in the first stage.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
integration of $\int_{1}^{\infty } \,\left(\frac{2x^{2}+bx\text{+}a}{x(2x+a)} -1\right) \, dx=1$ i need help for this problem;
Find values of a and b $$\int_{1}^{\infty} \left( \frac{2x^{2}+bx+a}{x(2x+a)} -1\right) \, dx=1$$
I very appreciate your comments and suggestions.
| For
\begin{align}
I = \int_{1}^{\infty} \left( \frac{2 x^2 + b x + a}{x(2x+a)} - 1 \right) \, dx = 1
\end{align}
then it is seen that
\begin{align}
I &= \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a}{2x+a} + \frac{a}{x(2x+a)} \right) \, dx = \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a-2}{2x+a} + \frac{1}{x} \right) \, dx \\
&= \lim_{p \to \infty} \, \left[ \frac{b-a-2}{2} \, \ln(2x+a) + \ln(x) \right]_{1}^{p} \\
&= \lim_{p \to \infty} \ln\left[ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right] = \ln(e) = 1
\end{align}
for which
\begin{align}
\lim_{p \to \infty} \left\{ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right\} = e.
\end{align}
The limit is evaluated as follows:
\begin{align}
e &= \lim_{p \to \infty} \left\{ p \left(\frac{2p+a}{2+a}\right)^{\frac{b-a-2}{2}} \right\} \\
&= \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \, \left( 1 + \frac{a}{2p}\right)^{\frac{b-a-2}{2}} \right\} = \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \right\}
\end{align}
If $b=a$ then the limit is reduced to $1$. This yields
\begin{align}
e = \frac{a + 2}{2}
\end{align}
leading to $a = b = 2(e - 1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the series: $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? does $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? I think yes, it does, because the $a_n$ in the series converges to zero. but I'm trying to prove this by the help of the fact that:
$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
any suggestions?
| Notice that
$$ \frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{n^2+2n+1-n^2}{n^2(n+1)^2}, $$
so the series telescopes:
$$ \sum_{n=1}^m \frac{2n+1}{n^2(n+1)^2} = 1-\frac{1}{2^2} + \frac{1}{2^2} -\dotsb - \frac{1}{m^2} + \frac{1}{m^2} - \frac{1}{(m+1)^2} = 1- \frac{1}{(m+1)^2} \to 1 $$
as $m \to \infty$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the exact value of $\eta(6i)$? Let $\eta(\tau)$ be the Dedekind eta function. In his Lost Notebook, Ramanujan played around with a related function and came up with some of the nice evaluations,
$$\begin{aligned}
\eta(i) &= \frac{1}{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(2i) &= \frac{1}{2^{11/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(3i) &= \frac{1}{2\cdot 3^{3/8}} \frac{1}{(2+\sqrt{3})^{1/12}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(4i) &= \frac{1}{2^{29/16}} \frac{1}{(1+\sqrt{2})^{1/4}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(5i) &= \frac{1}{2\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(6i) &=\; \color{red}{??}\\
\eta(7i) &= \frac{1}{2\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(8i) &= \frac{1}{2^{73/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(16i) &= \frac{1}{2^{177/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\end{aligned}$$
with the higher ones $>4$ added by this OP. (Note the powers of $2$.)
Questions:
*
*Similar to the others, what is the exact value of $\eta(6i)$?
*Is it true that the function,
$$F(\sqrt{-N}) = \frac{\pi^{3/4}}{\Gamma\big(\tfrac{1}{4}\big)}\,\eta(\sqrt{-N}) $$
is an algebraic number only if $N$ is a square?
P.S. It seems strange there is a function that yields an algebraic number for square input $N$ and a transcendental number for non-square $N$. (Are there well-known functions like that?) For an example of non-square $N$, we have,
$$\eta(\sqrt{-3}) = \frac{3^{1/8}}{2^{4/3}} \frac{\Gamma\big(\tfrac{1}{3}\big)^{3/2}}{\pi} = 0.63542\dots$$
and $F(\sqrt{-3})$ seems to be transcendental.
| After persevering with a Mathematica session, I found that $F(6i)$ is the root of $96$-deg eqn (no wonder it was hard to find!) but could be prettified as,
$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \left(\frac{5-\sqrt{3}}{2}-\frac{3^{3/4}}{\sqrt{2}}\right)^{1/6}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$
However, the second question is still open.
$\color{blue}{Update}$: Four years after this question was asked, Giuseppe Manco found a more elegant factorable formulation in this long post,
$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \big(2-\sqrt3\big)^{1/24}\, \big(\sqrt2-\sqrt[4]3\big)^{1/4}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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"answer_id": 2
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How to get the given equality? I have the following sum ($n\in \Bbb N)$:
$$ \frac {1}{1 \times 4} + \frac {1}{4 \times 7} + \frac {1}{7 \times 10} +...+ \frac {1}{(3n - 2)(3n + 1)} \tag{1} $$
It can be proved that the sum is equal to
$$ \frac{n}{3n + 1} \tag{2}$$
My question is, how do I get the equality? I mean, if I hadn't knew the formula $(2)$, how would I derive it?
| Hint:
Use partial fraction of the ratio
$\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$
You will see mass cancellation such as below:
$\frac{1}{3}\left[1-\frac{1}{4}\right]$
$\frac{1}{3}\left[\frac{1}{4}-\frac{1}{7}\right]$
..
$\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$
Summing all you get $\left[\frac{1}{3}[1-(\frac{1}{3n+1})\right]$
Simplifying you get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the value of $x$ in $x^5=32$ I understand that $2^5=32$ but how would one go about finding it without doing any guessing (what if the numbers were much greater)?
| USING COMPLEX ANALYSIS
We have
$$z^5=2^5\implies z=2e^{i2n\pi/5}$$
for $n=0, \pm 1, \pm 2$.
$$\bbox[5px,border:2px solid #C0A000]{\text{Thus the five roots are} \,\, 2,\, 2e^{\pm i2\pi/5},\,2e^{\pm i4\pi/5}}\tag 1$$
FACTORORING A POLYNOMIAL
We have $x^5=32\implies x^5-2^5=0$.
We can then factor $x-2$ from the left-hand side and write
$$x^5-2^5=(x-2)(x^4+2x^3+4x^2+8x+16)=0 \tag 1$$
The quartic expression in $(1)$ can be factored into the product of two quadratics as
$$x^4+2x^3+4x^2+8x+16=(x^2+ax+4)(x^2+bx+4) \tag 2$$
where by matching coefficients in $(2)$, we find $a+b=2$ and $ab=-4$.
Solving for $a$ and $b$ reveals that $a=1+\sqrt{5}$ and $b=1-\sqrt{5}$ whence the original polynomial can be written as
$$x^5-2^5=(x-2)(x^2+(1+\sqrt{5})x+4)(x^2+(1-\sqrt{5})x+4) \tag 3$$
Finally, the quadratic terms in $(3)$ can easily be factored as
$$x^2+(1+ \sqrt{5})x+4=\left(x-\frac{(1+\sqrt{5})+i\sqrt{10-2\sqrt{5}}}{2}\right)\left(x-\frac{(1+\sqrt{5})-i\sqrt{10-2\sqrt{5}}}{2}\right)$$
and
$$x^2+(1- \sqrt{5})x+4=\left(x-\frac{(1-\sqrt{5})+i\sqrt{10+2\sqrt{5}}}{2}\right)\left(x-\frac{(1-\sqrt{5})-\sqrt{10+2\sqrt{5}}}{2}\right)$$
Thus the five roots of $x^5-2^5=0$ are
$$\bbox[5px,border:2px solid #C0A000]{2,\, \frac{(1+ \sqrt{5})\pm i\sqrt{10- 2\sqrt{5}}}{2},\,\frac{(1- \sqrt{5})\pm i\sqrt{10+ 2\sqrt{5}}}{2}} \tag 4$$
where the roots in $(4)$ are the rectangular coordinate form of the polar roots in $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find the maximum possible area of a certain right triangle I want to find the maximum possible area of a right triangle with hypotenuse $=10$.
My approach so far: let $x,y$ be the lengths of the two sides adjacent to the right angle; then $$100=x^2+y^2$$
Area $=\frac{xy}{2}$, so by the substitution method I got equality. But my result is wrong (probably I made a mistake in the equality), could someone show what to do?
| Let the non-hypotenuse sides of the triangle be $(a,b)$. Then the hypotenuse is $\sqrt{a^2+b^2}$ and the area is $A = \frac{ab}{2}$. So
$$a^2+b^2 = 100\\
A^2 = \frac14 a^2b^2 = \frac14 a^2 (100-a^2) \\
$$
Maximizing $A^2$ maximizes $A$, and
$$
\frac{d(A^2)}{da} =\frac14( 200a -4a^3)
$$
which is zero at $a = \sqrt{50}$ making $b=\sqrt{50}$ as well and the area $A = \frac12\sqrt{50}^2 = 25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337696",
"timestamp": "2023-03-29T00:00:00",
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Is the maximum of $\sum\limits_{\mathrm{cyc}} \frac{1}{3a + 5b + 7c}$ equal to $\frac{\sqrt{3}}{5}$?
Let $a,b,c>0$ such $ab+bc+ac=1$. Show that
$$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{5}.$$
(Note: $\mathrm{LHS} = \mathrm{RHS}$ when $a = b = c = 1/\sqrt 3$.)
since dear Mac sir,he solve with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$
use same methods,I think
$$(3a+5b+7c)^2\ge 75=75(ab+bc+ac)?$$
I would appreciate any help. Thanks in advance.
| You can use Lagrange multipliers. Let's denote
$$L(a, b, c) = \frac{1}{3a+5b+7c} + \frac{1}{3b+5c+7a} + \frac{1}{3c+5a+7b}-\lambda(ab+ac+bc-1);$$
we have system
$$\left\{\frac{\partial L}{\partial a}=0, \frac{\partial L}{\partial b}=0, \frac{\partial L}{\partial c}=0\right\},$$
or
$$
\left\{
\begin{aligned}
\frac{3}{(3a+5b+7c)^2} + \frac{7}{(3b+5c+7a)^2} + \frac{5}{(3c+5a+7b)^2}+\lambda(b+c)=0\\
\frac{5}{(3a+5b+7c)^2} + \frac{3}{(3b+5c+7a)^2} + \frac{7}{(3c+5a+7b)^2}+\lambda(a+c)=0\\
\frac{7}{(3a+5b+7c)^2} + \frac{5}{(3b+5c+7a)^2} + \frac{3}{(3c+5a+7b)^2}+\lambda(b+c)=0
\end{aligned}
\right.
$$
It's linear system for squares in denominators; solution is
$$
\left\{
\begin{aligned}
\frac{1}{(3a+5b+7c)^2}=\frac{90}{\lambda}(11c-4b-19a)\\
\frac{1}{(3b+5c+7a)^2}=\frac{90}{\lambda}(11a-4c-19b)\\
\frac{1}{(3c+5a+7b)^2}=\frac{90}{\lambda}(11b-4a-19c)
\end{aligned}
\right.,
$$
or if you prefer
$$
\left\{
\begin{aligned}
(3a+5b+7c)^2(11c-4b-19a)=\lambda/90\\
(3b+5c+7a)^2(11a-4c-19b)=\lambda/90\\
(3c+5a+7b)^2(11b-4a-19c)=\lambda/90
\end{aligned}
\right..
$$
We can solve this system in Maple, or to find solution in a form $a=b=c$:
$$
3c^2=1\Longrightarrow a=b=c=1/\sqrt3,
$$
and $\lambda = -243000c^3$ from the last system (other solutions are complex). Substituting it into your function, we have
$$
\frac{1}{3a+5b+7c} + \frac{1}{3b+5c+7a} + \frac{1}{3c+5a+7b}\le\frac{3}{15c}=\frac{\sqrt3}{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Product limit with exponentials Find an explicit formula for the limit:
$$\lim_{n \rightarrow \infty} n \prod_{k=2}^{n} (2 - e ^ {\frac 1 k})$$
I am not asking for convergence proof since I know the sequence is decreasing and bounded.
| This is going to be a very primitive attempt to answer this question. Yet I think it is worth posting since this kind of calculations appear many times in calculus. We take the log of the limit and then we Taylor expand the following function:
\begin{equation}
\log\left(2 - e^{\frac{1}{k}}\right) = \log\left(1 - \sum\limits_{p=1}^\infty \frac{1}{p!} \frac{1}{k^p}\right) = -\frac{1}{1} \sum\limits_{p=1}^\infty \frac{1}{p!} \frac{1}{k^p} - \frac{1}{2} \sum\limits_{p=2}^\infty \frac{2^p-2}{p!} \frac{1}{k^p}-\frac{1}{3} \sum\limits_{p=3}^\infty \frac{3^p-3 \cdot 2^p+3}{p!} \frac{1}{k^p}-\frac{1}{4} \sum\limits_{p=4}^\infty \frac{4^p-4 \cdot 3^p+6\cdot 2^p-4}{p!} \frac{1}{k^p}-\cdots
\end{equation}
Now if we denote the unknown limit by $g$ we clearly have:
\begin{equation}
\log(g) = \log(n) + \sum\limits_{k=2}^n \log(2-e^{\frac{1}{k}}) = \log(n) - (H_n-1) - \frac{1}{1} \sum\limits_{p=2}^\infty \frac{1}{p!} (\zeta(p)-1)) - \frac{1}{2} \sum\limits_{p=2}^\infty \frac{2^p-2}{p!} (\zeta(p)-1) -\frac{1}{3} \sum\limits_{p=3}^\infty \frac{3^p-3 \cdot 2^p+3}{p!} (\zeta(p)-1) -\cdots
\end{equation}
Now $\log(n) - (H_n-1) \rightarrow 1 - \gamma$ when $n\rightarrow \infty$ and the remaining sums all converge . The final result is therefore:
\begin{equation}
g = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} \sum\limits_{l=1}^p \frac{1}{l} \sum\limits_{j=0}^{l-1} \binom{l}{j} (l-j)^p (-1)^j \right) = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} \sum\limits_{l=1}^p \frac{1}{l} l! S_2(p,l)\right) = \exp\left(1-\gamma - \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{1}{p!} (-1)^p Li_{1-p}(2)\right) \simeq 0.5335376801314199077153...
\end{equation}
where $S_2(p,l)$ are Stirling numbers of the second kind and $Li_n(x)$ is the polylogarithmic function.
The question remains is it possible to further simplify the result..
Note that the original sequence converges very slowly. It is only for $n> 60000000$ that $g_n < 0.53353768$. On the other hand the series in the exponential in the last formula converge very fast. I truncated the sum over $p$ at $p=150$ and I obtain an accuracy of twenty three digits.
Now, the question remains is it possible to further simplify the result. Let us take the sum in the exponential:
\begin{equation}
S = \sum\limits_{p=2}^\infty (\zeta(p)-1) \frac{(-1)^p}{p!} Li_{1-p}(2) = \sum\limits_{k=2}^\infty \sum\limits_{p=2}^\infty \frac{(-1/k)^p}{p!} Li_{1-p}(2) = \sum\limits_{k=2}^\infty \sum\limits_{p=2}^\infty \frac{(-1/k)^p}{p!} \frac{d^{p-1}}{d \epsilon^{p-1}} \left.\left(\frac{2 e^{\epsilon}}{1-2 e^{\epsilon}}\right)\right|_{\epsilon=0} = \sum\limits_{k=2}^\infty \int\limits_{0}^{-1/k} \left(\frac{2 e^\xi}{1-2 e^\xi} - \frac{2}{1-2}\right) d\xi = \sum\limits_{k=2}^\infty \left( \log\frac{1}{2 e^{-1/k}-1} - \frac{2}{k}\right) = \sum\limits_{k=2}^\infty \left(-\log(2-e^{1/k})-\frac{1}{k}\right) = - \log(g) + \log(n) - \left(H_n-1\right)
\end{equation}
Taking the limit $n\rightarrow \infty$ we get $S = -\log(g) + 1- \gamma$ which is equivalent to $g = \exp(1-\gamma-S)$. As we can see this was going in circles. But at least, as a sanity check, we made sure that the result is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find remainder of $\frac{17^{235}}{ 23}$ I need to find remainder of $\frac{17^{235}}{ 23}$.
This is supposed to be solved using the following method:
*
*$\varphi(23) = 22$
*${17}^{235} = (({17}^{22})^{10})\cdot {17}^{15}$
*${17}^{22}\equiv 1 \pmod{23}$
*Then, according to Euler theorem, $17^{235} \equiv 17^{15} \pmod {23}$
and then i get stuck.
| $17^{15}= 17^1\times 17^2\times17^4\times 17^8$.
In the rhs every term is the square of the previous term.
$17^2=289=230+69-10\equiv \mathbf{-10}\pmod{23}$.
$17^4\equiv(-10)^2=100\equiv\mathbf{8}\pmod{23} $
$17^8\equiv 8^2=64\equiv\mathbf{-5}\pmod{23} $
Now multiply $17.(-10).8.(-5)\equiv\mathbf{-8}\pmod{23}$
The answer is 15.
The answer is 15.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $ $$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $$
How can I solve this one, I mean I get something like this:
$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$
This equation seems rather hard to solve from here, any tips or other ways to come to an solution?
| I shall use $\sum \cos$ when angles are in arithmetic progression whose usefulness is more apparent if the number of summands is more than three.
If $\sin x=0,x=n\pi\implies\cos2x+\cos4x+\cos6x=1+1+1\ne0$
So, $\cos2x+\cos4x+\cos6x=0\implies\sin x\ne0$
$$\dfrac{2\sin x\cos2x+2\sin x\cos4x+2\sin x\cos6x}{2\sin x}=\dfrac{\sin7x-\sin x}{2\sin x}$$
Now $\sin7x=\sin x\implies7x=m\pi+(-1)^mx$ where $m$ is any integer
If $m$ is even $=2r$(say) $7x=2r\pi+x\iff x=\dfrac{r\pi}3$
If $m$ is odd $=2r+1$(say) $7x=(2r+1)\pi-x\iff x=\dfrac{(2r+1)\pi}8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
| see that $a_n = (2n-1)/((n*(n+1)*(n+2)=-1/2*(1/n)+3*(1/(n+1))-5/2*(1/(n+2)) = 1/2* (-1/n+1/(n+1)+5*(1/(n+1)-1/(n+2))$, so $S_n = 1/2*(-1+5+1/(n+1)-5/(n+2)) = 1/2*(4n^2+8n+5)/((n+1)*(n+2))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| $[6\equiv\color\red2\pmod4]\wedge[7^{32}\equiv\color\green1\pmod4]\implies[6\cdot7^{32}\equiv\color\red2\cdot\color\green1\equiv\color\purple2\pmod4]$
$[7\equiv\color\red3\pmod4]\wedge[9^{45}\equiv\color\green1\pmod4]\implies[7\cdot9^{45}\equiv\color\red3\cdot\color\green1\equiv\color\orange3\pmod4]$
$[6\cdot7^{32}\equiv\color\purple2\pmod4]\wedge[7\cdot9^{45}\equiv\color\orange3\pmod4]\implies[6\cdot7^{32}+7\cdot9^{45}\equiv\color\purple2+\color\orange3\equiv1\pmod4]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Find the missing digit in the number 23104*791
Find the missing digit in the number $23104*791$ if
(i) it is divisible by $11$,
(ii) it is divisible by $13$,
(iii) it is divisible by $63$.
(i) $23104*791=231 (990+10)^2+4*(990+10)+11\times 72-1 \implies 11|4*0-1 \implies 11|400+*0-1 \implies 11|3+*0 \implies 11|*3 \implies *=3$
(ii) & (iii) Unable to adjust. Please help me to solve.
| For the second, since we have
$$10^3\equiv (-3)^3\equiv -27\equiv -1\pmod{13},$$
we have
$$(231)\cdot (10^3)^2+(40+*)\cdot 10^3+791\equiv 0\pmod{13}$$
$$\Rightarrow 10\cdot 1+(1+*)\cdot (-1)+11\equiv 0\pmod{13}$$
$$\Rightarrow *\equiv 10-1+11\equiv 7\pmod{13}$$
Hence, we have $*=\color{red}{7}$. (This is sufficient.)
For the third, it is divisible by $9$, so we have
$$2+3+1+0+4+*+7+9+1\equiv 0\pmod 9\Rightarrow *\equiv 0\pmod{9}$$
Also, it is divisible by $7$. Since we have
$$10^3\equiv 3^3\equiv 27\equiv -1\pmod{7},$$
we have
$$(231)\cdot (10^3)^2+(40+*)\cdot 10^3+791\equiv 0\pmod{7}$$
$$\Rightarrow 0+(5+*)\cdot (-1)+0\equiv 0\pmod 7$$
$$\Rightarrow *\equiv 2\pmod 7$$
Hence, we have $*=\color{red}{9}$. (This is sufficient.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove a sequence converges using sub-sequences
Let there be a sequence $a_n$
The following sub-sequences converge: $a_{n^3},a^3_{2n+3}-a^3_{2n+4},a^2_{2n+3}-a^2_{2n+4},a_{2n+15}$
Prove: $a_n$ converges
I think it has something to do with binomial due to the given sub-sequences
for example: $(a_{2n+3}-a_{2n+4})\cdot (a_{2n+3}+a_{2n+4})=a^2_{2n+3}-a^2_{2n+4}$ so can I say something about the components of $a^2_{2n+3}-a^2_{2n+4}$?
| The odds converge, since $a_{2n+15}$ converges. Let their limit be $x$ and let the limit of $a_{2n+3}^3 - a_{2n+4}^3$ be $y$. Then, by algebra of limits,
$$a_{2n+4} = (a_{2n+3}^3 - (a_{2n+3}^3 - a_{2n+4}^3))^{1/3} \rightarrow (x^3 - y)^{1/3}.$$
Thus the evens also converge. Let $b_n = a_{n^3}$, and $z$ be its limit. Consider the subsequences $b_{2n}$ and $b_{2n+1}$. These are also subsequences of the odd and even terms of $a_n$, hence they converge respectively to $x$ and $(x^3 - y)^{1/3}$. But, they are subsequences of $b_n$, so they also converge to $z$. By uniqueness of limits, $x = z = (x^3 - y)^{1/3}$. So, the odd and even terms of $a_n$ converge to the same limit, so the entire sequence converges to that limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the general values of $x$ satisfying the trigonometric equation
Find the general values of $x$ satisfying
$$
\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}
$$
It seems to me just some equality case of an inequality. But I am unable to find the inequality. Thanks.
| For $$\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}$$
it can be seen that the values of $x$ that satisfy this result are $4 x_{n} = (2n+1) \pi$, $n \geq 0$. For each value of $x_{n}$ the values $\sin x_{n} = \pm \frac{1}{\sqrt{2}}$ and $\cos x_{n} = \pm \frac{1}{\sqrt{2}}$. Now,
\begin{align}
\phi_{n} &= \frac{\tan^2 x_{n} \, \sin^2 x_{n}}{1-\sin^2 x_{n} \, \cos(2x_{n})}+\frac{\cot^2 x_{n} \, \cos^2 x_{n}}{1-\cos^2 x_{n} \, \cos(2x_{n})}+\frac{2\sin^2 x_{n}}{\tan^2 x_{n} + \cot^2 x_{n}} \\
&= \frac{\frac{1}{2}}{1 - \frac{1}{2} \cdot 0} + \frac{\frac{1}{2}}{1 - \frac{1}{2} \cdot 0} + \frac{2 \cdot \frac{1}{2}}{1 + 1} \\
&= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}.
\end{align}
Hence
\begin{align}
x_{n} = \frac{(2n+1) \, \pi}{4} \hspace{5mm} n \geq 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is wrong in my $f'(x)$? We have $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\frac{x^2-x+1}{x^2+x+1}$ and we need to find $f'(x)$.
Here is all my steps:
$$\begin{align}f'(x)&=\frac{(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}\\&=\frac{(2x-1)(x^2+1)-(2x+1)(x^2+1)}{(\cdots)^2}\\&=\frac{(x^2+1)(2x-1-2x-1)}{(\cdots)^2}\\&=\frac{2(1-x)(1+x)}{(\cdots)^2},\forall x\in\mathbb{R}\end{align}$$
But in my book they say that $f'(x)=\frac{2(x-1)(x+1)}{(x^2+x+1)^2}$. What is wrong in my method ?
| $$\begin{align}
(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1) & =(2x-1)(x^2+1)+2x^2-x-((2x+1)(x^2+1)-2x^2-x)\\
& = (x^2+1)(2x-1-2x-1)+2x^2-x+2x^2+x\\
& = -2(x^2+1)+4x^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Least Common Denominator: $ \frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}} $ $$ \frac{\sqrt{x}}{x}+\frac{\ln \ x}{2\sqrt{x}} $$
I have tried combining these two fractions; however, I keep getting stuck.
$$\frac{2\sqrt{x}}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}}\cdot \frac{x}{x}$$
So that we get
$$\frac {2x}{2\sqrt{x}\cdot(x)}+\frac {x\ln\ x}{2\sqrt{x}\cdot(x)}$$
What am I doing wrong? The answer is $$ \frac{2+\ln\ x}{2\sqrt{x}}$$
| The square root $\sqrt{x}$ seems to be the thing that bothers you: pretend it's not there! That is, write temporarily $\sqrt{x}=z$ so that $x=z^2$; then
$$
\frac{\sqrt{x}}{x}+\frac{\ln x}{2\sqrt{x}}
=
\frac{z}{z^2}+\frac{\ln(z^2)}{2z}
$$
Now, for avoiding misunderstandings, write temporarily $\ln(z^2)=\ln x=L$ so you have
$$
\frac{z}{z^2}+\frac{L}{2z}=\frac{1}{z}+\frac{L}{2z}=
\frac{2+L}{2z}
$$
Substitute back to reinstate $x$:
$$
\frac{2+L}{2z}=\frac{2+\ln x}{2\sqrt{x}}
$$
This is longer, but safer. With some experience, you'll recognize more easily that
$$
\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}
$$
but the method of temporarily renaming objects to make them simpler works in several other cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
solve $\sqrt{x+7}
solve $\sqrt{x+7}<x$
I tried $\sqrt{x+7}<x\\
x+7<x^2\\
x^2-x-7>0\\
x\in \left(-\infty, \dfrac{1-\sqrt{29}}{2}\right) \cup \left( \dfrac{1+\sqrt{29}}{2},+\infty\right) $
I m not sure, if this is correct method and if the solution is correct .
I look for a simple and short way.
I have studied maths up to $12$th grade.Thanks.
| Hint:
Note that $\sqrt{x+7}$ is a real number only if $x+7\ge 0$ and, in this case,$\sqrt{x+7}$ is a positive number so that we must have $x \ge 0$.
From all these conditions you find :
$$
\sqrt{x+7}>0 \iff
\begin {cases}
x\ge 0\\
x+7<x^2
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\int _0^\infty \frac{\ln x}{(x^2+1)^2}dx$ Calculate $$\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx.$$
I am having trouble using Jordan's lemma for this kind of integral. Moreover, can I multiply it by half and evaluate $\frac{1}{2}\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$?
| \begin{align}
I&=\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{\ln x}{(x^2+1)^2}dx+\int _1^\infty \dfrac{\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{\ln x}{(x^2+1)^2}dx-\int _0^1 \dfrac{x^2\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{(1-x^2)\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \sum_{j=0}^{\infty}(-1)^j (2 j+1) x^{2 j}\ln xdx\\
&=\sum_{j=0}^{\infty}(-1)^j (2 j+1)\int _0^1 x^{2 j}\ln xdx\\
&=-\sum_{j=0}^{\infty}(-1)^j \frac{2 j+1}{(2j+1)^2}\\
&=-\sum_{j=0}^{\infty}(-1)^j \frac{1}{2j+1}\\
&=-\frac{\pi}{4}
\end{align}
where at last you may recall the expansion for $\arctan$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
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Find all numbers that have 30 factors and have 30 as one of their factors. Find all numbers that have 30 factors and have 30 as one of their factors.
Thank you.
Note: please show way if possible.
| We write the number as $2^{1+a}3^{1+b}{5^{1+c}}n$ with $(n,30)=1$.
The number of divisors of the number is $30=(2+a)(2+b)(2+c)\tau(n)$. Hence $\tau(n)=1\implies n=1$ ($\tau(n)$ is the number of divisors of $n$).
So $2+a,2+b,2+c$ are $2,3,5$ in some order.
There are $6$ solutions:
$2\cdot3^2\cdot5^4=11250$
$2\cdot3^4\cdot5^2=4050$
$2^2\cdot3\cdot5^4=7500$
$2^2\cdot3^4\cdot5=1620$
$2^4\cdot3\cdot5^2=1200$
$2^4\cdot3^2\cdot5=720$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Strategies For Summing Harmonic Numbers Lately, I have found several interesting problems involving Harmonic numbers such as \begin{equation*}\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}=\frac{7\pi^4}{360}\end{equation*}
I am not familiar with computing sums involving Harmonic numbers. Is there a general strategy for tackling such problems?
How can this series be evaluated by operating on the generating function for $H_n^{(2)}$?
| You may evaluate it without using generating functions.
$$S=\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2} = \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{n}\frac{1}{n^2k^2}$$
By changing the order of summation, you may write it as:
$$\begin{align}S &= \sum\limits_{k=1}^{\infty}\sum\limits_{n=k}^{\infty}\frac{1}{n^2k^2}\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\sum\limits_{n=k}^{\infty}\frac{1}{n^2}\right)\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{k-1}\frac{1}{n^2}\right) \\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\zeta(2) - \sum\limits_{n=1}^{k-1}\frac{1}{n^2}\right)\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\zeta(2) + \frac{1}{k^2} - H_k^{(2)}\right)\\&= \zeta(2)\sum\limits_{k=1}^{\infty}\frac{1}{k^2} + \sum\limits_{k=1}^{\infty}\frac{1}{k^4} - \sum\limits_{k=1}^{\infty}\frac{H_k^{(2)}}{k^2} \\&= \zeta^2(2)+\zeta(4) - S\end{align}$$
Hence, $\displaystyle S = \frac{\zeta^2(2)+\zeta(4)}{2}$.
In general for any sequence $(a_n)_{n \ge 1}$ such that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges absolutely,
We have $$\sum_{n=1}^{\infty}\left(a_n\sum_{k=1}^{n}a_k\right) = \frac{1}{2}\left(\sum_{n=1}^{\infty}a_n^2 + \left(\sum_{n=1}^{\infty}a_n\right)^2\right)$$ by using the same method as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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} |
Real part of $\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}$ How can I compute the real part of
\begin{equation*}
\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}, \quad \text{where}\ \theta \in \mathbb{R}?
\end{equation*}
Maybe it's a silly question, but I'm feeling rusty with complex numbers...
| So the solution I've found is:
$$ \frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1} = $$
$$ \frac{\cos{(n+1)\theta} -1 +i\sin{(n+1)\theta}}{(\cos\theta - 1)^2 + \sin^2 \theta}(\cos \theta -1 -i\sin\theta)= $$
$$ \frac{\cos{n\theta} - \cos(n+1)\theta - \cos\theta + 1}{2-2\cos\theta} -i \frac{\sin{n\theta} - \sin(n+1)\theta + \sin\theta}{2-2\cos\theta}= $$
$$ \frac{\sin{(n+\frac{1}{2})\theta} + \sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}} +i \frac{\sin(n+1)\frac{\theta}{2} \sin(n\frac{\theta}{2})}{\sin\frac{\theta}{2}}$$
Now it's easy to distinguish real and imaginary parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
When solving $\tan(3x) = \cot(4x)$, how to formulate the answer? when I solve the following equation:
$\tan(3x) = \cot(4x)$
I get the following solution:
$x = \frac{\pi}{14} + \frac{\pi n}{7}, n \in \mathbb{Z}$
But as x must be $\neq \frac{\pi}{6} + \frac{\pi k}{3}$ and $\neq \frac{\pi k}{4}$, with $k \in \mathbb{Z}$, there are values for n for which $x = \frac{\pi}{14} + \frac{\pi n}{7}, n \in \mathbb{Z}$ gives wrong solutions, e.g.:
for
$n = 3, x = \frac{\pi}{14} + \frac{3\pi}{7} = \frac{7\pi}{14} = \frac{\pi}{2}$
Which is not a solution because $\tan(3(\frac{\pi}{2}))$ is not defined (cosine is 0)
So how do I merge the solution $x = \frac{\pi}{14} + \frac{\pi n}{7}, n \in \mathbb{Z}$ with the domain? What is the algorithm?
Or, can I just write:
$x \in \{\frac{\pi}{14} + \frac{\pi n}{7}: n \in \mathbb{Z}\} \setminus \{\frac{\pi}{6} + \frac{\pi k}{3} : k \in \mathbb{Z} \} \setminus \{\frac{\pi k}{4} :k \in \mathbb{Z}\}$
Which is a bit ugly, I think.
What is the right way to do that?
Thanks for the attention!
|
You get $\cos(7x)=0$ and $\cos(3x)\sin(4x) \ne 0$.
As you've already figured out
$\cos(7x) = 0 \implies x = \dfrac{2j+1}{14}\pi \quad (j \in \mathbb Z)$
$\cos(3x) = 0 \implies x = \dfrac{2k+1}{6}\pi \quad (k \in \mathbb Z)$
$\sin(4x) = 0 \implies x = \dfrac{\ell}{4} \pi \quad (\ell \in \mathbb Z)$
We look for coincidence of points
\begin{align}
\dfrac{2j+1}{14}\pi &= \dfrac{2k+1}{6}\pi \\
6j+3 &= 14k+7 \\
3j - 7k &= 2 \\
\hline
j &= 7n + 3\\
k &= 3n + 1\\
\hline
\theta &\ne \dfrac{2n+1}{2}\pi
\end{align}
\begin{align}
\dfrac{2j+1}{14}\pi &= \dfrac{\ell}{4} \pi \\
4j+2 &= 7 \ell \\
7 \ell - 4j &= 2 \\
\hline
j &= 7n + 3\\
\ell &= 4n + 2\\
\hline
\theta &\ne \dfrac{2n+1}{2}\pi
\end{align}
So we have to require $j \ne 7n + 3$, which we can write as
$j \not \equiv 3 \pmod 7$. In summary
$x = \dfrac{2j+1}{14}\pi\;$ where $\; j \not \equiv 3 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Definite integration. $$\int _0^1 \arctan(x^2-x+1)\,dx$$
ATTEMPT:
$\int \arctan(x^2-x+1)\,dx$
Let
$\arctan(x^2-x+1)=u$ and $dx=dv$:
$$du=\frac{2x−1}{(x^2−x+1)^2+1}dx= \frac{2x−1}{(x^2+1)(x^2-2x+2)}$$ and $v=x$.
Now from Integration by parts $I=uv-$$\int vdu.$
$$I=\arctan(x^2-x+1)x-\int \frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}.$$
Now using partial fractions
$$
\int\frac{x(2x−1)}{(x^2+1)(x^2-2x+2)}=\int \frac{x}{(x^2+1)}-\frac{x}{x^2-2x+2} $$
Integrating term by term and finally arranging the result:
$$I=x\arctan(x^2−x+1)+\frac{1}{2}\ln(∣x^2+1∣)−\frac{1}{2}\ln(∣x^2−2x+2∣)−\arctan(x−1).$$
Now substituting the limit i got $I=\ln2.$
Is there any other way to approach this problem like by using the properties of definite integration as this method is long and tedious but though it works!!
| We have $$\int_0^1\arctan(x^2-x+1)dx =\int_0^1\left (\arctan(\frac1x)+\arctan(x-1) \right)dx=
\\\\ \left [x\cot^{-1}(x)+\frac12 \log(1+x^2)\right] \Biggr|_0^1+\left[(1-x)\tan^{-1}(1-x)-\frac12 \log(1+(x-1)^2)\right] \Biggr|_0^1=\log2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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What is the non-inductive proof of this inequality? $$\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{1}{\sqrt{3n+1}}.$$
However I've non-inductive proof of $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{2n+1}}$, but I can't prove it for $3n+1$. It is obvious is to see $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{3n+1}}$ is stronger inequality.
Let $S=\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{2 \cdot 4 \cdot 6 \cdots(2n)}{3 \cdot 5 \cdot 7 \cdots(2n+1)}$ [$\frac{1}{2}<\frac{2}{3} $ and so on.]
So, $S<\dfrac{1}{S(2n+1)}$ Implies, $S<\dfrac{1}{\sqrt{2n+1}}$
| $$S = \prod_{k=1}^n \frac{2k - 1}{2k} = \prod_{k=1}^n \frac{(2k)(2k - 1)}{(2k)^2} = \frac{(2n)!}{2^{2n}(n!)^2} = \frac{1}{4^n} {2n \choose n} $$
According to Wikipedia, ${2n \choose n} \le \frac{4^n}{\sqrt{3n + 1}}$
And the result follows.
(This is a proof by Wikipedia, rather than induction).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find coefficients of polynomials $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$ $(a,b,c,d \in \mathbb{R})$ Roots of polynomial $f(x)=x^2+ax+b$ are cubes of the roots of polynomial $g(x)=x^2+cx+d$. Sum and product of roots of polynomial $g(x)$ are equal. Find coefficients $a,b,c,d$ so that polynomial $f(x)$ has double root, and $g(x)$ doesn't have double root.
From the condition that $f(x)$ has double root, discriminant of quadratic equation $x^2+ax+b=0$ must be zero, $$D=a^2-4b=0,b=\frac{a^2}{4}$$
Sum and product of roots of polynomial $g(x)$ are equal, $$x_1^3+x_2^3=x_1^3x_2^3$$
Vieta's formulas for two polynomials,
$$x_1+x_2=-a$$
$$x_1x_2=b$$
$$x_1^3+x_2^3=-c$$
$$x_1^3x_2^3=d$$
If we cube first equation, $$(x_1^3+x_2^3)+3x_1x_2(x_1+x_2)=-a^3$$
$$-c-3ba=-a^3$$
From $b=\frac{a^2}{4}$ $$a^3-4c=0,c=\frac{a^3}{4}$$
From $x_1^3+x_2^3=x_1^3x_2^3$ implies $d=-c$
If we cube second equation, $$x_1^3x_2^3=b^3=\frac{a^6}{64}=d=-c$$
I don't know how to find these coefficients.
Thanks for replies.
| The roots of $f$ are equal, and the coefficients of $f$ are real. So we must have $f(x)=(x-\alpha)^2$ for some real $\alpha$.
The roots of $g$ are all cube roots of $\alpha$, so they must be drawn from the set $\{\sqrt[3]{\alpha},\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}\}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is a primitive cube root of unity. Note that $\sqrt[3]{\alpha}$ is the only real number in this set; since the roots of $g$ are distinct, it follows that they cannot all be real. Since $g$ has real coefficients, they must be complex conjugates; thus $g$ has roots $\omega\sqrt[3]{\alpha},\omega^2\sqrt[3]{\alpha}$.
Now, as the sum and product of the roots of $g$ is equal, we have
$$(\omega+\omega^2)\sqrt[3]{\alpha}=\omega^3\sqrt[3]{\alpha^2}\\
-\sqrt[3]{\alpha}=\sqrt[3]{\alpha^2}$$
and since $\alpha$ is real this implies that $\alpha^2=-\alpha$, and so $\alpha=-1$. That is, $f(x)=(x+1)^2=x^2+2x+1$, and $g(x)=(x+\omega)(x+\omega^2)=x^2-x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Finding the lowest value y can have in $y = \frac{1}{2}(e^x - e^{-x}) + \frac{n}{2}(e^x + e^{-x }) $ How can I find the lowest value $y$ can have when $n$ is greater than or equal to $2$ using only algebra?
$$y = \frac{1}{2}(e^x - e^{-x}) + \frac{n}{2}(e^x + e^{-x })$$
| Solution without differentiation.
Given is
$$
y = \frac{1}{2} \Big( \exp(x) - \exp(-x) \Big)
+ \frac{n}{2} \Big( \exp(x) + \exp(-x) \Big).
$$
Step 1:
$$
y = \frac{n+1}{2} \exp(x) + \frac{n-1}{2} \exp(-x).
$$
Step 2:
$$
y = \frac{n+1}{2} \exp(-a) \exp(x+a) + \frac{n-1}{2} \exp(a) \exp(-[x+a]).
$$
Step 3:
Find $a$ such that
$$
\frac{n+1}{2} \exp(-a) : \frac{n-1}{2} \exp(a) = 1,
$$
so solve
$$
\frac{n+1}{n-1} \exp(-2a) = 1 \Rightarrow a = \log \sqrt{ \frac{n+1}{n-1} }.
$$
Step 4:
$$
y = \frac{\sqrt{n^2-1}}{2}
\Big[ \exp(x+a) + \exp(-[x+a]) \Big]
$$
Step 5:
$$
y = \sqrt{n^2-1}
\Bigg[ 1 + \Big[ \tfrac{1}{2} \exp(\tfrac{1}{2}[x+a])
- \tfrac{1}{2}\exp(-\tfrac{1}{2} [x+a]) \Big]^2 \Bigg]
$$
You can read out the minimum, as $y$ has the form $y = a ( 1 + \xi^2)$.
The minimum is $a$ for $\xi=0$.
The minimum is given by
$$
\sqrt{n^2-1}
$$
for
$$
x = \frac{1}{2} \log \frac{n-1}{n+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that there exist infinitely many integers $(n^{2015}+1)\mid n!$
I conjecture that there exist infinitely many integers $n$ such
that $$(n^{2015}+1)\mid n!.$$
I have seen a simpler problem that there exist infinitely many integers $n$ such that $(n^2+1)\mid n!$.
Alternatively, I considered the Pell equation
$n^2+1=5m^2$, $2m<n$, but for $2015$ I can't figure it out.
| Modest progress. There are infinitely many integers $n$ such that $n^3+1\mid n!$.
We always have $n^3+1=(n+1)(n^2-n+1)$. Let $n=k^2+1$. Then
$$
n^2-n+1=(1+k+k^2)(1-k+k^2).
$$
Assume further that $k\equiv1\pmod3$. In that case $1+k+k^2$ and $n+1=2+k^2$ are both divisible by $3$. For all sufficiently large $k\equiv1\pmod3$ we thus have
$$
(k^2+1)^3+1=3^2\cdot\frac{k^2+2}3\cdot\frac{k^2+k+1}3(k^2-k+1)
$$
that is clearly a factor of $(k^2+1)!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
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"answer_id": 1
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If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then.. If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then
(A) $a=c$
(B) either $a=c$ or $a+b+c+d=0$
(C) $a+b+c+d=0$
(D) $a=c$ and $b=d$
I solved $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ and got $a(a+b+d)=c(c+b+d)$ and so I thought that (A) is the correct option. But the correct answer is (B).
I'm how $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ if $a+b+c+d=0$. Please help.
| On the right hand side, you should have $c(c + b + d)$, not $c(a + b + d)$. Check your work on that again.
It might be easier if you leave things fully expanded, then bring all terms to one side and then factor at the end.
| {
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"url": "https://math.stackexchange.com/questions/1368787",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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} |
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
| The following proof is far to complicated, but it is a new one and I think it is somewhat funny too.
For $x\in\left[0;1\right)$ we have:
$$
\frac{1}{\left(1-x\right)^2}=\frac{1}{1-\left(2x-x^2\right)}=\sum_{k=0}^{\infty}\left(2x-x^2\right)^k=\sum_{k=0}^{\infty}{\sum_{r=0}^{k}\binom{k}{r}(-1)^r2^{k-r}x^{k+r}}=\sum_{k=0}^{\infty}c_kx^k
$$
We have $c_{2n}=\sum_{s=0}^{n}{\binom{n+s}{n-s}(-1)^{n-s}2^{2s}}$ and $c_{2n+1}=\sum_{s=0}^{n}{\binom{n+s+1}{n-s}(-1)^{n-s}2^{2s+1}}$. Applying the identity $\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}$ we obtain:
$$
c_{2n+2}=\sum_{s=0}^{n+1}{\binom{n+1+s}{n+1-s}(-1)^{n+1-s}2^{2s}}=\sum_{s=0}^{n+1}{\binom{n+s}{n-s}(-1)^{n+1-s}2^{2s}}+\sum_{s=0}^{n+1}{\binom{n+s}{n-s+1}(-1)^{n+1-s}2^{2s}}=-c_{2n}+2c_{2n+1}
$$
$$
c_{2n+3}=\sum_{s=0}^{n+1}{\binom{n+s+2}{n-s+1}(-1)^{n+1-s}2^{2s+1}}=\sum_{s=0}^{n+1}{\binom{n+s+1}{n-s}(-1)^{n+1-s}2^{2s+1}}+\sum_{s=0}^{n+1}{\binom{n+s+1}{n-s+1}(-1)^{n+1-s}2^{2s+1}}=-c_{2n+1}+2c_{2n+2}=3c_{2n+1}-2c_{2n}
$$
Therefore:
$$
c_{2n+3}-c_{2n+2}=3c_{2n+1}-2c_{2n}+c_{2n}-2c_{2n+1}=c_{2n+1}-c_{2n}=…=c_1-c_0=1
$$
Thus:
$$
c_{2n+2}=-c_{2n}+2\left(c_{2n}+1\right)=c_{2n}+2
$$
$$
c_{2n+3}=3c_{2n+1}-2\left(c_{2n+1}-1\right)=c_{2n+1}+2
$$
Together with $c_0=1$ and $c_1=2$ we obtain $c_n=n+1$. Thus, for $x\in\left[0;1\right)$:
$$
\frac{1}{\left(1-x\right)^2}=\sum_{k=0}^{\infty}(k+1)x^k
$$
By observing, that:
$$
\sum_{k=0}^{\infty}(k+1)(-x)^k+\sum_{k=0}^{\infty}(k+1)x^k=\sum_{k=0}^{\infty}(4k+2)x^2k=\frac{4}{\left(1-x^2\right)^2}-\frac{2}{1-x^2}
$$
We get the analogous result for $x\in\left(-1;0\right]$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
*given $$ \frac {1} {a} + \frac {1} {b} + \frac {1} {c} + \frac {1} {d} = 65, \frac {1} {a^2} + \frac {1} {b^2} + \frac {1} {c^2} + \frac {1} {d^2} = 209$$
find $$\frac {1} {ab} + \frac {1} {ac} + \frac {1} {ad} + \frac {1} {bc} + \frac {1} {bd} + \frac {1} {cd}$$
(A) $2006$ (B) $2007$ (C) $2008$ (D) $2009$ (E) $2010$
I've use the direct way (make them become 1 fraction)
$$\frac {abc+abd+acd+bcd} {abcd} = 65$$
$$\frac {a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2} {a^2b^2c^2d^2} = 209$$
$$\frac {a^2b^2c^2d^2(ab+ac+ad+bc+bd+cd)} {a^3b^3c^3d^3}$$
$$\frac {(ab+ac+ad+bc+bd+cd)} {abcd}$$
but it doesn't make sense with these power $abcd$ thing
| $$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2=\\\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+\left(\frac{1}{d}\right)^2+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\65^2=209+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)\\\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd}\right)=\frac{1}{2}\left(65^2-209\right)=2008$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$ $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$
I let, $x = \frac{1}{n}$, then as
$\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$
L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$
Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
| note that :in line 4 you missed something ,that is $$y=(1+x)^{\frac{1}{x}} ,y' =?\\\ln y=\ln(1+x)^{\frac{1}{x}} \\\ln y=\frac{1}{x} \ln(1+x)\\\frac{y'}{y}=\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\\y'= (1+x)^{\frac{1}{x}}\cdot \bigg(\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove an equality If $a+b+c=0$
prove that
$\frac {(a^4 +b^4 +c^4)}{2}=\frac {(a^2+b^2+c^2)}{2^2}^2$
I have expanded the right side and have got this far:
$a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)$
I need $a^2=b^2=c^2$ to prove the equality.
Any ideas?
| $$\dfrac{\sum a^4}2-\left(\dfrac{\sum a^2}2\right)^2=\dfrac{\sum a^4-\sum2b^2c^2}4$$
$$\sum a^4-\sum2b^2c^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$(a^2+b^2-c^2)^2-(2ab)^2=(a^2+b^2-c^2-2ab)(a^2+b^2-c^2+2ab)$$
$$a^2+b^2-c^2+2ab=(a+b)^2-c^2=(a+b+c)(a+b-c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| $$6-\sqrt{x} = t^3$$
then
$$6+\sqrt{x} = t^3+2\sqrt{x}$$
The expression becomes:
$$t + \sqrt[3]{t^3+2\sqrt{x}} = \sqrt[3]{3}$$
Substitute back $\sqrt{x}$ in terms of t, rewrite so the cube root is alone. Raise to 3, solve 3rd degree polynomial equation. Test all roots.
Fun fact: The valid $t$ solution happens to become closely related to the golden ratio: $t =\sqrt[3]{3}\left(\frac{1+\sqrt{5}}{2}\right) = \sqrt[3]{3}\varphi$, where $\varphi = \frac{\sqrt{5}+1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Sine/cosine series $$\frac{\sin²(1°) + \sin²(2°) + \sin²(3°) + .. + \sin²(90°)}{\cos²(1°) + \cos²(2°) + \cos²(3°) + .. + \cos²(90°)} = ?$$
I tried to use multiple identities but I couldn't simplify the expression. Where should I start?
| $$\begin{align}
&\frac{\sin^2 1+\sin^2 2+\sin^2 3+\dots+\sin^2 44+\sin^2 45+\sin^2 46+\dots+\sin^2 90}{\cos^2 1+\cos^2 2+\cos^2 3+\dots+\cos^2 44+\cos^2 45+\cos^2 46+\dots+\cos^2 90}\\
&=\frac{\sin^2 1+\dots+\sin^2 44+\sin^2 45+\sin^2 (90-44)+\dots+\sin^2 (90-1)+\sin^2 90}{\cos^2 1+\dots+\cos^2 44+\cos^2 45+\cos^2 (90-44)+\dots+\cos^2 (90-1)+\cos^2 90}\\
&=\frac{\sin^2 1+\dots+\sin^2 44+\frac{1}{2}+\cos^2 44+\dots+\cos^2 1+1}{\cos^2 1+\dots+\cos^2 44+\frac{1}{2}+\sin^2 44+\dots+\sin^2 1+0}\\
&=\frac{\overbrace{1+1+\dots+1}_{44\,\text{times}}+\frac{1}{2}+1}{\overbrace{1+1+\dots+1}_{44\,\text{times}}+\frac{1}{2}+0}\\
&=\frac{44+\frac{1}{2}+1}{44+\frac{1}{2}}\\
&=\frac{88+1+2}{88+1}=\frac{91}{89}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.
$$\int \sqrt{16-9x^2}\,dx$$
I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
| $\frac{16}{3}\int(cos^2\theta)$
= $\frac{16}{3}\int\frac{1-cos2\theta}{2}$
= $\frac{16}{6}\int(1-cos2\theta)$
= $\frac{16}{6}(\int(1) - \int(cos2\theta))$
= $\frac{16}{6}\theta$ - $\frac{16}{6}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}(\frac{1}{2})\int(cos(v))$, substitute $v = 2\theta$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(v)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(2\theta)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}(2) sin(\theta) cos
(\theta)$
= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{8}{3}(\frac{3x}{4})$$\frac{sqrt (16-9x^2)}{4}$
= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{1}{2}x[sqrt (16-9x^2)]$ + C
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 3
} |
Show that $(1+p/n)^n$ is a Cauchy sequence for arbitrary $p$ It is a generalization of this question. I am looking for a similar derivation as in here.
Can we prove that $(1+p/n)^n$ is a Cauchy sequence for any $p \in [a, b]$ by showing that
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{m}\right)^m \Bigg| \leq f(n)$$
where $f(n)$ is something that tends to zero as $n$ goes to infinity?
Here is an attempt.
Let $m=n+1$. Then,
$$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{n+1}\right)^{n+1} \Bigg| = \Bigg| \sum_{k=0}^{n} \binom{n}{k}\left( \frac{p}{n} \right)^{k} - \sum_{k=0}^{n+1} \binom{n+1}{k}\left( \frac{p}{n+1} \right)^{k} \Bigg| = \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg| \leq
\sum_{k=0}^{n} \frac{|p|^k}{k!} \left[ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \right] + \Bigg| \frac{p^{n+1}}{(n+1)^{n+1}} \Bigg|
$$
Now intuitively, $\sum_{k=0}^{n} \frac{|p|^k}{k!}$ tends to the constant $e^{|p|}$ while the expression in square brackets tends to zero as $n$ goes to infinity. I was trying to show that this expression is less than a constant divided by $n^2$ (because we will sum up the consecutive terms for $m >n$, and the sum should converge which $\frac{C}{n^2}$ would provide).
Can we also find an estimate of the constant $C$?
Thanks to all for good answers. Meanwhile, I'm still quite interested in working it out algebraically. My next idea was to use the triangle inequality. Let us ignore the last term in the inequality above, it obviously tends to zero as $n$ goes to infinity. We want the sum to go to zero fast enough. So,
$$ \Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) \Bigg| \leq \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg| + \\
\Bigg| \sum_{k=0}^{n} \frac{p^k}{k!} \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \sum_{k=0}^{n} \frac{p^k}{k!} \Bigg|
$$
Each product is evidently positive and smaller than one. So we end up having kind of a usual exponential series, but weighted. Otherwise, we could try to plug something instead of the products, what would, when subtracted from the products, give something which decreases fast enough. I feel this should be possible since we have all the freedom to plug in whatever we want.
| Sequence is Cauchy
Bernoulli's Inequality shows that the sequence is monotonically increasing.
$$
\begin{align}
\left.\left(1+\frac p{n+1}\right)^{n+1}\middle/\left(1+\frac pn\right)^n\right.
&=\frac{n+p}{n}\left(\frac{n+p+1}{n+1}\frac{n}{n+p}\right)^{n+1}\\
&=\frac{n+p}{n}\left(1-\frac{p}{(n+1)(n+p)}\right)^{n+1}\\
&\ge\frac{n+p}{n}\left(1-\frac{p}{n+p}\right)\\[4pt]
&=1\tag{1}
\end{align}
$$
If $p\le0$, then $\left(1+\frac pn\right)^n\le1$, so $\left(1+\frac pn\right)^n$ is bounded above.
If $p\ge0$, the Binomial Theorem shows that the sequence is bounded above.
$$
\begin{align}
\left(1+\frac pn\right)^n
&=1+\frac n1\frac pn+\frac{n(n-1)}{2!}\left(\frac pn\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac pn\right)^3+\cdots\\
&=1+\frac nn\frac p1+\frac{n(n-1)}{n^2}\frac{p^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{p^3}{3!}+\cdots\\
&\le1+p+\frac{p^2}{2!}+\frac{p^3}{3!}+\cdots\\[6pt]
&=e^p\tag{2}
\end{align}
$$
Since $\left(1+\frac pn\right)^n$ is increasing and bounded above, it is convergent, hence Cauchy.
If we are willing to use the inequality $1+x\le e^x$, then we can simplify and extend the proof of $(2)$ to $1+\frac pn\le e^{p/n}$ and raise both sides to the $n^{\text{th}}$ power. This shows that $\left(1+\frac pn\right)^n\le e^p$ for all $p\gt-n$ (and since we are looking at large $n$, this is fine).
Regarding $\ \boldsymbol{f}$
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left[\log(1+x)-x+\frac{x^2}2\right]
=\frac{x^2}{1+x}
\ge0\tag{3}
$$
Thus, for $x\ge0$,
$$
\log(1+x)\ge x-\frac{x^2}2\tag{4}
$$
Therefore, for $p\ge0$, we have
$$
n\log\left(1+\frac pn\right)\ge p-\frac{p^2}{2n}\tag{5}
$$
Thus,
$$
\begin{align}
e^p
&\ge\left(1+\frac pn\right)^n\tag{6}\\
&\ge e^pe^{-\frac{p^2}{2n}}\tag{7}\\[3pt]
&\ge e^p\left(1-\frac{p^2}{2n}\right)\tag{8}
\end{align}
$$
Explanation:
$(6)$: apply $(2)$
$(7)$: apply $\exp$ to $(5)$
$(8)$: $e^x\ge1+x$
Thus, for $m\gt n$, we have
$$
\begin{align}
\left(1+\frac pm\right)^m-\left(1+\frac pn\right)^n
&\le e^p-\left(1+\frac pn\right)^n\\
&\le e^p\frac{p^2}{2n}\tag{9}
\end{align}
$$
Therefore,
$$
f(n)=e^p\frac{p^2}{2n}\tag{10}
$$
works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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