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First of all the question is a little bit ambigous. We don't know whether the table is a circle of a row, but from your reasoning it can be concluded that it's a circle, although not for certain. Also are there $n$ men and $n$ women total or $n$ men and $n$ women plus John and Mary? Considering the textbook answer I would go with the first option. Anyway from now on I will accept the mentioned assumptions as true.
Note that there have to be an empty space between each men. Or to make it more simple if there are $2n$ numbered seats man can sit in the even or odd seats. So we have to put $n$ men in $n$ seats around a round table. The number of possibilities is $(n-1)!$
Now as Mary has to sit next to John we have two options for her, left of John or right of him. Next the remaining $n-1$ women can be seated in $n-1$ seats in $(n-1)!$. Eventually as the events are not related we have that the total number of combinations is $2\left((n-1)!\right)^2$
I guess this explanation will fill the "holes" in your solution.
• It's probably a circle since the question says "around". – hypergeometric Nov 1 '16 at 18:24
The rule that men and women alternate should also apply to John and Mary. Hence only your 1st case applies, i.e. $[(n-1)!]^2$, and not your 2nd case.
However, you should also consider the "mirror configuration" of the 1st case, i.e. where John is seated on Mary's left/right.
Hence the number of configurations should be multiplied by $2$, giving $2[(n-1)!]^2$.
NB - this is not the same as beads on a ring where you can flip over the ring and consider the configuration to be the same - here you can't flip over the dinner table! | {
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# Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$
Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.
Thus, any help to get me out of this rift is more than welcome!
• Possibly related: math.stackexchange.com/questions/576110/… – abiessu Nov 26 '13 at 18:16
• The numerical value is 1.2722249619362552835210450521628613228181075332403 , according to PARI. Using the inverse symbolic calculator, I found no closed expression. – Peter Nov 26 '13 at 18:28
• n=10000;u=(1/n)^(1/n);while(n>2,n=n-1;u=(u+1/n)^(1/n));print(u) – Peter Nov 26 '13 at 18:51
• You could see if Landau's Algorithm (for denesting radicals) is of any help, but my guess is there isn't a "nice" expression for what you have here... – Benjamin Dickman Nov 26 '13 at 20:06
• I guess its limit is 1. Because it is increasing sequence which is bounded(maybe) by a number less than 2. But I don't know how to prove it! – Hamid Shafie Asl Jun 11 '14 at 11:20
I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$. Now, we can write ,
$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c$
Squaring on both sides,
$\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c^2$
Which is nothing but,
$\frac{1}{2}+c = c^2$
Solving above quadratic expression, value of $c$ will be $\dfrac{1+\sqrt{3}}{2}$ . Thus we get slightly improved upper bound for $L$ as $L < \dfrac{1+\sqrt{3}}{2}$ | {
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• That's even better! I wonder if it is possible to get even tighter bound... – Aron D'souza Feb 20 '16 at 11:51
• The next bound is approximately $L<1.272871$ – Yuriy S Feb 20 '16 at 11:54
• Add it as an answer then. – Aron D'souza Feb 20 '16 at 11:55
Using Aron D'souza's idea further we can get:
$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$
$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$
$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$
To find the next bound we will need to solve:
$$c^4-c-\frac{1}{4}=0$$
The exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:
$$\sqrt[4]{\frac{1}{4}+\sqrt[4]{\frac{1}{4}+\dots}}=1.0723501510383$$
The bound will become:
$$L<\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.0723501510383}}=1.272871$$
Which is three correct digits of the numerical value of the limit.
To make my answer more complete, the exact value of $c_4$ is:
$$c=\frac{1}{2} \left( b+\sqrt{\frac{2}{b}-b^2} \right)$$
$$b=\sqrt{ \sqrt[3]{ \frac{a}{18} }-\sqrt[3]{ \frac{2}{3a} } }$$
$$a=9+\sqrt{93}$$
And solving the quintic equation:
$$c^5-c-\frac{1}{5}=0$$
We get the upper bound for the limit with four correct digits:
$$L<1.272282$$
Taking into account the corresponding lower boundary:
$$L>\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\sqrt[5]{\frac{1}{5}+\sqrt[6]{\frac{1}{6}}}}}}=1.271035$$
We see that truncating the limit gives less accurate solutions than the method in this answer.
However, truncating at $\frac{1}{7}$ we can finally get very good boundaries:
$$1.27207<L<1.27228$$
This is a partial result:
The underlying sequence is increasing and upper bounded by $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=\dfrac{1+\sqrt{5}}{2}=\phi$. Thus the limit exists and it is less than $\phi$.
Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size. | {
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First we notice that:
$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$
This is not a proof, but the fact is well known. Now let's consider the following:
$$1< \left(\frac{1}{n-1}+1 \right)^{\frac{1}{n-1}}<1+\frac{1}{(n-1)^2}$$
$$1< \left(\frac{1}{n-2}+1+\frac{1}{(n-1)^2} \right)^{\frac{1}{n-2}}<1+\frac{1}{(n-2)^2}+\frac{1}{(n-2)(n-1)^2}$$
On the next step we get:
$$\dots 1+\frac{1}{(n-3)^2}+\frac{1}{(n-3)(n-2)^2}+\frac{1}{(n-3)(n-2)(n-1)^2}$$
In the end we obtain the following inequality:
$$\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots}}<1+\sum^{\infty}_{k=2} \frac{1}{k ~k!}=Ei(1)-\gamma=1.3179$$
We can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):
$$1+2\sum^{\infty}_{k=3} \frac{1}{k ~k!}=1+2(Ei(1)-\gamma-1-1/4)=1.135804$$
$$L < \sqrt{\frac{1}{2}+1.135804}=1.27899$$
$$1+2\cdot 3 \sum^{\infty}_{k=4} \frac{1}{k ~k!}=1+6(Ei(1)-\gamma-1-1/4-1/18)=1.074080$$
$$L < \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.074080}}=1.27305$$
Now for the lower boundary the better estimation would be:
$$L > \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1}}=1.26517$$
This is not ideal, but much more accurate than just truncating the sequence. | {
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# Do both versions (invariant and primary) of the Fundamental Theorem for Finitely Generated Abelian Groups hold at the same time?
So there are the two versions of the Fundamental Theorem for Finitely Generated Abelian Groups (FTFGAG). I take the following from A First Course in Abstract Algebra by Fraleigh. The first is as follows:
FTFGAG 1: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \cong \mathbb{Z}_{p_1^{r_1}} \times \cdots \times \mathbb{Z}_{p_n^{r_n}} \times \mathbb{Z} \times \cdots \times \mathbb{Z}$$ where $p_i$ are primes (not necessarily distinct) and $r_i$ are positive integers.
But then we also have the second version:
FTFGAG 2: Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form $$G \cong \mathbb{Z}_{m_1} \times \cdots \times \mathbb{Z}_{m_r} \times \mathbb{Z} \times \cdots \times \mathbb{Z}$$ where $m_1 | m_2 | \cdots | m_r$.
My question is whether these two hold at all times? So say for $\mathbb{Z}_{20}$, do we have $\mathbb{Z}_{20} \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_5 \cong \mathbb{Z}_4 \times \mathbb{Z}_5 \cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$, even though $10$ is not a power of a prime (though of course $2|10$)?
I've also seen statements of the Chinese Remainder Theorem (CRT) which say that $\mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m$ if and only if $\gcd(n,m)=1$. Does this not contradict $\mathbb{Z}_{20} \cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$ from above? Or is treating $\mathbb{Z}_n$ and $\mathbb{Z}_m$ as rings in the CRT what makes the situation different? I guess what I'm asking in a nutshell is: why don't the primary and invariant forms of the structure theorems for abelian groups (and also modules over PIDs, etc) contradict each other? | {
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Edit: So it seems that I didn't see this question which basically answers mine. CRT gives us a way of decomposition but FTFGAG only tells us that some decomposition is always possible. So for FTFGAG 1, $\mathbb{Z}_4 \times \mathbb{Z}_5$ suffices, for FTFGAG 2, $\mathbb{Z}_{20}$ suffices, and for the CRT $\mathbb{Z}_{20} \cong \mathbb{Z}_4 \times \mathbb{Z}_5$ works.
• Yes they are both correct theorems so they both hold for all finitely generated abelian groups. May 27, 2018 at 13:02
• Thanks, would you be able to shed some light on why the stuff in the CRT paragraph is/isn't a contradiction? May 27, 2018 at 13:04
• It is not true that $\mathbb{Z}_{20}\simeq\mathbb{Z}_2\times\mathbb{Z}_{10}$. May 27, 2018 at 13:09
• @MichaelBurr Indeed, I have just seen the (~duplicate) question which I linked in my edit. May 27, 2018 at 13:12
For your example, $\mathbb{Z}_{20}$,
• The first decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{2^2}\times\mathbb{Z}_5$.
• The second decomposition gives you $\mathbb{Z}_{20}\simeq\mathbb{Z}_{20}$. In other words, this one doesn't break up the abelian group at all.
• You can note that $\mathbb{Z}_{20}\not\simeq\mathbb{Z}_2\times\mathbb{Z}_{10}$ since $\mathbb{Z}_{20}$ has an element of order $20$ while the maximum order of an element of $\mathbb{Z}_2\times\mathbb{Z}_{10}$ is $10$.
The idea of the statements is that it is possible to write it in this form, not that all products of the given form are isomorphic to the given group.
A more interesting example might be given by $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5.$$
• The first decomposition gives you $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq \left(\mathbb{Z}_{2^2}\right)\times\left(\mathbb{Z}_2\times\mathbb{Z}_3\right)\times\mathbb{Z}_5\simeq \mathbb{Z}_2\times\mathbb{Z}_{2^2}\times\mathbb{Z}_3\times\mathbb{Z}_5.$$ | {
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• On the other hand, the second decomposition gives you $$\mathbb{Z}_4\times\mathbb{Z}_6\times\mathbb{Z}_5\simeq\mathbb{Z}_2\times\mathbb{Z}_{2^2\cdot 3\cdot 5}=\mathbb{Z}_2\times\mathbb{Z}_{60}.$$ The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction.
• "The group on the right combines the highest powers of each prime that appear in the fully expanded product of the first decomposition, then you work your way down by induction." I really like that way of thinking about it, thanks! May 27, 2018 at 13:23
You are correct about the Chinese remainder theorem indeed $\mathbb{Z}_{20} \not\cong \mathbb{Z}_2 \times \mathbb{Z}_{10}$. The problem with applying the 2nd version in this manner is the rank uniquely determines the group. So for example $\mathbb{Z}_{20}$ is rank 1, that is it already is written in that form (trivially) | {
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# Calculus / find the value of $x$ so that $f ''(x)=0$
Let $f(x)= 10xe^x$
$(a)$ Find the exact value of $x$ so that $f ''(x) = 0$.
I tried:
\begin{align}f'(x)& = 10e^x\\f''(x)&= e^x\end{align} but at that point, the $f''(x)$ would never be a zero. So what is my mistake?
$(b)$ For what interval is $f(x)$ concave up?
I wonder how to know the concavity after knowing the equation
• One mistake is not using the product rule to evaluate $(10xe^x)'$. – David Mitra Jun 18 '14 at 23:45
• Once you have the correct second derivative function, $\ f(x) \$ is "concave upward" wherever $\ f \ ''(x) \ > \ 0 \$ and "concave downward" wherever $\ f \ ''(x) \ < \ 0 \$ . Keep in mind that the exponential factor $\ e^x \$ is always positive. – colormegone Jun 18 '14 at 23:49
• @DavidMitra thanks for the reminder – John Jun 18 '14 at 23:50
(a) $f'(x)=10e^x + 10xe^x$.
$f''(x)=10e^x + 10e^x + 10xe^x = 20e^x + 10xe^x =10e^x(2+x)$
So, at $x=-2$ you have $f''(x)=0$.
(b) Hint: If $f''(x)>0$, $f$ is concave up at $x$, and if $f''(x)<0$, $f$ is concave down at $x$.
• Thaanks !What did you do to go from f'(x) to f''(x)? – John Jun 18 '14 at 23:49
• I computed it using the product rule. – Twink Jun 18 '14 at 23:50
• @John, derivation using the product rule. $f''(x) = (10e^x + 10xe^x)' = 10 (e^x)' + 10(x)'e^x + 10x(e^x)' = 10e^x + 10 e^x + 10xe^x = 10(2+x)e^x$ – Graham Kemp Jun 18 '14 at 23:52
$(a)$ Since $f(x)=10x\cdot e^x$ we will use the product rule to obtain $f'(x)$ and $f''(x)$. So $f'(x)=10x\cdot e^x+10e^x$ and $f''(x)=10x\cdot e^x+20e^x$. Set $f''(x)=0$. So $10x\cdot e^x+20e^x=0$ implies that $10e^x(x+2)=0$. We know that $10e^x$ is never $0$ and so $x=-2$ will make $f''(x)=0$.
$(b)$ If $x<-2$, then the function is concave down. If $x>-2$, then the function is concave up. | {
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# Arbitrary union definition in set theory
I am reading Enderton's "Elements of Set Theory". He defines the union operation as $$\cup A = \{ x \;|\; x \text{ belongs to some member of } A\} = \{x \;|\; (\exists b \in A) x \in b\}$$ Maybe I am missing a subtle point from earlier on in the text, but what if the set in question is not a "set of sets"? For example, if $$A = \{1,2,3\}$$, then is $$\cup A = \emptyset$$?
• In systems of set theory such as ZF, every set is "a set of sets". – Lord Shark the Unknown Mar 30 at 19:11
• So then is my $A$ really $\{\{1\},\{2\},\{3\}\}$? – theQman Mar 30 at 19:12
• No, it's really $\{1,2,3\}$. – Lord Shark the Unknown Mar 30 at 19:13
• See Enderton's chapter on "Natural Numbers" to see the usual way of representing positive integers as sets. – Lord Shark the Unknown Mar 30 at 19:20
• Even without the "everything-is-a-set" context, we can still make sense of this by carefully writing down the definition: see this earlier question. – Noah Schweber Mar 30 at 19:32
When you're working in a system built out of the set theory, everything is a set. I assume the book will cover this later on, but one way of "building natural numbers" out of sets is called Von Neumann ordinals and the construction is as follows:
Let $$0 \equiv \emptyset$$. Let the number $$n \equiv n - 1 \cup \{ n - 1\}$$. That is, under this system: \begin{align*} &1 \equiv 0 \cup \{ 0 \} = \emptyset \cup \{\emptyset\} = \{\emptyset\} \\ &2 \equiv 1 \cup \{1 \} = \{\emptyset\} \cup \{\{\emptyset\}\} = \{\emptyset, \{\emptyset\}\} \\ &3 \equiv 2 \cup \{2\} = \dots = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} \end{align*}
The point of the representation is that 0 is a set with zero elements, 1 is a set with 1 element, 2 is a set with 2 elements and so on. So we can define the "size of a set" as "the natural number it is in bijection with". | {
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This goes down a rabbit hole, and we can then ask "well, now that I have the natural numbers, how do I define addition? Multiplication? What about the integers? rationals? reals?"
We can construct all of these things, and a set theory book usually describes these encodings in one of the later chapters.
But the point is that at this level, all the "stuff of math" is sets, and you can always write expressions such as $$1 \ \cup 2 \cup 3 = \{\emptyset\} \cup \{ \emptyset, \{\emptyset\}\} \cup \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\} \} = \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\}\} = 3$$!
• +1.... In the most-favored system of set theory, not only is everything (that can be said to exist) a set, but in the formal language there is no word or definition of "set". Things ( with various properties that $can$ be stated in the language ) are merely asserted to exist or not exist. – DanielWainfleet Mar 30 at 19:39
• Indeed, thanks! much appreciated – Siddharth Bhat Mar 30 at 20:18 | {
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As Lord Shark the Unknown indicates, in nearly all systems of set theory, every object under the sun is a set. In particular, the elements of any set are, in turn, sets themselves. For the specific example you give, i.e. $$\cup \{ 1,2,3 \},$$ we need to be able to describe the natural numbers as sets. On page 67 of Enderton's text, the construction is outlined. He defines \begin{align} 0 &= \varnothing \\ 1 &= \{0\} = \{ \varnothing \} \\ 2 &= \{0,1\} = \{ \varnothing, \{\varnothing\} \} \\ 3 &= \{0,1,2\} = \{ \varnothing, \{\varnothing\}, \{ \varnothing, \{\varnothing\} \}\}, \end{align} and so on. The "and so on" is explained in somewhat more detail in chapter 4 of the text (starting on page 66). Note that we could write $$2 = \{\varnothing, \{\varnothing\} \}$$ over and over again, but it is likely easier to see what is going on if we work one level of abstraction higher. That is, $$2 = \{0,1\}$$. Once we accept this abstraction, we have $$\cup\{1,2,3\} = 1 \cup 2 \cup 3 = \{0\} \cup \{0,1\} \cup \{0,1,2\} = \{0,1,2\} = 3.$$
• I see. Maybe I am getting ahead of myself since I am only in chapter 2. But I guess the construction you provided is pretty natural since at this point the only set that we can use as a building block is $\emptyset$ (Empty Set Axiom)? Then we can use "pairing" to form the singleton $\{\emptyset\}$, which is defined to be $\{\emptyset, \emptyset\}$? – theQman Mar 31 at 2:09 | {
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Question
# If two concentric circles are drawn with centre in first quadrant, one touches the X-axis and Y-axis and the other passes through the origin, the radius of the smaller circle is $$'r'$$, then the centre is
A
(2r,r)
B
(r,2r)
C
(r,r)
D
none of these
Solution
## The correct option is C $$(r,r)$$Let $$(a,b)$$ be the center of the circles Given that one of the circles touches both the axes and the other passes through origin. Smaller circle touches the axes and Bigger circle passes through $$O$$. Given radius of smaller circle is $$r$$ and $$X,Y$$ axes are tangents Which implies radius = perpendicular distance from center to tangent $$(y =0 , x = 0)$$ $$r = \dfrac{|a|}{\sqrt{1^2 + 0}} , r = \dfrac{|b|}{\sqrt{1^2 + 0}}$$ $$\implies a = r, b = r$$ $$\implies (a, b) = (r,r) = center$$Maths
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## The Plane in R^3
Prerequisites: Euclidean Space and Vectors
### Suppose we have a plane in ${\Bbb R}^{3}$ which contains the point $(0,0,0)$ but does not intersect the axes at any other point. How many octants does the plane intersect?
Let's start by analyzing the 2-dimensional analog, then come back to the 3-dimensional problem with an algebraic approach and a geometric interpretation. Finally, we'll try to generalize the answer to higher dimensions.
#### Lines in the Plane ${\Bbb R}^{2}$
The 2-dimensional analog of a plane in ${\Bbb R}^{3}$ containing the origin is a line in ${\Bbb R}^{2}$ containing the origin. For those familiar with linear algebra, the former is a vector subspace of ${\Bbb R}^{3}$ of dimension 2 (i.e. co-dimension 1), and the latter is a vector subspace of ${\Bbb R}^{2}$ of dimension 1 (still co-dimension 1). If you aren't familiar with linear algebra, ignore that sentence and just keep reading.
The equation of a line in ${\Bbb R}^{2}$ is $y = mx + b$, where $m$ is the slope ("rise over run," or change in $y$ per change in $x$), and $b$ is the $y$-intercept (where the line hits the $y$-axis). A point $(x_{0},y_{0})$ is on the line if it satisfies the equation, i.e. if the equality $y_0 = mx_0 + b$ is true.
A line passing through the origin has a $y$-intercept of 0, thus $b=0$ and the equation is simply $y=mx$. Now, if $m$ is 0, then the line would just go along the $x$-axis, so if it doesn't touch the axis, we must have $m \neq 0$, so either $m > 0$ or $m < 0$. In the case where $m>0$, a positive $x$ value gives a positive $y$, and a negative $x$ gives a negative $y$, so the line would pass through quadrants 1 and 3 (top-right and bottom-left). In the case where $m<0$, the same analysis shows that the line passes through quadrants 2 and 4 (top-left and bottom-right). So the line passes through 2 quadrants. | {
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The $y = mx + b$ formulation works fine for this question, but if we want to put $x$ and $y$ on more even footing (this will come in handy in the higher-dimensional cases so that we don't always have to solve for one of the variables), we can use the other form of the equation of a line, $ax + by = c$, where $a,b,c \in {\Bbb R}$ are constants. In order to have the origin on the line, we must have $c=0$, because the point $(0,0)$ is on the line, and thus $a(0)+b(0) = 0 = c$. so the equation is simply $ax+by=0$. If either $a$ or $b$ were zero, the line would just be one of the axes, so we must have $a,b \neq 0$. Whether they are positive or negative, you can work out by plugging in positive or negative $x$ and $y$ values that the line will either pass through quadrants 1 and 3 or 2 and 4.
For example, if $a,b>0$, then if $x>0$, we must have $y<0$ in order to have $ax+by=0$. Similarly, if $x<0$, then $y$ would have to be positive. So the line passes through quadrants 2 and 4. We get the same result in the case where $a,b<0$. If $a$ and $b$ have opposite signs, then the line will pass through quadrants 1 and 3.
Notice that the line does not pass through the quadrant containing the point $(a,b)$. The vector $(a,b)$ is actually perpendicular to the line $ax+by=0$, and we can see that from the fact that the equation can be rewritten as ${\bf n} \cdot {\bf x} = 0$ where ${\bf n} = (a,b)$ and ${\bf x} = (x,y)$ (recall that two vectors are perpendicular if and only if their dot product is zero).
#### The Plane in Space ${\Bbb R}^{3}$ | {
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#### The Plane in Space ${\Bbb R}^{3}$
Let's go back to the ${\Bbb R}^{3}$ case, building off of the above discussion. The general equation of a plane is $ax+by+cz=d$, where $a,b,c,d \in {\Bbb R}$ are constants. In order for the plane to contain $(0,0,0)$, we must have $d=0$, so the equation is now just $ax+by+cz=0$. If any of the constants is 0, then the plane will actually look like the equation of a line from above. For example, if $c=0$, then we'd have $ax+by=0$, which would be a line in the $xy$-plane extended vertically up and down in the positive and negative $z$ directions, and in fact containing the entire $z$-axis. Can you see why (hint: show that the points on the $z$-axis, i.e. points of the form $(0,0,z)$, all satisfy the equation of the plane)?
We can also interpret the equation geometrically as follows: the equation $ax+by+cz=0$ is equivalent to ${\bf n} \cdot {\bf x} = 0$, where ${\bf n} = (a,b,c)$ and ${\bf x} = (x,y,z)$. Note that here, ${\bf n}$ and ${\bf x}$ are vectors, and their dot product is a scalar, so the $0$ on the right is a scalar zero, not the zero vector ${\bf 0} = (0,0,0)$.
As mentioned above, the dot product of two vectors is zero if and only if the vectors are perpendicular. Therefore, this equation is saying that any vector ${\bf x}$ that is perpendicular to ${\bf n}$ is on the plane. For this reason, ${\bf n}$ is called the plane's normal vector (normal is a synonym for perpendicular, as is orthogonal, which is also used frequently). In the example above where $c=0$, the plane's normal vector is $(a,b,0)$, which lies in the $xy$-plane. Thus, the $z$-axis, being orthogonal to the $xy$-plane, is contained in our plane.
Now, in order to answer the geometric question of which vectors are orthogonal to ${\bf n}$, we can look at the algebraic equation $ax+by+cz=0$. | {
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Since the plane does not touch the axes except at the origin, we must have $a,b,c \neq 0$. As an example, let's look at the case where $a,b,c>0$. Then we can have the following combinations for $(x,y,z)$ in order to have $ax+by+cz=0$:
$$(+,+,-) \\ (+,-,+) \\ (+,-,-) \\ (-,-,+) \\ (-,+,-) \\ (-,+,+)$$ The remaining two combinations, $(+,+,+)$ and $(-,-,-)$, do not work, because then the left side of the equation would have to be positive or negative (respectively) and thus not zero.
If we grind through the algebra of the other 7 combinations for $(a,b,c)$, we see that we get 6 possibilities each time, so the plane intersects 6 of the 8 octants, and we have the answer to the problem. I'm not going to go through all the cases, because that would be quite boring, but you can see that there is a certain symmetry in the plane's equation between $(a,b,c)$ and $(x,y,z)$. Once you've solved it for the case $a,b,c>0$, you've pretty much solved it for all the cases. Can you see why? So we've got the answer- it's 6.
#### The Hyperplane in ${\Bbb R}^{n}$
${\Bbb R}^{n}$ is the $n$-dimensional analog of ${\Bbb R}^{3}$ and is the set of ordered $n$-tuples of real numbers: ${\Bbb R}^{n} = \{ (x_1,x_2,x_3,...,x_n) \ \colon \ {\scr each} \ x_i \in {\Bbb R} \}$. We can't picture this $n$-dimensional space, but we can use the same types of algebraic equations that work in ${\Bbb R}^{3}$ to analyze ${\Bbb R}^{n}$.
${\Bbb R}^{n}$ is divided into $2^n$ orthants, also known as hyperoctants or $n$-hyperoctants, based on the signs, positive or negative, of the $n$ components of a point. A 2-hyperoctant is a quadrant in ${\Bbb R}^{2}$ and a 3-hyperoctant is an octant in ${\Bbb R}^{3}$. The $x_i$-axis is the set of points where all coordinates except possibly the $i^{\scr th}$ are zero. | {
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A hyperplane in ${\Bbb R}^{n}$ is a set $P$ of points (equivalently, vectors) that are orthogonal to a normal vector ${\bf n} = (a_1, a_2, ... a_n)$. In symbols, $P = \{ {\bf x} \in {\Bbb R}^{n} \ \colon \ {\bf n} \cdot {\bf x} = 0 \}$. For those familiar with linear algebra, the hyperplane containing the origin is a vector subspace of ${\Bbb R}^{n}$ of dimension $n-1$, i.e. co-dimension 1. A hyperplane in ${\Bbb R}^{2}$ is a line, and a hyperplane in ${\Bbb R}^{3}$ is a plane.
### How many $n$-hyperoctants does a hyperplane $P \subset {\Bbb R}^{n}$ intersect, given that it contains the origin, but does not intersect the axes at any other point?
To answer this question, we can use the discussion above from the $n=2$ and $n=3$ cases and generalize the results. We can then prove the answer is correct using induction.
When we went from $n=2$ to $n=3$, we took the equation $ax+by=0$ (i.e. the line in ${\Bbb R}^{2}$ with normal vector $(a,b)$), extended it into 3-dimensional space to make the plane whose normal vector is $(a,b,0)$, and then added a non-zero third coordinate to the normal vector to "tilt" the plane off of the $z$-axis.
Now, a hyperplane (including the line and plane in the $n=2$ and $n=3$ cases) is orthogonal to its normal vector ${\bf n}$ as well as the negative of the normal vector, $-{\bf n}$. In fact, the hyperplane is orthogonal to any scalar multiple of ${\bf n}$, but my point in mentioning $-{\bf n}$ is that the hyperplane won't intersect the $n$-hyperoctants that contains ${\bf n}$ or $-{\bf n}$. | {
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Let's look at the case where ${\bf n}$ lies in the first $n$-hyperoctant, i.e. has all positive coordinates. As mentioned above, the other cases are pretty much the same because of the symmetries of the equation ${\bf n} \cdot {\bf x} = \sum_{i=1}^{n}{a_i x_i} = 0$, so the number of $n$-hyperoctants the hyperplane intersects is the same in all cases. In the case that the $a_i$ are positive, the hyperplane doesn't intersect the first $n$-hyperoctant or the one with all negative coordinates (whatever number we want to assign to that one).
In the ${\Bbb R}^{2}$ case, the line intersects quadrants 2 and 4. When we extended ${\bf n} = (a,b)$ to ${\bf n} = (a,b,0)$ in ${\Bbb R}^{3}$, we got a plane that contained the entire $z$-axis. The intersection of this plane with the $xy$-plane is the line $ax+by=0$, which remains the case regardless of the third coordinate of ${\bf n}$. Now, this plane intersects the octants $(+,-,+)$, $(+,-,-)$, $(-,+,+)$, and $(-,+,-)$. We took the original quadrants 2 and 4 and multiplied them by 2 to get 4 octants.
When we add a non-zero third coordinate to ${\bf n}$ (let's assume it's positive), the new plane also intersects two additional octants: $(+,+,-)$ and $(-,-,+)$. The first two coordinates of these two would have not been included in the 2-d case, but the third coordinate allows us to use those combinations and still get the equation $ax+by+cz$ to equal zero. $(+,+,+)$ and $(-,-,-)$ still don't work though.
The same logic works when going from ${\Bbb R}^{n}$ to ${\Bbb R}^{n+1}$ when $n>2$, and we can prove it by induction.
Thoerem: For $n \geq 2$, a hyperplane in ${\Bbb R}^{n}$ containing the origin, but not intersecting the coordinate axes at any other point, intersects $2^{n}-2$ $n$-hyperoctants.
The proof is a bit lengthy, but basically just formalizes the idea of extending the line in ${\Bbb R}^{2}$ into a plane in ${\Bbb R}^{3}$ and then tilting it off the $z$-axis
Proof: The base case of $n=2$ was already shown above. | {
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Proof: The base case of $n=2$ was already shown above.
For the induction step, assume the theorem is true for ${\Bbb R}^{n-1}$, and consider a hyperplane $P = \{{\bf x} \in {\Bbb R}^{n} \ \colon \ {\bf n} \cdot {\bf x}=0 \}$ where ${\bf n} = (a_1,a_2,...,a_n)$.
The equation of $P$ is $\sum_{i=1}^{n}{a_i x_i} = \sum_{i=1}^{n-1}{a_i x_i} + a_n x_n = 0$. By the induction hypothesis, the solutions to the equation $\sum_{i=1}^{n-1}{a_i x_i} = 0$ intersect $2^{n-1}-2$ $(n-1)$-hyperoctants. Let's call those solutions $P_{n-1}$, which is a hyperplane in ${\Bbb R}^{n-1}$
Take a point ${\bf x}_{0, n-1} = (x_{0,1}, x_{0,2},...,x_{0,n-1}) \in {\Bbb R}^{n-1}$ which satisfies the equation of $P_{n-1}$. If $x_{0,1}>0$, then $x_{0,1}+\epsilon>0$ as well, where $\epsilon = \frac{1}{2}|x_{0,1}|$. Similarly, if $x_{0,1}<0$, then $x_{0,1}+\epsilon<0$ as well, so the point ${\bf x}_{1,n-1} = (x_{0,1}+\epsilon, x_{0,2},...,x_{0,n-1})$ is in the same $(n-1)$-hyperoctant as ${\bf x}_{0, n-1}$. By a similar argument, so is the point ${\bf x}_{2,n-1} = (x_{0,1}-\epsilon, x_{0,2},...,x_{0,n-1})$.
Define the points ${\bf x}_{1} = (x_{0,1}+\epsilon, x_{0,2},..., x_{0,n-1}, -\frac{a_1}{a_n}\epsilon), \ {\bf x}_{2} = (x_{0,1}-\epsilon, x_{0,2},..., x_{0,n-1}, \frac{a_1}{a_n}\epsilon) \in {\Bbb R}^{n}$. Then \begin{align} {\bf n} \cdot {\bf x}_{1} &= a_1 (x_{0,1}+\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} + a_n (-\frac{a_1}{a_n}\epsilon) \\[2mm] &= a_1 (x_{0,1}+\epsilon -\epsilon) + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} \\[2mm] &= a_1 x_{0,1} + a_2 x_{0,2} + ... + a_{n-1} x_{0,n-1} = 0 \end{align}
with the final equality being true because ${\bf x}_{0,n-1} \in P_{n-1}$. | {
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This shows that ${\bf x}_{1} \in P$. Similarly, ${\bf x}_{2} \in P$. The first $n-1$ coordinates of these two points are in the same $(n-1)$-hyperoctant as ${\bf x}_{0,n-1}$, and the $n^{\scr th}$ coordinates of ${\bf x}_{1}$ and ${\bf x}_{2}$ have opposite sign. This shows that we have kept the $2^{n-1}-2$ $(n-1)$-hyperoctants of the $(n-1)$-hyperplane when we extended it into ${\Bbb R}^{n}$ and actually multiplied them by 2 (by adding both positive and negative $n^{\scr th}$ coordinates) to get $2(2^{n-1}-2)$ = $2^{n}- 4$ $n$-hyperoctants.
We just need to show that we've also added two more $n$-hyperoctants. These are the ones where the first $n-1$ coordinates all have the same sign or all have the opposite sign as the first $n-1$ coordinates of ${\bf n}$, just like when we went from $n=2$ to $n=3$ above. Examples of solutions to the equation of $P$ that are in those 2 $n$-hyperoctants are $(a_1,a_2,...,a_{n-1},-\dfrac{1}{a_n}\sum_{i=1}^{n-1}{a_i^2})$ and $(-a_1,-a_2,...,-a_{n-1},\dfrac{1}{a_n}\sum_{i=1}^{n-1}{a_i^2})$.
So now we are up to $2^{n}-4+2 = 2^{n}-2$, so $P$ intersects at least that many $n$-hyperoctants. There are only 2 more $n$-hyperoctants, and those are the ones that contain $\pm {\bf n}$, but we already know that points in those $n$-hyperoctants cannot satisfy the equation of ${\bf n} \cdot {\bf x} = 0$, so $P$ intersects exactly $2^{n}-2$ of the $2^n$ $n$-hyperoctants, and the theorem is proved.
$\square$
Here's a diagram illustrating the objects described in the proof in the case where $n=3$ and $n-1=2$. Apologies for the low quality (I made it in MS Paint), but note that the bottom of the red plane, $P$, comes out towards the viewer, in front of the blue plane, and the top half of $P$ is behind the blue plane. The points ${\bf x}_{0,n-1}$, ${\bf x}_{1,n-1}$, and ${\bf x}_{2,n-1}$ are in in the $x_{1}x_{2}$-plane, with ${\bf x}_{1,n-1}$ in front of $P$ and ${\bf x}_{2,n-1}$ behind $P$. | {
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2. A hypersurface is a generalization of a hyperplane in the context of manifolds. I'm not sure the question about octants would make sense in that context since a manifold need not be in ${\Bbb R}^{n}$ (though it has a map to ${\Bbb R}^{n}$ in a neighborhood of each point). Maybe there's a way to reformulate the question to apply in that context... | {
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IntMath Home » Forum home » Methods of Integration » Taking e to both sides
# Taking e to both sides [Solved!]
### My question
Your Integration Chapter: The Basic Logarithmic Form, example 4.
Can you show how you took e to both sides to arrive at your result please?
### Relevant page
2. Integration: The Basic Logarithm Form
### What I've done so far
After integrating, t = ln 20 - ln (20-v).
You applied log laws to get t = ln(20/(20-v)).
You took "e to both sides" to get e^t = 20/20-v.
We thought by taking e to both sides the result would be the exponents equated:
t = 20/20-v.
X
Your Integration Chapter: The Basic Logarithmic Form, example 4.
Can you show how you took e to both sides to arrive at your result please?
Relevant page
<a href="https://www.intmath.com/methods-integration/2-integration-logarithmic-form.php">2. Integration: The Basic Logarithm Form</a>
What I've done so far
After integrating, t = ln 20 - ln (20-v).
You applied log laws to get t = ln(20/(20-v)).
You took "e to both sides" to get e^t = 20/20-v.
We thought by taking e to both sides the result would be the exponents equated:
t = 20/20-v.
## Re: Taking e to both sides
@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".
Writing the full details would be:
t = ln (20 /(20-v))
e^t = e^[ln (20/(20-v))]
e^t = 20/(20-v)
e^t = e^[(20/(20-v))] (without "ln")
t = 20/(20-v)
but that's not what we started with.
A similar situation is the case where "raising to power 2" is the opposite of "square root".
y = sqrt(x+2)
Squaring both sides gives us:
y^2 = (sqrt(x+2))^2 = x+2
The "power 2" has "undone" the sqrt operation.
Similarly, the "e to the power" operation undoes the "ln" expression in the example you are asking about.
Hope it makes sense.
X
@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".
Writing the full details would be:
t = ln (20 /(20-v))
e^t = e^[ln (20/(20-v))] | {
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Writing the full details would be:
t = ln (20 /(20-v))
e^t = e^[ln (20/(20-v))]
e^t = 20/(20-v)
e^t = e^[(20/(20-v))] (without "ln")
t = 20/(20-v)
but that's not what we started with.
A similar situation is the case where "raising to power 2" is the opposite of "square root".
y = sqrt(x+2)
Squaring both sides gives us:
y^2 = (sqrt(x+2))^2 = x+2
The "power 2" has "undone" the sqrt operation.
Similarly, the "e to the power" operation undoes the "ln" expression in the example you are asking about.
Hope it makes sense.
## Re: Taking e to both sides
It does now. Thank you.
X
It does now. Thank you. | {
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# Law of Sines in Triangle
On the side BC of the triangle ABC we construct towards the exterior a square BCDE. Denote the intersection between AE and BC by M. Use the law of sines to prove that
$$\frac{BM}{CM}=\frac{\cos \measuredangle B\cdot \sin \measuredangle C}{\sqrt{2} \cdot \sin \measuredangle B \cdot \sin(\measuredangle C+45°)}$$
If someone could please help me prove this problem. I do not have a similar problem to work off of. I am unclear of where the $\sin(\measuredangle C+45^{\circ})$ comes into play and the $\sqrt{2}$.
• the end got cut off...i am unlcear where the sin(<C+45) and \sqrt{2} come into play. – user423388 Mar 8 '17 at 13:42
• Provide the figure – Nick Pavini Mar 8 '17 at 14:09
Let $\measuredangle B=\beta$, $\measuredangle C=\gamma$, $\measuredangle BAE=\alpha_1$, $\measuredangle CAE=\alpha_2$. By applying the sine rule to triangles $BAM$ and $CAM$ one gets: $${BM\over\sin\alpha_1}={AM\over\sin\beta},\quad {CM\over\sin\alpha_2}={AM\over\sin\gamma},\quad \hbox{whence:}\quad {BM\over CM}={\sin\alpha_1\over\sin\alpha_2}{\sin\gamma\over\sin\beta}.$$ By applying then the sine rule to triangles $BAE$ and $CAE$ one gets: $${\sin\alpha_1\over BE}={\sin(\beta+90°)\over AE},\quad {\sin\alpha_2\over CE}={\sin(\gamma+45°)\over AE}.$$ From that, taking into account that $CE=\sqrt2 BE$, one can readily obtain $\displaystyle{\sin\alpha_1\over\sin\alpha_2}$ and thus the desired result.
• fully understood this !! thank you so much ! – user423388 Mar 8 '17 at 16:34
Using sine rule at $ACE$ we have:
$$\frac{AE}{\sin(C+45°)}=\frac{AC}{\sin \alpha}$$
Using sine rule at $ABE$ we have:
$$\frac{AE}{\sin(B+90°)}=\frac{AB}{\sin \beta}$$
Dividing both equations we have:
$$\frac{\cos B}{\sin (C+45°)}=\frac{AC}{AB}\cdot \frac{\sin \beta}{\sin \alpha} \quad (1)$$
Using sine rule at $ABC$ we get
$$\frac{AC}{AB}=\frac{\sin B}{\sin C}$$
So, from $(1)$
$$\frac{\cos B\cdot \sin C}{\sin B\cdot \sin (C+45°)}=\frac{\sin \beta}{\sin \alpha}\quad (2)$$ | {
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$$\frac{\cos B\cdot \sin C}{\sin B\cdot \sin (C+45°)}=\frac{\sin \beta}{\sin \alpha}\quad (2)$$
but $\alpha + \beta=45°$ so
$$\frac{\sin \beta}{\sin \alpha}=\frac{\sin \beta}{\sin (45°-\beta)}=\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}\quad (3)$$
Finaly we can use, from the triangle BME, that
$$\tan \beta = \frac{BM}{BE}=\frac{BM}{BM+CM}\to \frac{BM}{CM}=\frac{\tan \beta}{1- \tan \beta} \quad (4)$$
Pluging $(3)$ and $(4)$ at $(2)$ we get what we want
• can you explain (3)? i dont understand how you switched from sine to tangent... – user423388 Mar 8 '17 at 16:17
• @erica: $\frac{\sin \beta}{\sin (45°-\beta)}=\frac{\sin \beta}{(\sqrt{2}/2)(\cos \beta-\sin \beta)}=$, now divide the numerator and denominator by $\cos \beta$ and get $\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}$ – Arnaldo Mar 8 '17 at 16:33
In triangle $ABM$, using the law of sines implies $$\frac{BM}{AM}=\frac{\sin\measuredangle BAM}{\sin\measuredangle B}$$ similarly, in $ACM$: $$\frac{CM}{AM}=\frac{\sin\measuredangle CAM}{\sin\measuredangle C}$$ combining these two yields: $$\frac{BM}{CM}=\frac{\sin\angle BAM}{\sin\angle CAM}\cdot\frac{\sin\angle C}{\sin\angle B}\label{*}\tag{*}$$ Now, in triangle $ABE$ note that $\measuredangle ABE=\measuredangle B+90^{\circ}$.
Thus $\sin\measuredangle ABE=\cos\measuredangle B$ and the law of sines implies $$\frac{\sin\measuredangle BAM}{\cos\measuredangle B}=\frac{BE}{AE}$$ and for triangle $ACE$: $$\frac{\sin\measuredangle CAM}{\sin(\measuredangle C+45)}=\frac{CE}{AE}$$ Note that $CE=\sqrt{2}BE$. Now combining the two equations implies: $$\frac{BE}{CE}=\frac 1{\sqrt 2}=\frac{\sin\measuredangle BAM}{\sin\measuredangle CAM}\cdot\frac{\sin(\measuredangle C+45)}{\cos\measuredangle B}$$ And finally, you can use $\eqref{*}$ to get the desired result.
• OOps. I am late to the party – polfosol Mar 8 '17 at 15:00
• can you explain how $\text{sin}\angleABE = \text{cos}\angleB$? – user423388 Mar 8 '17 at 15:58 | {
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## Unique Path I
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
## Analysis:
This is an easy problem. From the description we know that, the robot can only move down or right, which means, if the robot is now in position (x,y), then the position before this step must be either (x-1,y) or (x, y-1). Since current position is only from these two previous positions, the number of possible paths that the robot can reach this current position (x,y) is the sum of paths from (x-1, y) and (x, y-1).
We want to get the number of paths on position (m,n), we need to know (m-1,n) and (m, n-1). For (m-1,n), we must know (m-2,n) and (m-1,n-1) ... until we back to the start position (1,1) ([0][0] in C++). Note that the boundary of the map, we can easily know that the top row and the first column of the map are all 1s. Use loop can solve the problem then.
### Code(C++):
class Solution {
public:
int uniquePaths(int m, int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int **arr = new int* [m];
for (int i=0;i<m;i++){
arr[i]= new int[n];
}
arr[0][0]=1;
for (int i=0;i<m;i++){
arr[i][0] = 1;
}
for (int i=0;i<n;i++){
arr[0][i] = 1;
}
for (int i=1;i<m;i++){
for(int j=1;j<n;j++){
arr[i][j] = arr[i][j-1] + arr[i-1][j];
}
}
return arr[m-1][n-1];
}
};
### Code(Python):
class Solution:
# @return an integer
def uniquePaths(self, m, n):
#define map and initialization
mp = [[0 for i in xrange(n)] for i in xrange(m)]
for i in xrange(m):
mp[i][0]=1
for j in xrange(n):
mp[0][j]=1
for i in xrange(1,m):
for j in xrange(1,n):
mp[i][j]=mp[i-1][j]+mp[i][j-1]
return mp[m-1][n-1]
1. I think most of your posts are great but sometimes it is very hard to understand your English though. | {
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1. Actually I have been refining both the analysis and code in all my posts recently. Thank you very much for your suggestion.
2. Does this algorithm have a name?
3. Does this algorithm have a name?
1. Dynamic Programming
4. I LOVE YOU YU!
5. why did you initialise by 1 for the borders ? | {
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# find length of rectangle given diagonal and area
• September 24th 2013, 11:08 AM
Bonganitedd
find length of rectangle given diagonal and area
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
• September 24th 2013, 11:18 AM
votan
Re: find length of rectangle given diagonal and area
Quote:
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Sketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.
• September 24th 2013, 11:30 AM
Bonganitedd
Re: find length of rectangle given diagonal and area
Its same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2
The area, LW =168
• September 24th 2013, 12:17 PM
Plato
Re: find length of rectangle given diagonal and area
Quote:
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method | {
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Have a look at this webpage.
• September 24th 2013, 01:15 PM
Soroban
Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Quote:
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
Factor: . $(W^2 - 49)(W^2 - 576) \:=\:0$
$\begin{array}{ccccccccc}W^2-49 \:=\:0 & \Rightarrow & W^2 \:=\:49 & \Rightarrow & W \:=\:7 \\ W^2 - 576 \:=\:0 & \Rightarrow & W^2 \:=\:576 & \Rightarrow & W \:=\:24 \end{array}$
$\text{Assuming }L > W\text{, we have: }\:W\,=\,7,\;\boxed{L \,=\,24}$
• September 24th 2013, 02:08 PM
HallsofIvy
Re: find length of rectangle given diagonal and area
Once you get to "[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the "fourth degree polynomial" bothers you, you can let $x= W^2$ so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.
• September 24th 2013, 07:02 PM
Bonganitedd
Re: find length of rectangle given diagonal and area
Not taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF. | {
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# Write $\cos^8(x)$ in terms of $\sin(x)$?
I am currently working on an integral using u substitution and have hit a wall at the above point. The integral in question is $\int$$\sin^7(x)$$\cos^9(x)$ using $u=sin(x)$.
After finding $du=cos(x) dx$, I am asked to express $\cos^8(x)$ in terms of $u$, at which point I am stuck. My current attempt was to reduce the power down from 8 to 2, and express it as $(1-u^2)$, but that gives me a wrong answer (this assignment checks for us before submission).
Any help is appreciated.
Thanks.
• What did you find the integral to be (your "wrong" answer)? – Henry Oct 18 '15 at 9:55
• I hadn't got to that point yet as the next few steps required it expressed in terms of u still. – Dwayne H Oct 18 '15 at 9:56
• OK, your "wrong" answer in terms of $u$? – Henry Oct 18 '15 at 9:57
• I'd only gotten to the point of $u^7(1-u^2)$, but seeing as the assignment said it was wrong I didn't try going ahead and solving for that. – Dwayne H Oct 18 '15 at 9:59
• $\displaystyle \int u^7(1-u^2)^4\,du$ with a power of $4$ seems more likely – Henry Oct 18 '15 at 10:00
If $$\cos^2 x = 1-\sin^2 x$$ then $$\cos^8 x = (1-\sin^2 x)^4.$$
Using $u=\sin x$, you have $du=\cos x\,dx$, and also $\cos^2x=1-u^2$, so $\cos^8x=(1-u^2)^4$. Thus your integral becomes $$\int u^7(1-u^2)^4\,du$$ | {
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Covariance Calculator With Probability | {
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In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Pre-trained models and datasets built by Google and the community. , the variables tend to show similar behavior), the covariance is positive. Their covariance Cov(X;Y) is de ned by. We assume that a probability distribution is known for this set. You can find formula used for calculation of covariance below the calculator. How can I calculate the covariance in R? I created two vectors x,y and fed them into cov(), but I get the wrong result. Recent research has pointed to the ubiquity and abundance of between-generation epigenetic inheritance. The term ANCOVA, analysis of covariance, is commonly used in this setting, although there is some variation in how the term is used. Each point in the x-yplane corresponds to a single pair of observations (x;y). Before you compute the covariance, calculate the mean of x and y. Application of this formula to any particular observed sample value of r will accordingly test the null hypothesis that the observed value comes from a population in which rho=0. Suppose, under uncertainty, the manager believes that a risky asset, for example, an equity, can bring him different results, which at the moment of portfolio formation can only be judged with some probability, as shown in Table. Covariance can be calculated by using the formula. If a jpd is over N random vari-ables at once then it maps from the sample space to RN, which is short-hand for real-valued vectorsof dimension N. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. The probability that a woman has all three risk factors, given that she has A and B, is 1/3. | {
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The probability that a woman has all three risk factors, given that she has A and B, is 1/3. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. Properties of variance and covariance (a) If and are independent, then by observing that. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. You COULD do a calculation patterned after covariance with 3 or more variables, but I don’t see it as meaning anything, and I don’t think it would be admissable as a valid statistical function. This binomial calculator can help you calculate individual and cumulative binomial probabilities of an experiment considering the probability of success on a single trial, no. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. It is found that the effect of estimator variability is significant to obtain VaR forecast with better coverage. An investor is facing two potential financial losses and with the following joint density function: Let be the total of these two losses. Example: Plastic covers for CDs (Discrete joint pmf) Measurements for the length and width of a rectangular plastic covers for CDs are rounded to the nearest mm(so they are discrete). The correlation coefficient is equal to the covariance divided by the product of the standard deviations of the variables. More about Covariance. Chapter 5: JOINT PROBABILITY DISTRIBUTIONS Part 2: Covariance and Correlation Section 5-4 Consider the joint probability distribution fXY(x;y). If instead, you want to get a step-by-step calculation of all descriptive statistics, you can try our descriptive statistics calculator. Let W have the exponential distribution with mean 1. Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Expected Return Formula Calculator. If A is a scalar, cov(A) returns 0. Covariance • Variance and | {
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Expected Return Formula Calculator. If A is a scalar, cov(A) returns 0. Covariance • Variance and Covariance are a measure of the "spread" of a set of points around their center of mass (mean) • Variance - measure of the deviation from the mean for points in one dimension e. Covariance is driving me nuts, is there any simple way to make the calculator do the work for you? Get your mind off your Level I results with a free 2020 Level II JumpStart package. 35 5% -3% Given The Above Information On Two Investments A And B, Calculate The Statistics Below. Please enter the necessary parameter values, and then click 'Calculate'. Suppose, under uncertainty, the manager believes that a risky asset, for example, an equity, can bring him different results, which at the moment of portfolio formation can only be judged with some probability, as shown in Table. A low value means there is a weak relationship. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: Σ ij = cov(x i,x j) = E p(x)[(x i −µ i)(x j −µ j)] (10) between variables x i and x j. • Probability and Statistics for Engineering and the Sciences, Covariance, Correlation, Sampling Distributions, Central Limit calculator or for any other. E(X1)=µX1 E(X2)=µX2 var(X1)=σ2 X1 var(X2)=σ2 X2 Also, we assume that σ2 X1 and σ2 X2 are finite positive values. Estimating the uncertainty of revenues and investment decisions. And, to calculate the probability of an interval, you take the integral of the probability density function over it. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The covariance matrix is a matrix that only concerns the relationships between variables, so it will be a k x k square matrix. Using the formulae above to compute covariance can sometimes be tricky. The covariance matrix can be calculated in NumPy using the cov() function. The sign (+ or -) of the correlation affects its interpretation. I can of course go | {
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cov() function. The sign (+ or -) of the correlation affects its interpretation. I can of course go ahead and derive all the elements of conditional covariance matrix. The Lognormal Probability Distribution Let s be a normally-distributed random variable with mean µ and σ2. The denominator is represented by (n-1), which is just one less than the number of data pairs in your data set. The converse. There is a 50 percent probability that after one year the stock's price will rise to $5 (yielding a 400 percent return). R Functions for Probability Distributions. The probability density function of the bivariate normal distribution is implemented as MultinormalDistribution[mu1, mu2, sigma11, sigma12, sigma12, sigma22] in the Wolfram Language package MultivariateStatistics. 128 CHAPTER 7. Variables are inversely. If the variables tend to show similar behavior, the covariance is positive. Uniform distribution Calculator - High accuracy calculation Welcome, Guest. This post provides practice problems to reinforce the concepts discussed in the preceding post on univariate probability distributions. That is, covariance matrices with small determinants denote variables that are redundant or highly correlated. Variables are positively related if they move in the same direction. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. This online calculator computes covariance between two discrete random variables. By default, this function will calculate the sample covariance matrix. The sign (+ or -) of the correlation affects its interpretation. 2 Covariance Covariance is a measure of how much two random variables vary together. Chapter 4 Variances and covariances Page 3 A pair of random | {
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much two random variables vary together. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. 1 Introduction. Joyce, Fall 2014 Covariance. [In our case, a 5×5 matrix. Also, it can be considered as a generalization of the concept of variance of two random variables. Covariance Calculator calculator, formula and. The Covariance Calculator an online tool which shows Covariance for the given input. Covariance gives you a positive number if the variables are positively related. Covariance[dist] gives the covariance matrix for the multivariate symbolic distribution dist. The formula defines covariance for discrete variables in Simon & Blume (1994): Mathematics for Economists, end of section A5. Linear Models in SAS (Regression & Analysis of Variance) The main workhorse for regression is proc reg, and for (balanced) analysis of variance, proc anova. In case the greater values of one variable are linked to the greater values of the second variable considered, and the same corresponds for the smaller figures, then the covariance is positive and is a signal that the two variables show similar behavior. If the variables tend to show similar behavior, the covariance is positive. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. If using percent form, the user must add the percent sign (%) at the end of the number. Covariance[v1, v2] gives the covariance between the vectors v1 and v2. In the matrix diagonal there are variances, i. (2005), Fundamentals of Probability with Stochastic Processes, Roussas, G. In addition, we may only be able to assess unconditional coverage probability for VaR forecast of the SVAR model. Problem 31B. I have a joint probability mass function of two variables X,Y like here. P function in Microsoft Excel. Joint Discrete Probability Distributions. | {
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variables X,Y like here. P function in Microsoft Excel. Joint Discrete Probability Distributions. Chapter 4 Variances and covariances Page 3 A pair of random variables X and Y is said to be uncorrelated if cov. I hope you found this video useful, please subscribe for daily videos! WBM Foundations: Mathematical logic Set theory Algebra: Number theory Group theory Lie groups Commutative rings Associative. By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. Covariance is a measure of how much two random variables vary together. When comparing data samples from different populations, two of the most popular measures of association are covariance and correlation. Applied to historical prices, covariance can help determine if stocks' prices tend. Video for finding the covariance and correlation coefficient by hand. TI-82 / TI-83 Graphing Calculator. Correlation values range from positive 1 to negative 1. 5 should display. For a discrete variable$\sum_{i=1}^nP(X=x_{i}|A)=1$, hence we immediately can fill in the missing values in the conditional distribution tables. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. Covariance & Correlation The covariance between two variables is defined by: cov x,y = x x y y Can I directly relate the free parameters to the covariance matrix? First calculate P(x) by marginalizing over y: P xˆexp{1 21 2 x x0 x 2} dyexp{1 21 2 [y y0 y 2 2 x x0 x y y0 y]} probability content is contained within a radius of s(2. Lecture 21: Conditional Distributions and Covariance / Correlation Statistics 104 Colin Rundel April 9, 2012 6. when the returns of one asset goes up, the. 92 and 202-205; Whittaker and Robinson 1967, p. For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if | {
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have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise. Discrete Random Variable Calculator. Now, we also need to be able to calculate the covariance and correlation for a joint probability function. It is considered the ideal case in which the probability structure underlying the categories is known perfectly. Although the covariance and variance are linked to each other in the above manner, their probability distributions are not attached to each other in a simple manner and have to be dealt separately. Descriptive Statistics which contains one variable and multivariable calculators for 20 descriptive statistics measures including: mean, variance, covariance, quantile, interquartile range, correlation and many more. There are various kinds of insurance. Recall that , and that is the normal density with mean and variance. In case the greater values of one variable are linked to the greater values of the second variable considered, and the same corresponds for the smaller figures, then the covariance is positive and is a signal that the two variables show similar behavior. Video and lecture notes from a tutorial on Probability and Statistics given at PyData NYC 2019. For each of the three factors, the probability is 0. when the returns of one asset goes up, the. This calculator is featured to generate the complete work with steps for any corresponding input values may helpful for grade school students to solve covaraince worksheet or homework problems or. The expected value of X is usually. Imagine how confusing it would be if people used degrees Celsius and degrees Fahrenheit interchangeably. 2 thoughts on " An Example on Calculating Covariance. Probability helps the company to calculate how many chances that insurance holders have to claim the insurance. it helps us to understand how two sets of data are related to each other. For example, if the number of desired outcomes divided by | {
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sets of data are related to each other. For example, if the number of desired outcomes divided by the number of possible events is. Use the theorem we just proved to calculate the covariance of X and Y. Sample Mean and Covariance Calculator. My ultimate objective is to calculate with a given set data the following ratio: Det[Covariance matrix for X]/Det[Conditional covariance matrix X|S] Any idea about how to achieve this?. Most articles and reading material on probability and statistics presume a basic understanding of terms like means, standard deviation, correlations, sample sizes and covariance. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: Σ ij = cov(x i,x j) = E p(x)[(x i −µ i)(x j −µ j)] (10) between variables x i and x j. - The probability that a company will pass the test given that it will subsequently survive 12 months, is. In the next section, read Problem 1 for an example showing how to turn raw data into a variance-covariance matrix. Covariance Calculator calculator, formula and work with steps to estimate the relationship (linear dependence) between two dataset in statistical experiments. WORKED EXAMPLES 3 COVARIANCE CALCULATIONS EXAMPLE 1 Let Xand Y be discrete random variables with joint mass function defined by f X,Y(x,y) = 1 4 Hence the two variables have covariance and correlation zero. I know the definition of covariance and I'm trying to solve some exercises. Rules for the Correlation Coefficient. Standard Deviation Calculator Variance Calculator Kurtosis Calculator Skewness Calculator. Using the formulae above to compute covariance can sometimes be tricky. com's Covariance calculator is an online statistics & probability tool to estimate the nature of association between two random variables X & Y in probability & statistics experiments. Before we demonstrate the function in Excel, a few observations on the covariance and correlation measures. The probability that a large earthquake | {
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few observations on the covariance and correlation measures. The probability that a large earthquake will occur on the San Andreas Fault in. If you see any typos, potential edits or changes in this Chapter, please note them here. function [probability] = comp_gauss_dens_val (mean, covarianceMatrix, givenMatrix ) The value assigned in lengthofM varibale is the length of the mean of the given matrix, which in our is one dimensional in our case. If an input is given then it can easily show the result for the given number. Joint Discrete Probability Distributions. A negative covariance means that the variables are inversely related, or that they move in opposite directions. If A is a vector of observations, C is the scalar-valued variance. All you can really tell from covariance is if there is a positive or negative relationship. Apart from the six-page limitation, originality, quality and clarity will be the criteria for choosing the material to be published in Statistics & Probability Letters. Sample covariance measures the …. The domain of t is a set, T , of real numbers. Provide an interpretation (10%). Covariance[m] gives the covariance matrix for the matrix m. Formulas that calculate covariance can predict how two stocks might perform relative to each other in the future. Click the Calculate! button and find out the covariance matrix of a multivariate sample. (i) The expected value measures the center of the probability distribution - center of mass. covariance calculator - step by step calculation to measure the statistical relationship (linear dependence) between the two sets of population data, along with. For example,. The correlation coefficient is also known as the Pearson Product-Moment Correlation Coefficient. pX Beta calculator. calculate and interpret covariance and correlation and interpret a scatterplot; calculate and interpret the expected value, variance, and standard deviation of a random variable and of returns on a portfolio; calculate and | {
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variance, and standard deviation of a random variable and of returns on a portfolio; calculate and interpret covariance given a joint probability function; calculate and interpret an updated probability using Bayes' formula;. In probability, we use 0. Adding a constant to a random variable does not change their correlation coefficient. is between and inclusive, which meets the first property of the probability distribution. ) S is said to have a lognormal distribution,. Full curriculum of exercises and videos. Small sample size problems. Different categories of descriptive measures are introduced and discussed along with the Excel functions to calculate them. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. To find the covariance matrix you need to calculate the Fisher information matrix. A covariance refers to the measure of how two random variables will change together and is used to calculate the correlation between variables. Obtaining covariance estimates between variables allows one to better estimate direct and indirect effects with other variables, particularly in complex models with many parameters to be estimated. In introductory finance courses, we are taught to calculate the standard deviation of the portfolio as a measure of risk, but part of this calculation is the covariance of these two, or more, stocks. Do October 10, 2008 A vector-valued random variable X = X1 ··· Xn T is said to have a multivariate normal (or Gaussian) distribution with mean µ ∈ Rn and covariance matrix Σ ∈ Sn 1. covariance calculator - step by step calculation to measure the statistical relationship (linear dependence) between the two sets of population data, along with. This calculator determines the following coin toss probability scenarios * Coin Toss Sequence such as HTHHT * Probability of x heads and y tails * Covariance of X | {
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scenarios * Coin Toss Sequence such as HTHHT * Probability of x heads and y tails * Covariance of X and Y denoted Cov(X,Y) * The. Is there a relationship between Xand Y? If so, what kind? If you’re given information on X, does it give you information on the distribution of Y? (Think of a conditional distribution). It's an online statistics and probability tool requires two sets of population data X and Y and measures of how much these data sets vary together, i. When the covariance is positive, X tends to be high when Y is high, and vice versa; when the covariance is negative, X tends to be high when Y is low, and vice versa. Consider two random variables$X$and$Y$. Here are some documents to help you use the TI-82 calculator. Enter X values (Separated by comma) Enter Y values (Separated by comma) Letter Arrangment Probability Calculator. Where Cov (A, B) – is covariance of portfolios A and B. Calculate the standard deviation of the returns using STDEV function; Finally, we calculate the VaR for 90, 95, and 99 confidence level using NORM. Or are they. The converse. 01 respectively for the VaR(90), VaR(95), and VaR(99). If the greater values of one variable mainly correspond with the greater values of the other variable, and the same holds for the lesser values, (i. S function is:. , the covariance of each element with itself. This suggests the question: Given a symmetric, positive semi-de nite matrix, is it the covariance matrix of some random vector?. The calculator will find the p-value for two-tailed, right-tailed and left-tailed tests from normal, Student's (T-distribution), chi-squared and Fisher (F-distribution) distributions. How to Calculate Expected Value. The probability that a large earthquake will occur on the San Andreas Fault in. After cleaning the data, the researcher must test the assumptions of ANOVA. Therefore, it is a straightforward exercise to calculate the correlation between X and. 2 Covariance Covariance is a measure of how much two random | {
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calculate the correlation between X and. 2 Covariance Covariance is a measure of how much two random variables vary together. More about Covariance. Welcome to StatCalculators. The sample variance, s², is used to calculate how varied a sample is. This calculator computes the variance from a data set: To calculate the variance from a set of values, specify whether the data is for an entire population or from a sample. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page. It is based on the probability-weighted average of the cross-products of the random variables’ deviations from their expected values for each possible outcome. (b) In contrast to the expectation, the variance is not a linear operator. Covariance is a statistical calculation that helps you understand how two sets of data are related to each other. And, we are given that the standard deviation of X is 1/2, and the standard deviation of Y is the square root of 1/2. Correlation between the two variables is a normalized version of the Covariance. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. described with a joint probability mass function. rameterized by a mean vector µ, and a variance-covariance matrix Σ, written X ∼ (µ,Σ). Given such information it is possible to calculate the marginal distributions and then the joint distribution follows. The function underlying its probability distribution is called a probability density function. Standard Deviation Calculator Variance Calculator Kurtosis Calculator Skewness Calculator. A zero covariance may indicate that the two assets are independent. This continues our exploration of the semantics of the inner product. CSC 411 / CSC D11 / CSC C11 Probability | {
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our exploration of the semantics of the inner product. CSC 411 / CSC D11 / CSC C11 Probability Density Functions (PDFs) The off-diagonal terms are covariances: Σ ij = cov(x i,x j) = E p(x)[(x i −µ i)(x j −µ j)] (10) between variables x i and x j. search(“distribution”). The Multivariate Gaussian Distribution Chuong B. Probability is the likelihood of something happening or being true. Both of these two determine the relationship and measures the dependency between two random. The covariance of two variables tells you how likely they are to increase or decrease simultaneously. 128 CHAPTER 7. Then m is the vector of means and V is the variance-covariance matrix. Online probability calculator to find expected value E(x), variance (σ 2) and standard deviation (σ) of discrete random variable from number of outcomes. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. Covariance correlations in collision avoidance probability calculations. Printer-friendly version Introduction. Begin your Level II studies with a FREE Schweser JumpStart Package. If you're behind a web filter, please make sure that the domains *. The TI-83 Graphing Calculator can facilitate the entry of ordered lists of data and perform some statistical analyses, but lacks a single command to calculate the covariance of two lists of numbers. The calculator based methods proposed above don’t account for the fact that a probability is associated with asset returns. Doing the calculation To calculate the beta coefficient for a we'll compare how the stock and the index move relative to each other with a covariance formula and then divide that result by the. 329) and is the covariance. ) q for "quantile", the inverse c. This calculator is featured to generate the complete work with steps for any corresponding input values may helpful for grade school students to solve. When a portfolio includes two risky assets, the Analyst needs to | {
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for grade school students to solve. When a portfolio includes two risky assets, the Analyst needs to take into account expected returns, variances and the covariance (or correlation) between the assets' returns. These usual, our starting point is a random experiment with a probability measure ℙ on an underlying sample space. Bayesian decision theory is a fundamental statistical approach to the problem of pattern classification. Correlation values range from positive 1 to negative 1. Returns population covariance, the average of the products of deviations for each data point pair in two data sets. Variance and Standard Deviation of a Random Variable. Hey Flashcop and welcome to the forums. That is, if one increases, the other increases. It's an online statistics and probability tool requires two sets of population data X and Y` and measures of how much these data sets vary together, i. If an input is given then it can easily show the result for the given number. In this section, we will study an expected value that measures a special type of relationship between two real-valued variables. The Covariance Calculator an online tool which shows Covariance for the given input. Input the matrix in the text field below in the same format as matrices given in the examples. I was concurrently taking a basic theoretical probability and statistics, so even the idea of variance was still vague to me. 4, and in Robert J. Applied to historical prices, covariance can help determine if stocks' prices tend. Descriptive Statistics which contains one variable and multivariable calculators for 20 descriptive statistics measures including: mean, variance, covariance, quantile, interquartile range, correlation and many more. J Matney a Collision-avoidance calculations between orbiting objects make use of covariance matrices to characterize the uncertainty of the orbital position in space and time. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. | {
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time. Byju's Covariance Calculator is a tool which makes calculations very simple and interesting. The Multivariate Gaussian Distribution Chuong B. We will go through a review of probability concepts over here, all of the review materials have been adapted from CS229 Probability Notes. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. 2 Covariance Covariance is a measure of how much two random variables vary together. 25, multiply the answer by 100 to get 25%. The uniform distribution is used to describe a situation where all possible outcomes of a random experiment are equally likely to occur. is the correlation of and (Kenney and Keeping 1951, pp. 12 that she has exactly these two risk factors (but not the other). In general, if there are n random variables, the outcome is an n-dimensional vector of them. Stock Correlation Calculator. All we'll be doing here is getting a handle on what we can expect of the correlation coefficient if X and Y are independent, and what we can expect of the correlation coefficient if X and Y are dependent. If the probability of the event changes when we take the first event into consideration, we can safely say that the probability of event B is dependent of the occurrence of event A. As you doubtless know, the variance of a set of numbers is defined as the "mean squared difference from the mean". This online calculator computes covariance between two discrete random variables. Let X 1 = number of dots on the red die X 2 = number of dots on the green die. The correlation coefficient is equal to the covariance divided by the product of the standard deviations of the variables. How to Calculate Covariance. Let's now look at how to calculate the risk of the portfolio. As a simple example of covariance we'll return once again to the Old English example of Section 2. This tool | {
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example of covariance we'll return once again to the Old English example of Section 2. This tool provides direct calculations for a variety of probability distributions. 5 Covariance and Correlation Covariance and correlation are two measures of the strength of a relationship be-tween two r. ] Before constructing the covariance matrix, it’s helpful to think of the data matrix as a collection of 5 vectors, which is how I built our data matrix in R. We construct a non-separable space-time covariance function based on a diffusive Langevin equation. If the two variables are dependent then the covariance can be measured using the following formula:. Although the covariance and variance are linked to each other in the above manner, their probability distributions are not attached to each other in a simple manner and have to be dealt separately. A positive covariance would indicate a positive linear relationship between the variables, and a negative covariance would indicate the opposite. The syntax of the Covariance. It is actually used for computing the covariance in between every column of data matrix. It is also helpful for insurance. They must then calculate the F-ratio and the associated probability value (p-value). By using this formula, after calculation, you can verify the result of such calculations by using our covariance calculator. If A is a matrix whose columns represent random variables and whose rows represent observations, C is the covariance matrix with the corresponding column variances along the diagonal. [In our case, a 5×5 matrix. It just creates confusion because they are not equivalent. In order to calculate the Student T Value for any degrees of freedom and given probability. Key Differences Between Covariance and Correlation. With the covariance option, correlate can be used to obtain covariance matrices, as well as correlation matrices, for both weighted and unweighted data. It's the statistics & probability functions formula reference sheet | {
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weighted and unweighted data. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. In my first machine learning class, in order to learn about the theory behind PCA (Principal Component Analysis), we had to learn about variance-covariance matrix. A distribution is described as normal if there is a high probability that any observation form the population sample will have a value that is close to the mean, and a low probability of having a value that is far from the mean. So, I wrote an add-in that used the matrix algebra functions to create a covariance matrix that would change if you changed the data. 05 Jeremy Orlo and Jonathan Bloom 1 Learning Goals 1. By default, this function will calculate the sample covariance matrix. Calculate the covariance of and. The function is new in Excel 2010 and so is not available in earlier versions of Excel. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. A Statistics Co-Variance Calculator An online Co-Variance Calculator to measure of two variables X and Y. Covariance and Correlation Definition: Covariance Let X and Y be two RV’s with means x and y, respectively. In statistical theory, covariance is a measure of how much two random variables change together. E(X1)=µX1 E(X2)=µX2 var(X1)=σ2 X1 var(X2)=σ2 X2 Also, we assume that σ2 X1 and σ2 X2 are finite positive values. 2 Covariance Covariance is a measure of how much two random variables vary together. Expectation of a Function of a Random Variable Suppose that X is a discrete random variable with sample space Ω, and φ(x) is a real-valued function with domain Ω. A sample is a randomly chosen selection of elements from an underlying population. Analysis of covariance (ANCOVA) allows to compare one variable in 2 or more groups taking into account (or to correct for) variability of other variables, called covariates. | {
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groups taking into account (or to correct for) variability of other variables, called covariates. It is important in security analysis to determine how much or how little price movements in two companies or industries are connected. These topics are somewhat specialized, but are particularly important in multivariate statistical models and for the multivariate normal distribution. Using the covariance formula, you can determine whether economic growth and S&P 500 returns have a positive or inverse relationship. Typically, the population is very large, making a complete enumeration of all the values in the population impossible. How does this covariance calculator work? In data analysis and statistics, covariance indicates how much two random variables change together. Consider two random variables$X$and$Y\$. Understand the meaning of covariance and correlation. If is the covariance matrix of a random vector, then for any constant vector ~awe have ~aT ~a 0: That is, satis es the property of being a positive semi-de nite matrix. The problem is solved by standardize the value of covariance (divide it by ˙. An investor is facing two potential financial losses and with the following joint density function: Let be the total of these two losses. It's the statistics & probability functions formula reference sheet contains most of the important functions for data analysis. S function is:. Population Standard Deviation The population standard deviation, the standard definition of σ , is used when an entire population can be measured, and is the square root of the variance of a given data set. Excel does such a great job in calculating correlation and covariance that it is not necessary to memorize the formulas of covariance and correlation, but here they are, along with examples worked out in Excel: Covariance of variables x and y from a known population = σ xy. The covariance for two random variates X and Y, each with sample size N, is defined by the expectation…. Calculate | {
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for two random variates X and Y, each with sample size N, is defined by the expectation…. Calculate the covariance of and. Discrete Random Variable Calculator. Properties of variance and covariance (a) If and are independent, then by observing that. A portfolio is the total collection of all investments held by an individual or institution, including stocks, bonds, real estate, options, futures, and alternative investments, such as gold or limited partnerships. On this page, we'll begin our investigation of what the correlation coefficient tells us. Consider the following example: Example. The covariance is defined as. Click the Calculate! button and find out the covariance matrix of a multivariate sample. A randomly selected day was a long commute. These usual, our starting point is a random experiment with a probability measure ℙ on an underlying sample space. For example, the probability of a two-dimensional case, in which the vector of random variables is X = [X, Y] T, can be calculated as. This matrix is not usually printed or saved. , the variables tend to show similar behavior), the covariance is positive. For instance, I have been given a discrete random variable X with probability function px(x) = 1/2 if x = -1, 1/4 if x = 0, 1/4 if x = 1, 0 otherwise. If the covariance is a large positive number, then we expect x i to be largerthanµ iwhenx j islargerthanµ j. Let's calculate the covariance for the example age and income data. In probability theory and statistics, covariance is a measure of the joint variability of two random variables. Covariance and correlation show that variables can have a positive relationship, a negative relationship, or no relationship at all. You COULD do a calculation patterned after covariance with 3 or more variables, but I don’t see it as meaning anything, and I don’t think it would be admissable as a valid statistical function. (The covariance of X and Y is sometimes written cov(X, Y). He also covers testing hypotheses, modeling | {
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covariance of X and Y is sometimes written cov(X, Y). He also covers testing hypotheses, modeling different data distributions, and calculating the covariance and correlation between data sets. | {
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# Kendall tau distance
The Kendall tau rank distance is a metric that counts the number of pairwise disagreements between two ranking lists. The larger the distance, the more dissimilar the two lists are. Kendall tau distance is also called bubble-sort distance since it is equivalent to the number of swaps that the bubble sort algorithm would take to place one list in the same order as the other list. The Kendall tau distance was created by Maurice Kendall.
## Definition
The Kendall tau ranking distance between two lists $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ is
$\displaystyle{ K(\tau_1, \tau_2) = |\{(i,j): i \lt j, ( \tau_1(i) \lt \tau_1(j) \wedge \tau_2(i) \gt \tau_2(j) ) \vee ( \tau_1(i) \gt \tau_1(j) \wedge \tau_2(i) \lt \tau_2(j) )\}|. }$
where
• $\displaystyle{ \tau_1(i) }$ and $\displaystyle{ \tau_2(i) }$ are the rankings of the element $\displaystyle{ i }$ in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ respectively.
$\displaystyle{ K(\tau_1,\tau_2) }$ will be equal to 1 if the two lists are identical and $\displaystyle{ -1 }$ (where $\displaystyle{ n }$ is the list size) if one list is the reverse of the other. The normalized Kendall tau distance therefore lies in the interval [-1,1].
Kendall tau distance may also be defined as
$\displaystyle{ K(\tau_1,\tau_2) = \begin{matrix} \sum_{\{i,j\}\in P} \bar{K}_{i,j}(\tau_1,\tau_2) \end{matrix} }$
where
• P is the set of unordered pairs of distinct elements in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$
• $\displaystyle{ \bar{K}_{i,j}(\tau_1,\tau_2) }$ = 0 if i and j are in the same order in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$
• $\displaystyle{ \bar{K}_{i,j}(\tau_1,\tau_2) }$ = 1 if i and j are in the opposite order in $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2. }$
Kendall tau distance can also be defined as the total number of discordant pairs. | {
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Kendall tau distance can also be defined as the total number of discordant pairs.
Kendall tau distance in Rankings: A permutation (or ranking) is an array of N integers where each of the integers between 0 and N-1 appears exactly once. The Kendall tau distance between two rankings is the number of pairs that are in different order in the two rankings. For example, the Kendall tau distance between 0 3 1 6 2 5 4 and 1 0 3 6 4 2 5 is four because the pairs 0-1, 3-1, 2-4, 5-4 are in different order in the two rankings, but all other pairs are in the same order.[1]
If Kendall tau function is performed as $\displaystyle{ K(L1,L2) }$ instead of $\displaystyle{ K(\tau_1,\tau_2) }$ (where $\displaystyle{ \tau_1 }$ and $\displaystyle{ \tau_2 }$ are the rankings of $\displaystyle{ L1 }$ and $\displaystyle{ L2 }$ elements respectively), then triangular inequality is not guaranteed. The triangular inequality fails in cases where there are repetitions in the lists. So then we are not dealing with a metric anymore.
## Example
Suppose one ranks a group of five people by height and by weight:
Person A B C D E
Rank by height 1 2 3 4 5
Rank by weight 3 4 1 2 5
Here person A is tallest and third-heaviest, and so on.
In order to calculate the Kendall tau distance, pair each person with every other person and count the number of times the values in list 1 are in the opposite order of the values in list 2.
Pair Height Weight Count
(A,B) 1 < 2 3 < 4
(A,C) 1 < 3 3 > 1 X
(A,D) 1 < 4 3 > 2 X
(A,E) 1 < 5 3 < 5
(B,C) 2 < 3 4 > 1 X
(B,D) 2 < 4 4 > 2 X
(B,E) 2 < 5 4 < 5
(C,D) 3 < 4 1 < 2
(C,E) 3 < 5 1 < 5
(D,E) 4 < 5 2 < 5
Since there are four pairs whose values are in opposite order, the Kendall tau distance is 4. The normalized Kendall tau distance is
$\displaystyle{ \frac{4}{5(5 - 1)/2} = 0.4. }$
A value of 0.4 indicates that 40% of pairs differ in ordering between the two lists.
## Computing the Kendall tau distance
A naive implementation in Python (using NumPy) is: | {
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## Computing the Kendall tau distance
A naive implementation in Python (using NumPy) is:
import numpy as np
def normalised_kendall_tau_distance(values1, values2):
"""Compute the Kendall tau distance."""
n = len(values1)
assert len(values2) == n, "Both lists have to be of equal length"
i, j = np.meshgrid(np.arange(n), np.arange(n))
a = np.argsort(values1)
b = np.argsort(values2)
ndisordered = np.logical_or(np.logical_and(a[i] < a[j], b[i] > b[j]), np.logical_and(a[i] > a[j], b[i] < b[j])).sum()
return ndisordered / (n * (n - 1))
However, this requires $\displaystyle{ n^2 }$ memory, which is inefficient for large arrays.
Given two rankings $\displaystyle{ \tau_1,\tau_2 }$, it is possible to rename the items such that $\displaystyle{ \tau_1 = (1,2,3,...) }$. Then, the problem of computing the Kendall tau distance reduces to computing the number of inversions in $\displaystyle{ \tau_2 }$—the number of index pairs $\displaystyle{ i,j }$ such that $\displaystyle{ i\lt j }$ while $\displaystyle{ \tau_2(i) \gt \tau_2(j) }$. There are several algorithms for calculating this number.
• A simple algorithm based on merge sort requires time $\displaystyle{ O(n \log n) }$.[2]
• A more advanced algorithm requires time $\displaystyle{ O(n\sqrt{\log{n}}) }$.[3]
Here is a basic C implementation.
#include <stdbool.h>
int kendallTau(short x[], short y[], int len) {
int i, j, v = 0;
bool a, b;
for (i = 0; i < len; i++) {
for (j = i + 1; j < len; j++) {
a = x[i] < x[j] && y[i] > y[j];
b = x[i] > x[j] && y[i] < y[j];
if (a || b)
v++;
}
}
return abs(v);
}
float normalize(int kt, int len) {
return kt / (len * (len - 1) / 2.0);
} | {
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# True or False : If $f(x)$ and $f^{-1}(x)$ intersect at an even number of points , all points lie on $y=x$
Previously I have discussed about odd number of intersect points (See : If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?) Now , I want to know the even condition . For example $f(x) = \sqrt{x}$ and $f^{-1}(x) = x^2 , x\ge 0$ intersects each other in $(0,0)$ and $(1,1)$ points and these points located on $y=x$ line
Edit : Consider $f$ is continuous function.
• How would any intersection point not lie on $y=x$? The graphs of the function and its inverse (if it exists) is symmetric about $y=x$, so any intersection must be on $y=x$ Feb 23 '17 at 15:58
• @user160738 No , for example consider $f(x) = \frac{1}{x}$ and $f^{-1}(x) = \frac{1}{x}$ Feb 23 '17 at 16:00
• @user160738 another example is $f(x) = -x^3$ and $f^{-1}(x) = -\sqrt[3]{x}$ . They intersect each other in $(1 , -1)$ that is not on $y=x$ line. See : math.stackexchange.com/questions/1452098/… Feb 23 '17 at 16:08
The continuous case:
If the domain is not an interval, we can still find a counterexample. Define $$f(x)=\cases{2x+2 & if x\in(0,2),\cr x-3 & if x\in(3,5),}$$ then $$f^{-1}(x)=\cases{x/2-1 & if x\in(2,6),\cr x+3 & if x\in(0,2),}$$ and intersections are $(1,4)$, $(4,1)$.
However, if the domain is connected, it turns out to be true.
Proof of the theorem in the case of a connected domain (interval)
For contradiction, let us suppose that there is an even number of intersections and an intersection is out of diagonal. | {
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1. We assume that $f^{-1}$ exists which means that $f$ is injective. Moreover, $f$ is continuous and its domain is interval, so $f$ is increasing or decreasing.
2. $f$ and $f^{-1}$ are symmetric by diagonal, therefore the set of intersections is symmetric by the diagonal.
3. Let denote $(x_1, x_2)$ an intersection out of diagonal. By symmetry, there is an intersection $(x_2, x_1)$. We can therefore WLOG assume that $x_1<x_2$.
4. $f(x_1) > x_1$ and $f(x_2) < x_2$, so by continuity, $f$ intersects the diagonal.
5. There is an even number of intersections and by symmetry, there is an even number of intersections out of diagonal. So there is even number of intersections on the diagonal. So there are at least two of them: $f(x_3) = x_3$, $f(x_4) = x_4$
6. $x_1<x_2$ and $f(x_1) > f(x_2)$, so $f$ is decreasing (on the whole, by 1).
7. $x_3<x_4$ and $f(x_3) < f(x_4)$, so $f$ is increasing.
• Okay but it's true in many cases in connected domain . Can you provide an example that it's wrong in connected domain ? Mar 3 '17 at 10:45
• Okay so why it's true for many examples ? Mar 3 '17 at 15:14
• You proved the theorem is wrong .okay , now provide some examples that shows theorem is false . Mar 3 '17 at 15:38
• Oh , I'm really sorry let me to read again . Mar 3 '17 at 15:40
• Sorry, I don't understand step 4 in your purported proof. Mar 4 '17 at 10:45
Assume that $f$ is a continuous and invertible real function with a connected domain. Then $f$ is either strictly decreasing or strictly increasing over its domain (or it would assume some value twice, and would not be one-to-one). Consider these two cases: | {
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1. $f$ is decreasing. Then, since $f(x)-x$ is also decreasing, $f$ can't cross $y=x$ more than once. If it never crosses $y=x$, then it lies entirely above or below that line, and its inverse lies entirely on the other side; they never meet, and the theorem holds vacuously. If, on the other hand, it crosses $y=x$ exactly once, then the total number of intersections between $f$ and $f^{-1}$ is odd, since off-diagonal intersections come in pairs. The theorem holds in this case too.
2. $f$ is increasing. Then it cannot intersect $f^{-1}$ at any point off the line $y=x$. Suppose it did, at a pair of points $(x,y)$ and $(y,x)$ with $x<y$: then we would have $f(x)=y > x=f(y)$, contradicting the fact that $f$ is increasing. In this case, then, all intersections are on $y=x$, and the theorem holds.
Since the theorem holds whether $f$ is increasing or decreasing, it is true in general.
• For all decreasing functions we can say $f(x) - x$ is also decreasing ? Is it right ? Mar 9 '17 at 6:03
• Sure. A decreasing function minus an increasing function decreases even faster. Mar 9 '17 at 20:48
• Okay and how you can conclude that it intersects at most one time ? Mar 9 '17 at 20:55
Your question, as it currently stands definitely does not hold, as you do not specify the domain, continuity or conditions on inverse. This answer does define an $f$ on the whole real line that has a proper inverse.
Consider the function: \begin{align} f(x) = \begin{cases} 1 &\text{if x = 0}\\ 0 &\text{if x = 1}\\ x^2 &\text{if x > 0 and x\neq 1}\\ -2x &\text{if x < 0} \end{cases} \end{align} which has inverse \begin{align} f^{-1}(y) = \begin{cases} 1 &\text{if y = 0}\\ 0 &\text{if y = 1}\\ \sqrt{x} &\text{if x > 0 and x\neq 1}\\ -\tfrac{x}{2} &\text{if x < 0} \end{cases} \end{align} and so $f(x) = f^{-1}(y)$ only at $x = 0$ and $x = 1$, but at these points $f(x) \neq x$. | {
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• Consider $f$ is continuous Feb 23 '17 at 17:24
• I will have a think @S.H.W, I imagine the answer might be that such a function cannot exist: if they intersect at two points then they are equal on an interval (that is my hunch anyway, YMMV). Feb 23 '17 at 21:17
• I think this statement is true but I can't prove it. Feb 23 '17 at 21:32
• Actually, I don't think that is quite true: consider $f(x) = x^2$ for $x \geqslant 0$ and $f(x) = -2x$ for $x < 0$, then $f^{-1}(x) = \sqrt{x}$ for $x \geqslant 0$ and $f^{-1}(x) = -\tfrac{x}{2}$ for $x < 0$ and so they only intersect at $x = 0$ and $1$. Feb 23 '17 at 21:59
• if $f(x) = x^2$ then $f^{-1}(x) = \sqrt{x} , x \ge 0$ and the intersection points are $(0,0)$ and $(1,1)$ which they are in $y=x$ line Feb 23 '17 at 22:03
For $f$ and $f^{-1}$ to intersect there must be a point(s) $(x,f(x)) = (x,f^{-1}(x))$. Assume $x \neq f(x)$. Because $f$ and $f^{-1}$ are symmetric around y=x then there must be a corresponding point(s) $(f(x),x) = (f^{-1}(x),x)$.
The line defined by the points $(x_0,f(x_0)$ and $(f(x_),x_0)$ is $$(y-f(x_0) = \frac{f(x_0)-x_0}{x_0-f(x_0)}(x-x_0)$$ and simplified becomes $$y=-x+x_0+f(x_0)$$ This line, with a slope of -1, is not parallel with $y=x$, therefore must cross $y=x$. Because the two points are symmetric around $y=x$, they must lie on different sides of $y=x$. By the intermediate value theorem both $f$ and $f^{-1}$ must cross $y=x$.
Hence there is at least one point of $f$ on $y=x$ | {
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# A particular vanishing integral
While dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation
$$\int_0^1\log\left(\frac{(x+1)(x+2)}{x+3}\right)\frac{\mathrm dx}{1+x}~=~0\tag1$$
The proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similar substitution - $$x\mapsto\frac{1-x}{1+x}$$ - to the auxiliary integral $$\mathcal I$$ given by
$$\mathcal I~=~\int_0^1\log\left(\frac{x^2+2x+3}{(x+1)(x+2)}\right)\frac{\mathrm dx}{1+x}$$
And the result follows. However, to consider precisely this integral seems highly unnatural to me (in fact, as I mentioned earlier, this integral was just a by-product while evaluating something quite different and I discovered $$(1)$$ when experimenting with various substitutions).
The crucial point to notice concerning $$\mathcal I$$ is the invariance of the polynomial $$f(x)=x^2+2x+3$$ regarding the self-similar substitution which allows us to deduce $$(1)$$. Additionally for myself I am quite surprised by the special structure of $$(1)$$ since we have factors of the form $$(x+1)$$, $$(x+2)$$ and $$(x+3)$$ combined which calls for a generalization (although I found none yet).
It there a more elementary approach, not relying on such an "accident" like examining the integral $$\mathcal I$$ for proving $$(1)$$? Additionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated! | {
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• The $\displaystyle\log$ argument numerator -in the $\displaystyle\mathcal{I}$ definition- must be $\displaystyle x^{2} + 3x + 2$ instead of $\displaystyle x^{2} + 2x + 3$. – Felix Marin Aug 27 '20 at 3:27
• @FelixMarin No. By my calculations $\left(\frac{1-x}{1+x}\right)^2+2\left(\frac{1-x}{1+x}\right)+3=2\frac{x^2+2x+3}{(1+x)^2}$ which is what the approach is all about. But $\left(\frac{1-x}{1+x}\right)^2+3\left(\frac{1-x}{1+x}\right)+2=2\frac{x+3}{(x+1)^2}$ which yields an interesting other identity but not the one the post is concerned with. – mrtaurho Aug 27 '20 at 3:42
That's quite an impressive method to show that the integral vanishes.
For the first part I'll show using a different approach that your integral vanishes. $$\mathcal J=\int_0^1 \ln\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm dx}{x+1}\overset{x+1=t}= \color{blue}{\int_1^2\ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}-\color{red}{\int_1^2 \frac{\ln t}{t}\mathrm dt}$$ Let's denote the blue integral as $$\mathcal J_1$$ then using the substitution $$\frac{2}{t}\to t$$ we get: $$\mathcal J_1=\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}=\int_1^2 \ln\left(\frac{2(t+1)}{t+2}\right)\frac{\mathrm dt}{t}$$ Adding both integrals from above gives us: $$\require{cancel} 2\mathcal J_1=\cancel{\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}+\int_1^2 \frac{\ln 2}{t}\mathrm dt+\cancel{\int_1^2 \ln\left(\frac{t+1}{t+2}\right)\frac{\mathrm dt}{t}}=\ln^2 2$$ $$\Rightarrow \mathcal J_1=\color{blue}{\frac{\ln^2 2}{2}}\Rightarrow \mathcal J=\color{blue}{\frac{\ln^2 2}{2}}-\color{red}{\frac{\ln^2 2}{2}}=0$$
As for the second part, a small generalization outcomes by experimenting with the blue integral. | {
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As for the second part, a small generalization outcomes by experimenting with the blue integral.
In particular, by the same approach we have: $$\int_1^a \ln\left(\frac{x+a}{x+1}\right)\frac{\mathrm dx}{x}=\int_1^a \frac{\ln x}{x}\mathrm dx$$ Which gives us a small generalization: $${\int_0^{a-1}\ln\left(\frac{x+a+1}{(x+1)(x+2)}\right)\frac{\mathrm dx}{x+1}=0}$$ Similarly, (with the substitution $$\frac{ab}{x}\to x$$) we get that: $$\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)=\int_a^b \ln\left(\frac{x}{a}\right)\frac{dx}{x}$$ And the following follows: $$\int_{a-1}^{b-1} \ln\left(\frac{a(x+b+1)}{(x+1)(x+a+1)}\right)\frac{dx}{x+1}=0$$ One might be interested in the following similar generalization too: $$\int_1^{t}\ln\left(\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\right)\frac{dx}{x}=0,\quad s\in R, t>1$$ | {
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• This is what I was looking for; I guess^^ (+1) anyway and I'm looking forward to see a generalization (if possible). I guess you know where I got this integral from :D – mrtaurho Aug 12 '19 at 15:17
• @mrtaurho I got there a small generalization for now. However since $\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)$ it might be possible to obtain a better one. (I'll try later to work with it). – Zacky Aug 12 '19 at 15:38
• It's hard to write out your name now :P But yes, this seems promising, I'm curious! As I noted above it was a rather strange by-product to discover this equality; and it was tedious to ran into it three times while hoping for something more helpful for solving the original task^^' – mrtaurho Aug 12 '19 at 15:44
• @mrtaurho just in case you missed it in winter I'll mention that $\mathcal I$ (before the self-similar sub was applied) originates from this generalization: math.stackexchange.com/a/3049039/515527. Aka: $$\int_1^{\sqrt{t}}\ln\left(\frac{x^4+sx^2+t}{x(x^2+sx+t)}\right)\frac{dx}{x}=0$$ – Zacky Aug 12 '19 at 16:06
• As I've upvoted both, the question you linked and your answer, I guess I've seen it at some point. But I'll take a look at it again :) – mrtaurho Aug 12 '19 at 16:23 | {
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I have used the substitution $$(x+1)(y+1)=2$$ before to good effect because $$\int_0^1f(x)\,\frac{\mathrm{d}x}{x+1}=\int_0^1f\!\left(\tfrac{1-y}{1+y}\right)\frac{\mathrm{d}y}{y+1}\tag1$$ If $$f(x)=\log\left(\frac{x+3}{(x+2)(x+1)}\right)$$, then $$f\!\left(\frac{1-y}{1+y}\right)=\log\left(\frac{(y+2)(y+1)}{y+3}\right)$$. Therefore $$\int_0^1\log\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm{d}x}{x+1}=\int_0^1\log\left(\frac{(y+2)(y+1)}{y+3}\right)\frac{\mathrm{d}y}{y+1}\tag2$$ and since the two sides of $$(2)$$ are negatives, they are both $$0$$.
We can generalize $$(1)$$ by letting $$(x+a)(y+a)=a(1+a)$$, then we get $$\int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3$$ | {
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# Moderate Averages Solved QuestionAptitude Discussion
Q. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?
✔ A. 57 ✖ B. 56.8 ✖ C. 58.2 ✖ D. 52.2
Solution:
Option(A) is correct
Let the average weight of the 59 students be $A$.
Therefore, the total weight of the 59 of them will be 59 $A$.
The questions states that when the weight of this student who left is added, the total weight of the class
= $59A + 45$
When this student is also included, the average weight decreases by 0.2 kgs.
$59A + \dfrac{45}{60} = A - 0.2$
$\Rightarrow 59A + 45 = 60A - 12$
$\Rightarrow 45 + 12 = 60A - 59A$
$\Rightarrow A = \textbf{57}$
Edit: For an alternative solution, check comment by Durga.
## (9) Comment(s)
Siddharth Sharma
()
60*0.2+45=57 thats the shortest method to do it.
Shivakumar
()
divided by 60 for 59A+45 not only for 45
Vijay
()
simply,
60*(x-.20)-59x=45;
=>x=45+12=57.
Sanajit Ghosh
()
@DURGA's ans is right
Zoha Amjad
()
answer is 56.8
0.2 kgs increase means
$0.2*59=11.8$
$11.8+45 = 56.8$
Shanu
()
The ans should be 56.8
it can be solved as
$60A-((A-0.2)*59)=45$
so $A(\text{avarage}) = 56.8$
Kiruthi
()
ans should be 56.8 because in question they have given increased by 200g but while solving they took it as decreased by 200g so 57
Durga
()
Hi ,
The answer is 57 only.
Let say the average of 60 students is $x$.
Total weight $=60*x$
If one man left average should increase 200gr
i.e $\dfrac{200}{1000} =0.2 \text{ kg}$
So, $60 \times x-45=59(x+0.2)$
$60x-59x=59*0.2+45$
$\Rightarrow x=56.8 \text{ kg}$
But here we need to calculate the average of 59 students
i.e $56.8+0.2= \textbf{57 kg}$
Elmira
()
Thank you for creating a very informative website, much appreciate it. | {
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Elmira
()
Thank you for creating a very informative website, much appreciate it.
Please review the solution once again, when multiplying the sides by 60, 59A was left without effect, therefore I think the response is wrong. Oh maybe I am missing something here.
Please explain.
Elmira | {
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# Find the matrix representation of a linear map
Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear map and consider the matrix representation $$M_{B}^{B} =\begin{pmatrix} 4 & 6\\ 6 & 3 \end{pmatrix}$$ with respect to the basis $B=\bigg\{ \begin{pmatrix} 2 \\ 2 \end{pmatrix} , \begin{pmatrix} 3\\ 2 \end{pmatrix}\bigg\}$. Find the matrix representation of $f$ with respect to the canonical basis.
MY ATTEMPT: Let $b_1=\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, $b_2=\begin{pmatrix} 3\\ 2 \end{pmatrix}$ and $e_j$ the j-th vector of the canonical basis. Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix} \Longrightarrow f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) =4e_1 + 6e_2$
$f(b_2)=\begin{pmatrix} 6\\ 3 \end{pmatrix} \Longrightarrow f(3e_1 +2e_2)=3 f(e_1) +2f(e_2) =6e_1 + 3e_2$
Solving this system I obtain that $f(e_1)=2e_1 -3e_2=\begin{pmatrix} 2\\ -3 \end{pmatrix}$ and $f(e_2)= 6e_2=\begin{pmatrix} 0\\ 6 \end{pmatrix}$. So the matrix rappresentation is $M_{E}^{E}=\begin{pmatrix} 2 & 0\\ -3 & 6 \end{pmatrix}$.
BUT my theacher says that the solution is $M_{E}^{E}=\begin{pmatrix} 13 & 10\\ 0 & 0 \end{pmatrix}$. Where I am wrong?
$$f(2e_1 +2e_2)=2 f(e_1) +2f(e_2) \color{red}{=4e_1 + 6e_2}$$
This is where you are wrong, $4$ and $6$ are coordinates in the basis $B$, so there is one more step to get $e_1$ and $e_2$ into the game. Otherwise your approach is nice.
• How can I do that? – fcoulomb Jun 1 '18 at 18:47
• @Besh00 write $\color{red}{=4b_1+6b_2}$ instead of the red part, and then write both $b_1$ and $b_2$ in terms of $e_1$, $e_2$. – Arnaud Mortier Jun 1 '18 at 18:50
• Why $f(b_1)=4b_1+6b_2$? – fcoulomb Jun 1 '18 at 21:10
• You wrote yourself "Then I have $f(b_1)=\begin{pmatrix} 4\\ 6 \end{pmatrix}$". These coordinates are with respect to the basis $B$ since they are taken from a matrix wrt $B,B$. – Arnaud Mortier Jun 1 '18 at 21:14 | {
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# How many $4$-digit numbers that do not have $5$ and have $7$ in the hundreds position?
I encountered this problem:
How many $4$-digit numbers are there that do not have $5$ and have $7$ in the hundreds position? Digits cannot be repeated.
My thinking: first, let's count numbers that do not have $0$. It's going to be $7\cdot 6 \cdot 5$ arrangements. Now, let's count numbers that have $0$ in the tens position. We have two positions open so there are $7 \cdot 6$ arrangements. We get the same number of arrangements for numbers that end with $0$. The total number of arrangements is $7\cdot 6 \cdot 5+ 2\cdot 7 \cdot 6=7\cdot 7 \cdot 6=294$.
However, my friend argues that there are $8$ choices for the last digit, $7$ choices for the tens digits and $6$ choices for the first digit so the total number should be $8\cdot 7 \cdot 6=336$.
I am asking who is right not to prove my friend wrong but to find the truth. Thanks for listening!
294 is correct.
First digit (most significant), can not be 7,0,5, so 7 choices; 3rd digit, cannot be 7,5, and the one chosen for 1st, so 7 choices; 4th digit, 6 choices;
$7\times7\times6=294$
• Welcome to the site. Great first answer! – B. Mehta Apr 14 '18 at 23:17
You are right and your friend is wrong. If you choose the units digit and then choose the tens digit (and are forced to use $7$ for the hundreds digit) then the number of choices for the thousands digit not always $6$: it is either $5$ or $6$, depending on if you've used $0$ yet or not.
For example, there are $5$ numbers satisfying the condition which have the form $x789$ (since $x$ can be any of $1,2,3,4,6$) but $6$ numbers satisfying the condition which have the form $x780$ (since $x$ can be any of $1,2,3,4,6,9$).
All this tells us is that the number of solutions is between $8\cdot7\cdot5 = 280$ and $8\cdot7\cdot6 = 336$, and for a more precise answer we do need to look at whether $0$ gets used. | {
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I think the key disagreement here is that your friend is thinking that $0$ is allowed to be the first digit (say, for a 4 digit passcode), while you are not. If $0$ can be the first digit, your friend is correct, but if not (as other posters explain), you are correct.
You don't have to choice the positions in order so don't worry about $0$s.
Do the Thousandth place first. It can not be $0$ and it can not be $7$ or $5$. So there are $7$ options. Then do either the tens or the ones, it doesn't matter which. It can be $0$ but it can't be $7$ or $5$ and it can't be what the thousandth place was. That is $7$ options as well. And the final position can't be $7$ or $5$ or either of the previous two positions. so that is $6$ options.
So there are $7*7*6 = 294$.
You were correct but you didn't have to work so hard.
Your freind is wrong in that $0$ is not an option for the first digit. | {
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# Telescoping Series Partial Fractions | {
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Telescoping Series and Partial Fractions. Use n = 3, since we're after the 3rd partial sum. You may want to review that material before trying these problems. 003+cdots#. The first part is 2/1, so thats 2. Ask a question. Geometric and telescoping series The geometric series is X1 n=0 a nr n = a + ar + ar2 + ar3 + = a 1 r provided jrj<1 (when jrj 1 the series diverges). be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. Determine convergence, absolute convergence and divergence of infinite series using the standard convergence tests. The partial fraction decomposition will consist of one term for the factor and three terms for the factor. You don't see many telescoping series, but the telescoping series rule is a good one to keep in your bag of tricks — you never know when it might come in handy. Now, we need to be careful here. Could anyone explain in plain english whats going on here? And clarify their formula. In this case, P a n = s. BC Sequences and series 2015. 4 Comparison Tests. Telescoping series. • Define what a power series is and be able to find the radius and interval of convergence for a given series. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Induction. a) Using partial fractions, , where A — n2—1 n —1 n +1. (a) Let sN be the Nth partial sum, sN = PN n=2 1 n2−n. It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,1381. (MCMC 2009I#4) Find the value of the in nite product 7 9 26 28 63 65 = lim n!1 Yn k=2 k3 1 k3 + 1 : Solution. Find the limit of a given sequence. We recover Theorem 1 from Theorem 6 as the case a1 = 1, ai = bj = 0 for all i > 1,j ≥ 1. Say we | {
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We recover Theorem 1 from Theorem 6 as the case a1 = 1, ai = bj = 0 for all i > 1,j ≥ 1. Say we have something we want to sum up, let's call it a k. partial fractions to rewrite it as a telescoping series. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. 10-2-43 telescoping series. Can the series be compared favorably to one of the special types?. 1A1: The nth partial sum is defined as the sum of the first n terms of a sequence. Laval Kennesaw State University Abstract This hand out is a description of the technique known as telescoping sums, which is used when studying the convergence of some series. Indicate a good method for evaluating the integral. Contractive sequences converge. Make sure you can correctly answer questions involving telescoping series and partial sums. There is no test that will tell us that we've got a telescoping series right off the bat. Write out the first three partial sums of 2 3 4 n n 4 b. Thus a telescoping series will only converge if bn approaches a finite limit. Series and Partial Sums. We explain calculus and give you hundreds of practice problems, all with complete, worked out, step-by-step solutions. Here, (using partial fractions) Partial sum: Thus the series converges and has sum = 5. Partial sums Value of infinite series as limit of sequence of partial sums Convergence vs. More examples can be found on the Telescoping Series Examples 1 page. Topic 9: Calculus Option Series Part 1 A series consists of C v+C w+C x…qK S=z T St_ qK s=z S T st_ It is denoted by: For a total sum For a partial sum The Divergence Test states:. Telescoping Series. can the integral test, the root test, or the ratio test be applied? 4. 10-3-33 sine series that diverges by nth term test. For example, is a partial fractions decomposition of. Telescoping Series - Sum. Improper integrals of type 2. Try this telescoping sum. Math 222, Calculus and Analytic Geometry 2, | {
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Improper integrals of type 2. Try this telescoping sum. Math 222, Calculus and Analytic Geometry 2, 2013-14 Notes for how to do partial fractions updated; This is some notes about infinite series and telescoping. 4 – Divergence & Integrals -Use the divergence test to determine whether a given series diverges (or whether the test is inconclusive). ALTERNATING SERIES Does an =(−1)nbn or an =(−1)n−1bn, bn ≥ 0? NO Is bn+1 ≤ bn & lim n→∞ YES n =0? YES an Converges TELESCOPING SERIES Dosubsequent termscancel out previousterms in the sum? May have to use partial fractions, properties of logarithms, etc. 1, etc), then:. The Partial Fraction Decomposition Calculator an online tool which shows Partial Fraction Decomposition for the given input. diverges c. Once it has been determined that a given series converges, it is usually a much more difficult problem to actually determine the sum of the series. Partial Fractions Telescoping Series, and Harmonic Series in this section. It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,1381. We know how to sum all these geometric series, so we get. The comparison tests are used to determine convergence or divergence of series with positive terms. And if the limit of the partial sum is nite, then it converges, and we. In the cases where series cannot be reduced to a closed form expression an approximate answer could be obtained using definite integral calculator. To be able to do this, we will use the method of partial fractions to decompose the fraction that is common in some telescoping series. Find the power series of functions, determine their radius and interval of convergence,. It is capable of computing sums over finite, infinite (inf) and parametrized sequencies (n). (1) and ask whether the sum is convergent. First, note that the telescoping series method only works on certain fractions. — Try Before you Buy To start a Maplet, click on its name. Here’s an example | {
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on certain fractions. — Try Before you Buy To start a Maplet, click on its name. Here’s an example series: Just like for sequences, we want to make explicit what the term is. If you update to the most recent version of this activity, then your current progress on this activity will be erased. In general, the series given by. Question is : Determine whether the series is convergent or divergent by expressing Sn as a telescoping sum. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Telescoping Series This next series is a clever series, called a telescoping series. Find the power series of functions, determine their radius and interval of convergence,. Find the sum of the series • Â k=0 4 k2 +3k+2 if it exists. 12 is an example of a telescoping series. Download this MAT 1348 class note to get exam ready in less time! Class note uploaded on Oct 14, 2019. 3 Telescoping Series. Even after 1,000,000 1,000,000 terms, the partial sum is still relatively small. Chapter 2 Infinite Series Page 1 of 7 Chapter 2 : Infinite Series Section B Telescoping and Harmonic Series By the end of this section you will be able to • use partial fractions to test for convergence and determine the sum of a series. We try This is an example of a telescoping series. Example diverve Determine if the following series serjes or diverges. For example, 1 + 1/2 + 1/3 is a partial sum of the first three terms. Welcome to Maplets for Calculus. 2) Expand first 5 terms. partial fractions to rewrite it as a telescoping series. Maclaurin and Taylor Series g. A telescoping series. Don't quite understand how to do that. 1 An introduction to series and sequences 245 261; 10. -To determine whether or not this series converges and to find what value it converges to, we need to use the telescoping series. Multiplying by a Constant Property. In mathematics, a telescoping series is an informal expression referring to a series whose sum can | {
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In mathematics, a telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with either a succeeding or preceding term. be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. The second part as it goes to infinity is 0, so the answer would be 2? I'm really confused and looking up examples of telescoping series and it is not helping. ) Telescoping series: A telescoping series is a series of the form P∞ n=1 (a n+i −a ) for some integer i ≥ 1. Necessary conditions for convergence. Geometric series: Converges if jrj< 1, diverges if jrj 1. Convergence Tests d. One of these tests is the telescoping series test. In this video from PatricJMT we look at why when evaluating a telescoping series, one typically finds an expression for a partial sum, and then takes the limit of this partial sum. A telescoping series (or telescoping sum) is one that "expands" in such a way that most of its terms cancel away. Use the Integral Test to decide whether the in nite series X1 n=1 2n n2 + 1 dx. Regardless, your record of completion wil. 4 e) Use partial fraction decomposition to decompose n n nn S S S S nn f f. In mathematics, a telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with either a succeeding or preceding term. [Hint: Telescoping - Use partial fraction decomposition] Solution We recognize the series as a telescoping series, so we want to construct a formula for the Nth partial sum, S N, and nd its limit as N !1. I know the answer is 3/4 but I can't figure out how to get to that answer. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. 1674 terms are needed for the partial sum to exceed 8. p-Series 4. Telescoping series. For example, 1 + 1/2 + 1/3 is a partial sum of the first three terms. The | {
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4. Telescoping series. For example, 1 + 1/2 + 1/3 is a partial sum of the first three terms. The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences. Calculus 2 Syllabus. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. More examples can be found on the Telescoping Series Examples 2 page. Then a partial fraction decomposition of is so that (This summation is a telescoping sum. The first part is 2/1, so thats 2. Improper integrals of type 2. Differentiating and Integrating Power Series. Also see pages 5 through 9 of the Class Notes Day 26 on Partial Fractions. 4 Comparison Tests: Test #1-in class. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. • Determine if a series is absolutely convergent. and B b) Determine whether the following telescoping series is con- vergent or divergent. The series in Example 8. I used partial fractions to get ((1/3)/n - (1/3)/(n(n+3))). Download this MAT 1348 study guide to get exam ready in less time! Study guide uploaded on Oct 14, 2019. In many cases it is possible at least to determine whether or not the series converges, and so we will spend most of our time on this problem. Find a formula for the nth partial sum s nof the series, and use the formula for s nto determine whether the series converges. Worksheet 2 – Special Series (Telescoping) Name _____ Find the partial fraction decomposition of each of the following. Ask a question. Scope and Sequence Unit 1 - Techniques of Integration This unit includes the chain rule, u-substitution, expanding, separating the numerator, completing the | {
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unit includes the chain rule, u-substitution, expanding, separating the numerator, completing the square, dividing, adding and subtracting terms, trig identities, integration by parts, trig integrals, trig substitution, partial fractions, L'Hopital's Rule, improper integrals, and Euler's Method. can the integral test, the root test, or the ratio test be applied? 4. In this case it’s , so we can write the series as (or just if we’re not too concerned with where the series starts). Telescoping series are not very common in math-ematics but are interesting to study. From Wikipedia, the free encyclopedia. Math 296: Calculus II. State the de nition of geometric series. FIND THE SUM OF TELESCOPING SERIES? Use partial fraction decomposition to write 1/ you can replace the single fraction with the sum of fractions and you have. The Partial Fraction Decomposition Calculator an online tool which shows Partial Fraction Decomposition for the given input. For n = 1, the series is a harmonic series 1 2 + 1 3 + 1 4 + 1 5 + which is divergent, and the formula 1=(n 1) would indicate that the series should be divergent. Z 8x p (telescoping series) 12. 3 cont'd Integral test for series with non-negative terms cont'd and estimations, Estimations of the value of the series. f ¦ diverges. The number and variety of exercises where the student must determine the appropriate series test necessary to determine convergence of a given series has been increased. Partial fractions - Integrating rational functions is an extremely entertaining activity. Before looking for the partial fraction decomposition of the ra-. Series Geometric Senes Telescoping Series Limaçons and Lemniscates Parametric Equations—Second Derivatives and Tangent Lines Partial Fractions Convergence and Divergence Series Indexing Arithmetic of Series Integration by Parts Il Vector Functions Implicit Differentiation Il Infinite Limits of Integration Partial Fractions Ill p-Series. Infinite Series, Geometric Series, Telescoping | {
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Limits of Integration Partial Fractions Ill p-Series. Infinite Series, Geometric Series, Telescoping Series; Integral Test, p-series, and Estimates of Sums; The Comparison Tests; Alternating Series and Estimates of Sums; Absolute and Conditional Convergence, Ratio and Root Tests; Strategy for Testing Series, Summary of Convergence Tests; Power Series; Representation of Functions as a Power Series. EXAMPLE 8:. The simplest way would be to use partial fractions, and then convert this into a telescoping series. These types of series frequently arise from partial fraction decompositions and lead to very convenient and direct summation formulas. Begin by using partial fraction decomposition to obtain 2 n(n+ 2) = A n + B n+ 2 = 1 n 1. 10-2-43 telescoping series. The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series. Telescoping series: Compute the nth partial sum, sn, and take the limit of sn as n goes to 1. The series in Example I(b) is a telescoping series of the form — 192) + (b2 — 173) + (b 3 — 125) + Telescoping series Note that b2 is canceled by the second term, 173 is canceled by the third term, and so on. Assignment for Day 26 and Answers. Geometric Series + 107. Here's another example that uses partial fractions. High School AP Calculus BC Curriculum. Example 4: Determine the series 1 1 n nn( 1) f ¦ is convergent or divergent. Theorem 4. A more general technique Efthirniou's technique can be generalized to series of the form CrI1unvnwhere it is convenient to write only vnas a Laplace transform integral. 12 is an example of a telescoping series. (Telescoping series) Find. if it is convergent, find its sum. First, note that the telescoping series method only works on certain fractions. Partial Fractions. This free online course offers you some Integral definitions, Explanations and Examples. If Partial fractions gives 1 k(k +1). Telescoping series. o q bASl BlB Zr niVg8hnt osS 5r8ewsXenrZv Yecdj. 2 | {
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fractions gives 1 k(k +1). Telescoping series. o q bASl BlB Zr niVg8hnt osS 5r8ewsXenrZv Yecdj. 2 Series A series is a sum of sequential terms. 2) Trigonometric Substitution (7. Harmonic series. Next, prove an important convergence theorem. com allows you to find the sum of a series online. 10-3-23 constant times a divergent series. Expanding the sum yields Rearranging the brackets, we see that the terms in the infinite sum cancel in pairs, leaving only the first and lasts terms. Download this MAT 1348 class note to get exam ready in less time! Class note uploaded on Oct 14, 2019. The value of the series is lim I lim Sn 553 Related 47-58 19—34. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. MCrawford (20:20:34). The second part as it goes to infinity is 0, so the answer would be 2? I'm really confused and looking up examples of telescoping series and it is not helping. Series Calculator computes sum of a series over the given interval. Substitution. Convergence b. How this Calculus 2 course is set up to make complicated math easy: This approximately 200-lesson course includes video and text explanations of everything from Calculus 2, and it includes more than 275 quiz questions (with solutions!) to help you test your understanding along the way. Too bad the book didn't do a good job on teaching how to find nth partial sum formula. That is, s k → -∞. Geometric and Telescoping Series In this worksheet, you will calculate the exact values of several series. The voice explains how to first plug in the numbers given for each variable in the fractions. Telescoping Series 1 1 2 3 3 Show that the telescoping series ln dive rges. Telescoping series is a series where all terms cancel out except for the first and last one. Absolute convergence. A telescoping series is a special type of series whose terms cancel each out in such a way that it is relatively easy to determine the exact value of its partial sums. In algebra, the partial | {
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that it is relatively easy to determine the exact value of its partial sums. In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. Exercise 6. The method of partial fractions may reveal these. Your explanation that the series is telescoping after written as a partial fraction is more concise. An in nite series can be represented as such: P 1 n=1 a n. z T BMAapdPeB wwMitEhL lIQnkfoimnBi\tieE rPCrvecWavlccfuxlluKsx. Telescoping Series Test Series: ∑∞𝑛=1 Ὄ 𝑛+1− 𝑛Ὅ Condition of Convergence: lim 𝑛→∞ 𝑛=𝐿 Condition of Divergence: None NOTE: 1) May need to reformat with partial fraction expansion or log rules. The typical example of telescoping series (for partial fractions) is. There is no way to actually identify the series as a telescoping series at this point. Math - Calculus II SERIES ( Partial Sums ) infinite series GEOMETRIC SERIES Otherwise, the series diverges ( that is, the series has no sum ). Let’s decompose it, using partial fractions: [math]\frac{1. The Telescoping Series! This type of infinite series utilizes the technique of Partial Fractions which is a way for us to express a rational function (algebraic fraction) as a sum of simpler fractions. Could anyone explain in plain english whats going on here? And clarify their formula. This is the main origin of the name telescoping series. If convergent find the sum, and if divergent enter DIV: 11. This is an excerpt from my full length lesson. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. be able to recognize a telescoping series, determine whether it is cvg or div and if cvg, to what. You don't see many telescoping series, but the telescoping series rule is a | {
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div and if cvg, to what. You don't see many telescoping series, but the telescoping series rule is a good one to keep in your bag of tricks — you never know when it might come in handy. \ B jArlnlA Er^iOgqhEtcsn srhemsNeKrkvre_dM. partial fractions to rewrite it as a telescoping series. Here’s an example series: Just like for sequences, we want to make explicit what the term is. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. After several hours of thinking about these type of problems I found out that I can separate the 1/[(n+1)(n+2)] into partial fractions, and I found out that it is a telescoping series. Telescoping Series with Partial Fractions Mathispower4u. The solution in the above example uses the method of partial fractions to try to calculate the sum of a telescoping series. Then evaluate lim n!1 S n to obtain the value of the series, or state that it diverges. The following exercises test your understanding of infinite sequences and series. • Use improper integrals to reinforce the notions of infinite series. A telescoping series is one whose partial sums eventually only have a fixed number of terms after cancellation. The result is a simple formula for the nth term of the sequence of partial sums. Her practical approach makes math manageable for anyone who's learning math for the first time, returning to school after a long break, or for anyone with a genuine curiosity who wants to dive deeper into the material. 1B3: If a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms has the same value. Now we must solve for A and B. In particular, in order for the fractions to cancel out, we need the numerators to be the same. (Of course, an infinite geometric series is a special case of a Taylor series. Note: For an example of a telescoping sums question, see question #2 in the Additional Examples section below. Berndt also states that there is | {
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question, see question #2 in the Additional Examples section below. Berndt also states that there is no “natural” way to obtain an expansion of H n in powers of m. Telescoping series: Expand out the sums to find a grouping pattern that can simplify in order to take the limit of S n *Note: Some series can be turned into subtraction via partial fractions or by log rules ∑Eg: ∑. Infinite Series Definition. This is an example of a telescoping series. Partial Fractions Introduction to Partial Fractions Linear Factors Irreducible Quadratic Factors Improper Rational Functions and Long Division Summary Strategies of Integration Substitution Integration by Parts Trig Integrals Trig Substitutions Partial Fractions Improper Integrals Type 1 - Improper Integrals with Infinite Intervals of Integration. If instructions say to determine if a series converges or diverges AND find the sum if it converges, then it must be an infinite geometric series or a telescoping. Improper integrals of type 2. If convergent find the sum, and if divergent enter DIV: 11. Geometric and Telescoping Series c. Look at the partial sums: because of cancellation of adjacent terms. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. Geometric Series + 107. If an input is given then it can easily show the result for the given number. How to get the partial fractions of higher degree numerators. An in nite series can be represented as such: P 1 n=1 a n. The Sigma notation for summation of series. 3 The Integral Test: The section includes the integral test and a discussion on the remainder estimate for the integral test. Review for Test 3 Math 1552, Integral Calculus The series is telescoping, so it converges and we can find its sum using partial fractions: ". Say we have something we want to sum up, let's call it a k. Be sure to review the Telescoping Series page before continuing forward. Partial Fractions. For the | {
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Be sure to review the Telescoping Series page before continuing forward. Partial Fractions. For the telescoping series X1 k=3 2 (2k 1)(2k + 1); nd a formula for the nth term of the sequence of partial sums fS ng. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. (1) and ask whether the sum is convergent. ~~~~~ AP Cal BC Students: Telescoping Series and Partial Fractions Tests. A telescoping series (or telescoping sum) is one that "expands" in such a way that most of its terms cancel away. ) (Now evaluate the limit. 2: p 2] EXAMPLES Does the series converge or diverge? 1. Definitions: Let{ } 1 n n a. Retrieved from " https: The following series, for example, is not a telescoping series despite the fact that we can partial fraction the series terms. Alternating Series Test cannot be applied 26. In nite Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. TELESCOPING SERIES Now let us investigate the telescoping series. ) = 1 - 0 = 1. So: s1=a1 s2=a1+a2 s3=a1+a2+a3 s4=a1+a2+a3+a4. This is a challenging sub-section of algebra that requires the solver to look for patterns in a series of fractions and use lots of logical thinking. Improper integrals of type 2. This time there will be a few more terms that do not cancel. Lecture 17: Series (II) Telescoping Series A telescoping series is a special type of series for which many terms cancel in the nth partial sums. (The first Maplet may take a little longer to open because it needs to start Java. Back to Thomas O'Sullivan's Homepage. Creating the telescoping effect frequently involves a partial fraction decomposition. Telescoping series. 3 Telescoping Series. You - ProProfs Discuss. We will now look at some more examples of evaluating telescoping series. If you are looking for more in partial | {
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look at some more examples of evaluating telescoping series. If you are looking for more in partial fractions, do check in: Partial fractions of lower degree numerators. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Partial fractions Fractions in which the denominator has a quadratic term Sometimes we come across fractions in which the denominator has a quadratic term which. Be able to determine the nth partial sum of any geometric series. Partial Fractions Stewart Chapter 7. Test for convergence of a series using an appropriate test: divergence, integral, comparison and limit comparison, ratio, root, or alternating series. High School AP Calculus BC Curriculum. For telescoping series, intermediate terms 'drop. For example, using partial fractions and cancelling a bunch of terms, we find that; An infinite series that arises from Parseval’s theorem in Fourier analysis. Most telescopic series problems involve using the partial fraction decomposition before expanding it and seeing terms cancel out, so make sure you know that very well before tackling these questions. Calculus 2 Syllabus. Note that the fundamental concepts of functions, graphs, and limits, which are studied at the beginning of courses in differential calculus, are often first introduced in earlier classes (most notably intermediate algebra and precalculus). Also note that it is possible to tell that this last series. Infinite series 1: Geometric and telescoping series Main ideas. The final answer is: Note that we have converted an infinite sum problem to adding up a finite number of fractions. 26, 29, 39, 40 21 – 42 Solve applications involving series. Exercise 6. 3) Cancel duplicates. If you do not own Maple, click Use MapleNET 12 at the top-right corner of this page. Note: is a telescoping series. Thus, as $$k$$ increases, the partial sum $$S_k$$ increases (the series is a sum of positive terms), but is always smaller than \(2\text{. Find the sum of the | {
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(the series is a sum of positive terms), but is always smaller than \(2\text{. Find the sum of the series •  k=0 4 k2 +3k+2 if it exists. That is, s k → -∞. Hence, Therefore, by the definition of convergence for infinite series, the above telescopic series converges and is equal to 1. Find the sum of the series •  k=1 1 k 1 k+2 if it exists. With geometric series, we carried out the entire evaluation process by finding a formu for the sequence of partial sums and evaluating the limit of the sequence. Infinite Series Chapter 1: Sequences and series Section 4: Telescoping series Page 2 We can use the method of partial fractions to rewrite the general term as: 4 1 1 2 2 2 2 ab n n n n n n §· ¨¸ ©¹ For your practice, check that this decomposition is correct! Now we can write any partial sum of this series as: 11 4 1 1 2 22 kk nn n n n n. This video is a great one on learning about evaluating fractions. Telescoping: Transform by partial fractions Procedure for Determining Convergence No Series Diverges nth-Term Test Yes or maybe Yes No Yes Nonnegative terms and/or absolute convergence No Yes No Can the Integral Test, the Ratio Test, or the Root Test be applied? Does the Integral Test apply? Yes Does the Ratio Test apply? No Yes Is Is Does the. (a) X1 n=3 1 2n 1 1 2n 4 As I stated in my email follow-up to this problem, this is not actually an easy series to nd a. Now pop in the first term (a 1) and the common ratio (r). The telescoping Series is a method for examining the convervence of infinite series of the form: This method, combined with partial fraction decomposition, is frequently effective. Be able to determine the nth partial sum of any geometric series. (a) X1 n=1 cos 1 n cos 1 n+ 1. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as: #3. Besides finding the sum of a | {
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find a fraction equivalent to any given repeating decimal such as: #3. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. According to Ferraro [4], Mengoli did so by making frequent use of the following. — Try Before you Buy To start a Maplet, click on its name. if integral of series reaches ∞ or DNE then series diverges. Telescoping Series and Partial Fractions. In this case, the series is convergent and the sum isS lim n→ sn b1 −lim n→ bn 1. See if you can figure it out. Welcome to Maplets for Calculus. 4 Comparison Tests. Can the series be compared favorably to one of the special types?. EXAMPLE 8:. 50 46 – 53 Evaluate telescoping series. telescoping series to find out. This is best demonstrated with an example. I work through an example of proving that a series converges and finding the sum of the series using Partial Fractions to create a Telescoping Series. Now let us investigate the telescoping series. (a)Sequences and Series, partial sums. Example: Determine whether the given series converge. Use comparison test if numerator fluctuates between two constants. From Wikipedia, the free encyclopedia. | {
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# Introduction
Although we have functions that can compute our $\gcd$easily it's important enough that we need to give and study an algorithm for it: the euclidean algorithm.
It's extended version will help us calculate modular inverses which we will define a bit later.
# Euclidean Algorithm
Important Remark
If $a = b \cdot q + r$and $d = \gcd(a, b)$ then $d | r$. Therefore $\gcd(a, b) = \gcd(b, r)$
## Algorithm
We write the following:
$a = q_0 \cdot b + r_0 \\ b = q_1 \cdot r_0 + r_1 \\ r_0 = q_2 \cdot r_1 + r_2 \\ \vdots \\ r_{n-2} = r_ {n-1} \cdot q_{n - 1} + r_n \\ r_n = 0$
Or iteratively $r_{k-2} = q_k \cdot r_{k-1} + r_k$ until we find a $0$. Then we stop
Now here's the trick:
$\gcd(a, b) = gcd(b, r_0) = gcd(r_0, r_1) = \dots = \gcd(r_{n-2}, r_{n-1}) = r_{n-1} = d$
If $d = \gcd(a, b)$ then $d$ divides $r_0, r_1, ... r_{n-1}$
Pause and ponder. Make you you understand why that works.
Example:
Calculate $\gcd(24, 15)$
$24 = 1 \cdot 15 + 9 \\ 15 = 1 \cdot 9 + 6 \\ 9 = 1 \cdot 6 + 3 \\ 6 = 2 \cdot 3 + 0 \Rightarrow 3 = \gcd(24, 15)$
Code
def my_gcd(a, b): # If a < b swap them if a < b: a, b = b, a # If we encounter 0 return a if b == 0: return a else: r = a % b return my_gcd(b, r)print(my_gcd(24, 15))# 3
Exercises:
1. Pick 2 numbers and calculate their $\gcd$by hand.
2. Implement the algorithm in Python / Sage and play with it. Do not copy paste the code
# Extended Euclidean Algorithm
This section needs to be expanded a bit.
Bezout's identity
Let $d = \gcd(a, b)$. Then there exists $u, v$ such that $au + bv = d$
The extended euclidean algorithm aims to find $d = \gcd(a, b), \text{ and }u, v$given $a, b$
# In sage we have the xgcd functiona = 24b = 15g, u, v = xgcd(a, b)print(g, u, v)# 3 2 -3 print(u * a + v * b)# 3 -> because 24 * 2 - 15 * 3 = 48 - 45 = 3 | {
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# 3.1 Functions and function notation (Page 4/21)
Page 4 / 21
$n$ 1 2 3 4 5 $Q$ 8 6 7 6 8
[link] displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.
Age in years, (input) 5 5 6 7 8 9 10 Height in inches, (output) 40 42 44 47 50 52 54
Given a table of input and output values, determine whether the table represents a function.
1. Identify the input and output values.
2. Check to see if each input value is paired with only one output value. If so, the table represents a function.
## Identifying tables that represent functions
Input Output
2 1
5 3
8 6
Input Output
–3 5
0 1
4 5
Input Output
1 0
5 2
5 4
[link] and [link] define functions. In both, each input value corresponds to exactly one output value. [link] does not define a function because the input value of 5 corresponds to two different output values.
When a table represents a function, corresponding input and output values can also be specified using function notation.
The function represented by [link] can be represented by writing
Similarly, the statements
represent the function in [link] .
[link] cannot be expressed in a similar way because it does not represent a function.
Input Output
1 10
2 100
3 1000
yes
## Finding input and output values of a function
When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. | {
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When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.
## Evaluation of functions in algebraic forms
When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function $\text{\hspace{0.17em}}f\left(x\right)=5-3{x}^{2}\text{\hspace{0.17em}}$ can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.
Given the formula for a function, evaluate.
1. Replace the input variable in the formula with the value provided.
2. Calculate the result.
## Evaluating functions at specific values
Evaluate $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+3x-4\text{\hspace{0.17em}}$ at
1. $2$
2. $a$
3. $a+h$
4. $\frac{f\left(a+h\right)-f\left(a\right)}{h}$
Replace the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the function with each specified value.
1. Because the input value is a number, 2, we can use simple algebra to simplify.
$\begin{array}{ccc}\hfill f\left(2\right)& =& {2}^{2}+3\left(2\right)-4\\ & =& 4+6-4\hfill \\ & =& 6\hfill \end{array}$
2. In this case, the input value is a letter so we cannot simplify the answer any further.
$f\left(a\right)={a}^{2}+3a-4$
3. With an input value of $\text{\hspace{0.17em}}a+h,\text{\hspace{0.17em}}$ we must use the distributive property.
$\begin{array}{ccc}\hfill f\left(a+h\right)& =& {\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ & =& {a}^{2}+2ah+{h}^{2}+3a+3h-4\hfill \end{array}$
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
$f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h-4$
and we know that
$f\left(a\right)={a}^{2}+3a-4$
Now we combine the results and simplify. | {
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and we know that
$f\left(a\right)={a}^{2}+3a-4$
Now we combine the results and simplify.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 | {
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# Fibonacci identity: $f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$
Consider this Fibonacci equation: $$f_{n+1}^2 - f_nf_{n+2}$$
The problem asked to write a program with given n, output the the result of this equation. I could use the formula $$f_n = \frac{(1+\sqrt{5})^n - ( 1 - \sqrt{5} )^n}{2^n\sqrt{5}}$$
However, from mathworld, I found this formula Cassini's identity $$f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$$
So, I decided to play around with the equation above, and I have:
$$\text{Let } x = n + 1$$ $$\text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1}$$ $$\Rightarrow -( f_{x-1}f_{x+1} - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}$$
So this equation either is 1 or -1. Am I in the right track?
Thanks,
Chan
-
Yes. $\mbox{}$ – Mariano Suárez-Alvarez Feb 8 '11 at 5:44
@Mariano Suárez-Alvarez: Thanks ^_^! Happy! – Chan Feb 8 '11 at 5:47
In your first expression, should the subscript on the square be "n+1" instead of "n-1"? If so, then you are on the right track. (Also, you'll want parentheses around your "-1"s.) – Blue Feb 8 '11 at 5:47
Note: You should write $(-1)^{n+2}$, and not $-1^{n+2}$; exponentiation takes precedence, so the former is either $-1$ or $1$ depending on the parity of $n$, but the latter is $-1^{n+2} = -(1^{n+2}) = -1$ always. – Arturo Magidin Feb 8 '11 at 5:50
@Day Late Don: nice catch. It was my typo. I was practicing using Latex:(. – Chan Feb 8 '11 at 5:51
We have the following (easily proved by induction):
$\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \\ \end{pmatrix}$
Equating the determinants of the matrices gives us the identity immediately.
-
NOTE $\$ This is precisely the same answer I gave a couple months ago in this duplicate thread – Bill Dubuque Feb 28 '11 at 4:06 | {
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Let us try to find gcd of $F_n$ and $F_{n+1}$ using Extended Euclidean algorithm. I will write the steps algorithm in a table; this table method was also explained in some Bill Dubuque's posts.
$\begin{array}{|l||c|c|} \hline F_{n+1} & 1 & 0 \\\hline F_{n} & 0 & 1 \\\hline F_{n-1} & 1 & -1 \\\hline F_{n-2} & -1 & 2 \\\hline F_{n-3} & 2 & -3 \\\hline \vdots & \vdots & \vdots \\\hline F_{n-k} & (-1)^{k+1}F_k & (-1)^kF_{k+1} \\\hline \end{array}$
After a few steps we can guess the $k$-th line, which gives us the following formula:
$F_{n-k}=(-1)^{k+1}F_kF_{n+1}+(-1)^kF_{k+1}F_n=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)$.
For $k=n-1$ we get Cassini's identity $F_1=(-1)^n(F_{n-1}F_{n+1}-F_n^2)$.
So the only thing we have to do is to verify the above formula, which can be done easily by induction on $k$.
Inductive step: We know that:
$F_{n-k}=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)=-(-1)^{k}(F_kF_{n+1}-F_{k+1}F_n)$
$F_{n-(k-1)}=(-1)^{k}(F_{k-1}F_{n+1}-F_{k}F_n)$
Since $F_{n-(k+1)}=F_{n-(k-1)}-F_{n-k}$, we get
$F_{n-(k+1)}=(-1)^{k}[(F_{k-1}+F_k)F_{n+1}-(F_k+F_{k+1})F_n]=(-1)^{k+2}(F_{k+1}F_{n+1}-F_{k+2}F_n)$
which completes the inductive step.
-
I tried to google a little, and I found almost the same derivation of Cassini's identity in the text Yiu: Recreational Mathematics. – Martin Sleziak Mar 16 '12 at 11:09 | {
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# Math Help - series
1. ## series
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
1)
$\displaystyle\huge\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}$
$=\displaystyle\huge\frac{n+1}{(n+1)n!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$
Therefore
$\displaystyle\huge\frac{1}{2!}+\frac{2}{3!}+......$
$=1-\displaystyle\huge\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+.......-\frac{1}{(n+1)!}$
$=1-\displaystyle\huge\frac{1}{(n+1)!}$
3. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman
Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2)
$\displaystyle\huge\frac{1}{(n+2)n!}=\frac{(n+1)}{( n+2)(n+1)n!}=\frac{n+1}{(n+2)!}$
$\displaystyle\huge\frac{n+2-1}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}$
$\displaystyle\huge\frac{n+2}{(n+2)(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$
Use this to form the telescope.
4. Thanks very much for the kind help.
May I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.
Lastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?
5. If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things: | {
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- an obvious way to apply partial fractions decomposition
- add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.
That is usually a good start. There is a lot of intuition involved in some examples and less in others.
6. Thanks very much for the reply.
Can I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.
How to apply telescope method in question 3?
I cannot wrap around my head to get started in problem3
Thanks
7. 3)
The sum can be written as...
$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \sum_{k=1}^{n} \frac{1}{k+5}$ (1)
In order to solve the second term of (1) we have to introduce the function digamma, defined as...
$\displaystyle \psi(x)= \sum_{j=1}^{\infty} \frac{x}{j\ (j+x)} - \gamma$ (2)
... where $\gamma$ is the 'Euler's constant'. From (2) You can derive the following basic identity...
$\displaystyle \psi(1+x) - \psi (x) = \frac{1}{1+x}$ (3)
Now setting in (3) $x=k+4$ You obtain...
$\displaystyle \psi(k+5) - \psi (k+4) = \frac{1}{k+5}$ (4)
... so that is...
$\displaystyle \sum_{k=1}^{n} \frac{1}{k+5} = \sum_{k=1}^{n} \{ \psi(k+5) - \psi (k+4) \} = \psi (n+5) - \psi(5)$ (5)
... and finally You can write...
$\displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \psi(n+5) + 5\ \psi(5)$ (6)
Digamma function - Wikipedia, the free encyclopedia
Kind regards
$\chi$ $\sigma$
8. That's pretty cool, chisigma, but I think there is an even simpler answer.
$\displaystyle\sum_{k=1}^n \frac{k}{k+5} = \sum_{k=1}^n \frac{k+5-5}{k+5} = \sum_{k=1}^n \left(\frac{k+5}{k+5}-\frac{5}{k+5}\right)$
$\displaystyle = \left(\sum_{k=1}^n 1\right)-5\left(\sum_{k=1}^n \frac{1}{k+5}\right) = n - 5\sum_{k=6}^{n+5} \frac{1}{k+5} = n - 5\left(H_n - \frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) = \frac{137}{12}+n-5H_n$ | {
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where [LaTeX ERROR: Convert failed] is the n-th Harmonic number. There is no simpler way to write this sum. | {
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What is the probability that the max of $n$ numbers from some distribution is a greater than another number from the same distribution?
Independently choose $n$ numbers $X_1, X_2, \ldots, X_n$ from some distribution. And Independently choose another number $X_{n+1}$ from the same distribution. Let $P(\max(X_1, X_2, \ldots,,X_n) > X_{n+1}))$ means the probability that the maximum of $X_1, X_2, \ldots, X_n$ is bigger than $X_{n+1}$. What is $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1}))$? I think that $P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = \frac{n}{n+1}$. Here is my proof.
The case that any one of $X_1, X_2, \ldots X_n, X_{n+1}$ is the maximum is symmetric. So $P(X_i\text{ is the maximum)} = \frac{1}{n+1}, (i = 1,\ldots,n+1)$.
$P(\max(X_1, X_2, \ldots,X_n) > X_{n+1})) = P(X_i \text{ is the maximum}, i =1,\ldots,n)= \sum_{i=1}^{n}P(X_i \text{ is the maximum}) = n \cdot \frac{1}{n+1} = \frac{n}{n+1}$
But I think that my proof is not rigorous. https://math.stackexchange.com/a/20841/16625 gives a rigorous proof for the $n=1$ case. Can anyone give a rigorous proof for the general case?
• You should get $\frac{n}{n+1}$. $\frac{1}{n+1}$ is the probability that $\max(X_i)<X_{n+1}$. – Thomas Andrews Jan 20 '16 at 3:53
• This only works for continuous random variables, of course. For instance, if the $X_i$ are die rolls, you will get a very different result. – Thomas Andrews Jan 20 '16 at 3:58
• @ThomasAndrews Yes. For discrete random variables, the case that random variables are equal can not be ignored. – Jingguo Yao Jan 21 '16 at 7:21
Probably easier to compute the probability that the max is less. Specifically, $P(\max(X_i)<x)=P(X<x)^n$. So if $p$ is the CDF for $X$ then the probability is:
$$\int p(x)^n\,dp(x) = \frac{1}{n+1}p(x)^{n+1}\Bigg{|}_{-\infty}^{\infty}=\frac{1}{n+1}$$ | {
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$$\int p(x)^n\,dp(x) = \frac{1}{n+1}p(x)^{n+1}\Bigg{|}_{-\infty}^{\infty}=\frac{1}{n+1}$$
This is sort of obvious - the probability that the last one is the largest is the same as the probability that any other is the largest. (If the pdf is discontinuous, it is possible, with non-zero probability, for two to be equal, of course. So we assume continuous.)
So the probability that you were seeking is $\frac{n}{n+1}$.
Here is a slightly more formal argument. Suppose $X_1, X_2, \ldots,$ are independent and identically distributed random variables. Then the maximum order statistic $X_{(n)}$ of a sample of size $n$ has CDF $$F_{X_{(n)}}(x) = \Pr[X_{(n)} \le x] = \prod_{i=1}^n \Pr[X_i \le x] = F_X(x)^n.$$ Suppose we are now interested in the random variable $Y = X_{(n)} - X_{n+1}$, and in particular, $$\Pr[Y > 0] = 1 - \Pr[Y \le 0] = 1- \int_{x = -\infty}^\infty F_{X_{(n)}}(x) f_X(x) \, dx = 1 - \int_{x = -\infty}^\infty F_X(x)^n f_X(x) \, dx.$$ With the substitution $$u = F_X(x), \quad du = f_X(x) \, dx, \quad F_X(-\infty) = 0, \quad F_X(\infty) = 1,$$ we arrive at $$\Pr[Y > 0] = 1 - \int_{u=0}^1 u^n \, du = 1 - \frac{1}{1+n} = \frac{n}{n+1},$$ as claimed.
Assuming that the $P(X_i=X_j)=0$ for $i\ne j$, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})$$ is the probability that $X_{n+1}$ is not the greatest of the $n+1$ samples. The probability that it is the greatest of $n+1$ samples is $$P(\max(X_1,X_2,\dots,X_n)\le X_{n+1})=\frac{n!}{(n+1)!}=\frac1{n+1}$$ because there are $n!$ arrangements where $X_{n+1}$ is greatest and $(n+1)!$ arrangements altogether.
Therefore, $$P(\max(X_1,X_2,\dots,X_n)\gt X_{n+1})=1-\frac1{n+1}=\frac n{n+1}$$
Your argument is correct if the $n+1$ observations are independent, an assumption that you did not state. | {
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## DEV Community is a community of 662,276 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
# Daily Challenge #230 - Beeramid
dev.to staff
The hardworking team behind dev.to ❤️
Let's pretend your company just hired your friend from college and paid you a referral bonus. Awesome! To celebrate, you're taking your team out to the terrible dive bar next door and using the referral bonus to buy, and build, the largest three-dimensional beer can pyramid you can. And then probably drink those beers, because let's pretend it's Friday too.
A beer can pyramid will square the number of cans in each level - 1 can in the top level, 4 in the second, 9 in the next, 16, 25...
Complete the beeramid function to return the number of complete levels of a beer can pyramid you can make, given the parameters of:
2) the price of a beer can
For example:
beeramid(1500, 2); // should === 12
beeramid(5000, 3); // should === 16
Tests:
beeramid(9, 2)
beeramid(21, 1.5)
beeramid(-1, 4)
This challenge comes from kylehill on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!
Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!
## Discussion (6)
Here's my C solution,
Notice that there is no need for any iteration because (taking number of cans = N and number of levels = L) | {
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$N = 1 + 2^2 + 3^2 + ... + L^2 = L(L+1)(2L+1)/6 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L^3/3 < N < (L+1)^3/3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L < (3N)^{1/3} < L+1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \newline \implies L = \lfloor (3N)^{1/3} \rfloor \quad or \quad L = \lfloor (3N)^{1/3} \rfloor - 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$
int beeramid(double bonus, double price) {
if (bonus < price) return 0;
int cans = floor(bonus / price);
int levels = floor(cbrt(3.0*cans));
return levels*(levels + 1)*(2*levels + 1) > 6*cans ? levels - 1 : levels;
}
Vidit Sarkar
Here is a Python Solution,
def beeramid(bonus, price):
canCount = int(bonus/price)
level = 0
while((level+1)**2 <= canCount):
level += 1
canCount -= level*level
return level
Test Cases,
print(beeramid(1500, 2)) # output -> 12
print(beeramid(5000, 3)) # output -> 16
print(beeramid(9, 2)) # output -> 1
print(beeramid(21, 1.5)) # output -> 3
print(beeramid(-1, 4)) # output -> 0
Michael Kohl • Edited
Ruby, using a lazy, infinite enumerator:
ROWS = (1..).lazy.map { |n| n * n }
def beeramid(bonus, price)
cans = bonus / price
ROWS.take_while { |n| (cans -= n) >= 0 }.count
end
Valts Liepiņš • Edited
Haskell solution using lazy, infinite list of cans:
import Data.List (findIndex)
beeramid :: Double -> Double -> Int
beeramid m p =
case findIndex (> cans) levels of
Just level -> level
Nothing -> 0
where
cans = m / p
levels = scanl1 (+) . fmap (^2) \$ [1..]
Paula Gearon • Edited
Clojure. Not very terse, but it gets a solution in constant time
(defn faul [n]
(let [n2 (* n n)
n3 (* n2 n)]
(/ (+ n3 n3 n2 n2 n2 n) 6))) | {
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} |
(defn faul [n]
(let [n2 (* n n)
n3 (* n2 n)]
(/ (+ n3 n3 n2 n2 n2 n) 6)))
(defn layers [n]
(let [est (int (Math/pow (* 3 n) (/ 1 3)))]
(first (filter #(>= n (faul %)) (range est 0 -1)))))
(defn beeramid [bonus price] (or (layers (/ bonus price)) 0))
Testing:
=> (beeramid 1500 2)
12
=> (beeramid 5000 3)
16
=> (beeramid 9 2)
1
=> (beeramid 21 1.5)
3
=> (beeramid -1 4)
0
Paula Gearon
To follow up on my solution, I borrowed from the Wikipedia article on Square Pyramidal Numbers. This provides a formula for the sum of squares as a special case of Faulhaber's formula:
$P_n = \sum_{k=1}^n k^2 = \frac {2n^3 + 3n^2 + n} 6$
So if we have n layers, then that will have a total of Pn cups.
As n becomes large then: 2n3 ≫ 3n2 + n
$P_n = \frac {2n^3} 6 + \Big( \frac {3n^2 + n} 6 \Big) \newline P_n > \frac {n^3} 3 \newline$
Since the layers is n and the cups in the pyramid is Pn then this gives a bound of:
$layers < \sqrt[3]{3 \cdot cups}$
Consequently, my code creates an estimate of the number of layers using this formula, and counts down from there applying the Faulhaber formula to check if the result is less than or equal to the provided number of cups.
Using this method, a small numbers of cups will require 2 guesses, but as the number of cups gets higher (and n3n2) then it is more likely to get it on the first guess. I told it to keep testing while the number of layers until 1, but it never actually gets beyond 2 attempts. | {
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# Math Help - definite integral
1. ## definite integral
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
2. Originally Posted by JGaraffa
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
Well, again, use substitution $z := x^3+5$, which gives that $dz=3x^2\,dx$ and hence you have
$\int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$
3. Originally Posted by JGaraffa
a little confused on this one
"solve the definite integral"
$\int^2_0 2x^2e^{x^3+5}$
What exactly are you confused about? Have you made an attempt at it?
Also the integral should be $\int^2_0 2x^2e^{x^3+5}dx$
I could tell you how to do it but i think i it's only fair that an attempt is made first
Hint: use substitution
4. ah perfect! i couldn't figure out how i could change that $2x^2$ to a $3x^2$ . thanks!
5. actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$
$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$
when x = 0, u = 5
when x = 2, u = 13
$\frac{2}{3} \int^{13}_5 e^u du$
$
\frac{2}{3} e^{13} - \frac{2}{3} e^5$
i'm getting 294,843.32 which makes me think i'm doing something wrong.
6. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$
$\frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$
when x = 0, u = 5
when x = 2, u = 13
$\frac{2}{3} \int^{13}_5 e^u du$
$
\frac{2}{3} e^{13} - \frac{2}{3} e^5$
i'm getting 294,843.32 which makes me think i'm doing something wrong.
No, that result is ok. Of course, $e^{x^3+5}$ is a fast growing function.
7. Originally Posted by JGaraffa
actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.
$\int^2_0 2x^2e^{x^3+5}$
u= $x^3 + 5$
h(u) = $3x^2$ | {
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"url": "http://mathhelpforum.com/calculus/143022-definite-integral.html"
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