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$1 = -1$
which is obviously wrong but I can't see where the error is
EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
Cosine has a zero every Pi units 2pi is too much see my above post.
9. Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.
You are right it only works for odd n.
10. Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.
yeah, I accounted for that when I gave the general solution as
$2x = \pm\frac{\pi}{2} + 2n\pi$
starting at $\frac{\pi}{2}$ and going up by $2\pi$ each time gives $\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...$
Starting at $-\frac{\pi}{2}$ and going up by $2\pi$ each time gives us $\frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, ...$
So all the zeros are covered by these two sets of solutions
(obviously n could be negative and we get zeros going the other way as well)
But how do I know which ones will hold true for the original equation (when I divide them by 2 to get x on its own)?
11. Originally Posted by TheEmptySet
You are really close you should have gotten
$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
So can any one explain why this only works for odd n? This problems is driving me up the wall!!
Regards,
Stonehambey | {
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# Need to determine function based on graph
1. Oct 18, 2014
### TheExibo
There is this graph: http://i.imgur.com/BdwhN3b.jpg where the equation of the function must be determined. Down below is some thinking I've done, where I've found that the oblique asymptote is -0.5x+2, and the v-asymptote=6. I know how to solve something similar with 2 x-int's, but I'm having difficulty understanding what to do with 3 x-int's. Can anyone help me please? Thanks!
2. Oct 19, 2014
### zoki85
plugin 3 values of (x,f(x)) from the graph of the eq. in order to get system of equations and solve it to find a,b,c.
Last edited: Oct 19, 2014
3. Oct 19, 2014
### NTW
I believe I'm saying the same as zoki85, but I'll try to elaborate a bit more...
It seems that you have come to the conclusion that your function has the form
y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)
In my opinion, and to find a, b, c you should take three values of x, y from the drawing, reasonably chosen among the left and right branches. For the left branch, one point. For the right branch, and always IMHO, one for the maximum, and another to the right, away from the maximum.
Thus, you'll have a 3 x 3 system and solve it for the coefficients a, b, c...
4. Oct 19, 2014
### TheExibo
I got it using only two of the intercepts: (-40,0) and (0.5,0) and putting them into -0.2x+50+(ax+b)/-(x+6)2
Graphing it in desmos.com, it shows the exact same graph. I must explain how I got this function, so how can I explain the third x-intercept? Is it true that I only need two points, one on each side of the asymptote, to find the exact function of the graph that is shown?
5. Oct 19, 2014
### NTW
I now see that you needed only to find two coefficients, a and b. I saw the numerator in x2, but it's true that I misquoted the denominator in your notes, writing (y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)
6. Oct 19, 2014
### TheExibo | {
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6. Oct 19, 2014
### TheExibo
But by doing that, I've come across the problem where I can't explain how the graph ended up with all 3 correct x-intercepts, when I only used the two. Anyone have any ideas?
7. Oct 20, 2014
### ehild
You also used the asymptote at infinity where the function behaves as 50-0.2x.
ehild
8. Oct 20, 2014
### tycoon515
Being that your graph has an oblique asymptote, an improper rational function should be the kind of function that comes to mind. By the looks of the graph, as we let $x$ tend toward either $\infty$ or $-\infty$, the function gets closer and closer in value to a linear function, so we should consider an improper rational function whose numerator is $1$ greater in degree than its denominator. Also, a rational function is equal to $0$ wherever its numerator, but not its denominator, is equal to $0$. Since you have three $x$-intercepts, the numerator should be a polynomial that is the product of three factors, each one corresponding to a different $x$-intercept. A preliminary construction might look like the following:
$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{?}$
The numerator, when expanded, appears to be a polynomial whose dominant term is $x^3$. Therefore, we should consider a denominator that is a polynomial of degree $2$. Also, the graph is undefined and has a vertical asymptote at only one value of $x$: $-6$. Notice also that the graph of $f(x)$ tends toward $-\infty$ as $x$ approaches $-6$ from either side. Therefore, the denominator should be a quadratic with at least one $(x+6)$ factor. To find the other factor, observe either that there is no sign change of the function across the vertical asymptote at $x=-6$, or that for any factor other than $(x+6)$ to be present in the denominator, the graph would have to have another vertical asymptote that isn't at $x=-6$, so the denominator clearly has repeated $(x+6)$ factors. Our construction now takes the following form: | {
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$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{(x+6)^2}=k\frac{(x+40)(\frac{2x-1}{2})(\frac{5x-1387}{5})}{(x+6)^2}=k\frac{(x+40)(2x-1)(5x-1387)}{10(x+6)^2}$
$=k\frac{(2x^2+79x-40)(5x-1387)}{10(x^2+12x+36)}=k\frac{10x^3-2379x^2-109773x+55480}{10x^2+600x+1800}$
A function of this form is EXACTLY equal to $0$ at all three of the intercepts shown on the graph provided. And, if we let $k=-\frac{1}{5}$, it has an oblique asymptote of $-\frac{1}{5}x+\frac{2379}{50}$, which has the same slope as the asymptote on the graph but NOT the same $y$-intercept. Also, the equation you came up with has the correct slope and $y$-intercept of the oblique asymptote, but it is NOT EXACTLY equal to $0$ when $x=277.4$, so one of two things must be going on here: either the $x$-intercepts displayed on the graph are slightly off (rounded), or the actual equation of the oblique asymptote is slightly different than $-\frac{1}{5}x+50$.
9. Oct 20, 2014
### TheExibo
Awesome! Thank you to everyone for the help! | {
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# Does the intersection of two finite index subgroups have finite index?
Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.
Can somebody help me out?
• Step 1: Show that $H \cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H \cap K][H\cap K : H]$. – Clive Newstead Apr 5 '12 at 22:41
• @MartQ. You don't have to show that it's a divisor. It suffices to show that $[G\colon(H\cap K)]\leq [G\colon H]\cdot[G\colon K].$ You can try to find an injection from a certain set to another. – user23211 Apr 5 '12 at 23:02
• A related question. – user23211 Apr 5 '12 at 23:06
• @CliveNewstead The formula $[G:H] = [G:H \cap K][H\cap K : H]$ seems incorrect to me. What is $[H\cap K : H]?$ Did you mean $[G:H\cap K]=[G:H][H:H\cap K]?$ – user23211 Apr 5 '12 at 23:23
• @ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment. – Clive Newstead Apr 7 '12 at 11:45
Proof $1$: $\quad[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:H\cap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $H\cap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/H\times G/K$. The latter is finite so the orbit of $H\times K$ is finite, and the stabilizer of it is simply $H\cap K$. Invoke orbit-stabilizer. | {
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• There's probably a thread in which your second equality is proved, too. I'll go digging. – Dylan Moreland Jul 14 '12 at 23:10
• – Dylan Moreland Jul 14 '12 at 23:15
• The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of $\{gK \mid g \in HK\}$. But then $g=hk$ and so any element of $\{gK \mid g \in HK\}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense? – Al Jebr Sep 23 '18 at 21:08
• @AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $h\in H$. – anon Sep 24 '18 at 0:29
Let $\{H_1,\dots,H_m\}$ be the left cosets of $H$, and let $\{K_1,\dots,K_n\}$ be the left cosets of $K$. For each $x\in G$ there are unique $h(x)\in\{1,\dots,m\}$ and $k(x)\in\{1,\dots,n\}$ such that $x\in H_{h(x)}$ and $x\in K_{k(x)}$. Let $p(x)=\langle h(x),k(x)\rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $H\cap K$, then $p(x)\ne p(y)$.
It follows immediately that $H\cap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $H\cap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}y\in H$.
• Please, professor, could you clarify what does the $p(x)$ mean? – Danilo Gregorin Afonso Sep 4 '17 at 1:20
• @Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values. – Shahab Sep 5 '17 at 15:32
• @Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else. – Danilo Gregorin Afonso Sep 5 '17 at 19:52 | {
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Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 \subset H$ and $N_2 \subset K$. Then $G/N_1 \times G/N_2$ is a finite group; do you see why this implies that $N_1 \cap N_2$ has finite index in $G$?
• I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal. – anon Jul 14 '12 at 22:59
• @anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind. – Dylan Moreland Jul 14 '12 at 23:03
$H, K$ be subgroups of $G$. Any $A \in \frac{G}{H \cap K}$ can be written as $A = B \cap C$, where $B \in G/H$ and $C \in G/K$ as follows. For any $g \in G$ $$g(H\cap K) = gH \cap gK$$.
Hence, $$[G:H \cap K] \leq [G:H] [G:K]$$.
Let $C$ be the set of left cosets of $S = H \cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C \to C_1 \times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| \leq |C_1| \cdot |C_2|$, which proves that $[G: S]$ is finite.
Let $$l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l \in G$$ and $$G$$ be partitioned as $$G=h_1H \cup ... \cup h_lH = k_1K \cup ... \cup k_mK$$
I will find $$a_1,...a_n \in G$$ s.t. $$G$$ is partitioned as $$G=a_1(H \cap K) \cup ... \cup a_n(H \cap K)$$ which means $$n=[G:H \cap K] < \infty$$. | {
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Let $$b_1 \in G$$. Then $$\exists i_1 \in \{1,...,m\}, j_i \in \{1,...,l\}$$ s.t. $$b_1 \in h_{i_1}H \cap k_{j_1}K=b_1H \cap b_1K = b_1(H \cap K)$$. Next, let $$b_2 \in G \ \setminus \ b_1(H \cap K)$$. Then $$b_2 \in b_2(H \cap K)$$ where $$b_2(H \cap K) \cap b_1(H \cap K) = \emptyset$$ because $$b_1H \cap b_2H = h_{i_1}H \cap h_{i_2}H = \emptyset = k_{j_1}K \cap k_{j_2}K = b_1K \cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 \in \{1,...,m\} \ \setminus \ \{i_1\}, j_2 \in \{1,...,l\} \ \setminus \ \{j_1\}$$ This process continues at most $$lm$$ times for $$a_p=b_p, p \in \{1,2,...,n\}$$ Thus, $$n \le lm < \infty.$$
I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $$H$$ and $$K$$ are finite index subgroups of $$G$$, we can consider the right cosets $$G/H$$ and this is finite by hypothesis. In particular $$G$$ acts (transitively) on $$G/H$$ by right multiplication, i.e. $$G/H \times G \to G, \qquad (Hx, g) \mapsto Hxg.$$ If we now restrict the action to $$K$$, and consider the $$K$$-orbit of $$H$$, call it $$\mathcal{O}_K(H)$$, then it must be finite since it's contained in $$G/H$$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality $$|\mathcal{O}_K(H)| = [K : \text{Stab}_K(H)],$$ where $$\text{Stab}_K(H)$$ is the stabiliser of $$H$$ under the $$K$$-action. Notice that $$\text{Stab}_K(H) = K \cap \text{Stab}_G(H)$$, and what's the stabiliser of $$H$$ for the $$G$$-action? But of course $$H$$ itself, so $$\text{Stab}_K(H) = K \cap H$$, and finally we conclude that $$[K : K \cap H]$$ is finite. Hence $$[G : H \cap K] = [G : K][K : K \cap H]$$ must be finite as well. | {
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# How to graph the equation: $y=\frac {x-2}{x+1}$?
the title says it all. I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?
"Graph the equation $y=\frac {x-2}{x+1}$"
I know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.
Any help would be appreciated. If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!
• Do you mean to show it? – Ofir Attia May 20 '13 at 8:36
• I just need to graph the equation, but an explanation on how to do it would be nice too :D – missiledragon May 20 '13 at 8:36
• $y$ can indeed equal 0, i.e, when $x = 2$. ;) – user49685 May 20 '13 at 8:38
• OH, I didn't realise that :P. Thanks @user49685 – missiledragon May 20 '13 at 8:40
When you want draw the graph of the function, you need to check a couple of things:
1. Check if there are any points in the domain where the function is undefined, you know there's something happening there. For instance here you saw that when $x=-1$ it is undefined.
2. Then, $y$ can be equal to $0$, there's no problem with that, and you should find all the $x$ such that $y=0$. In this case, when $x=2, y=0$. Thus you know that the function goes through the point $(2,0)$.
3. Now you have all the points where there's something special happening, you can check where the function is positive, and where it is negative. You need to look at all the intervals between the special points. In your case : $(-\infty,-1),\,(-1,2),\,(2,\infty)$. Take $(-\infty,-1)$ for example. On this interval, you know that the numerator is always negative and the denominator is always negative as well, so the value for $y\in(-\infty,-1)$ will always be positive. Do that for all the intervals. | {
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4. Finally, you need to check some limits to see what happens at those special points and also when you tend to infinity (in your case only $-1$ is a special point because you already now that the value of your function when $x=2$ is $0$). So you should check :
$$\lim_{x\rightarrow\infty}f(x),\; \lim_{x\rightarrow-\infty}f(x),\; \lim_{x\rightarrow-1^+}f(x),\; \lim_{x\rightarrow-1^-}f(x),\;$$
That's it, you have everything you need to graph your function! (If you want to go into further details you can also check the derivative for extremas and the second derivative for inflexion points. This means that you have to compute the derivatives and check for which values of $x$ they are equal to $0$).
• Thanks! that helped me a lot :D! – missiledragon May 20 '13 at 9:06
Rewrite as $y=1-3/(x+1)$, or $(y-1)= -3/(x+1)$
This is simply the regular recriprocal equation, with the axies rescaled. The standard $Y$ axis, representing $X=0$, is now marked $x=-1$. That is, the x-numbers are moved one to the right. The old X axis of 0,1,2,3 become -1, 0, 1, 2...
The old $X$ axis, or $Y=0$ becomes $y=1$, and we have to rescale by a factor of 3. So the old Y coordinates 0, 1, 2, 3... become 1, -2, -5, -8...
And that's it.
Full Function investigation, check of asymptotic (infinite and non setting points), extreme points and inflection points. nice tool to work with is desmos:
$y=(x−2)/(x+1)$ - Click to view
In addition : Investigating Functions - Click to view
hope its helped you. | {
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In addition : Investigating Functions - Click to view
hope its helped you.
Just to add to what Oliver and Wendy have said, To Draw any function:
1.)Check Domain and critical points.
2.)Check intercepts.By putting $x=0 ,y=0$
3.)Check Asymptodes around infinity and critical points by taking $$\lim_{x\rightarrow\infty}f(x),\; \lim_{x\rightarrow-\infty}f(x),\; \lim_{x\rightarrow-1^+}f(x),\; \lim_{x\rightarrow-1^-}f(x),\;$$ 4.)Check Symmetry about axis by putting $x = -x$ and $y=-y$
5.)Intervals of increase/ decrease by equating first derrivative to zero.
If you know all this information you can draw any graph manually.
Hope this helps :) | {
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# Big O Notation “is element of” or “is equal”
People are always having trouble with "big $O$" notation when it comes to how to write it down in a mathematically correct way.
Example: you have two functions $n\mapsto f(n) = n^3$ and $n\mapsto g(n) = n^2$
Obviously $f$ is asymptotically faster than $g$. Is it $f(n) = O (g(n))$ or is it $f(n) \in O(g(n))$?
My prof says that the first one is wrong but is a very common practice, therefore it is used very offten in books. Although the second one is the right one.
Why is that so?
• Formally, neither are. If you define $O(\cdot)$ as a set of functions (for asymptotics w.r.t. a given point), say $O(g) = \{ f : \exists N,C>0,\ \forall n\geq N\ f(n) \leq C\cdot g(n)\}$, then you would have to write $f\in O(g)$. $f(n)$ is not a function, it's a number. But most people are quite happy with the (slight) abuse of notation: $f(n) = O(g(n))$ for $f\in O(g)$, because there is no ambiguity anyway. – Clement C. Dec 20 '16 at 15:13
• Ah, because f(n) would be a specific value when f is meant generally, is that right? – Blnpwr Dec 20 '16 at 15:14
• Indeed. The same way that, technically, $f(x)$ is the value of the function $f$ when evaluated at the point $x$, not the function $f$ itself. – Clement C. Dec 20 '16 at 15:15
• For this particular case, you have in fact $g(n) = o(f(n))$ or $f(n) = \omega(g(n))$ because $\lim_n \frac{f}{g} = \infty$ – Alex Dec 20 '16 at 15:48
• For your example functions it's the other way around. $g(n)=O(f(n))$. I'm assuming we're talking about $n\to\infty$, which is hinted by the use of the letter $n$. – Meni Rosenfeld Dec 21 '16 at 9:48
I really like Wikipedia's note on this: | {
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I really like Wikipedia's note on this:
The statement “$f(x)$ is $O(g(x))$” […] is usually written as $f(x) = O(g(x))$. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading as it suggests a symmetry that this statement does not have. As de Bruijn says, $O(x) = O(x^2)$ is true but $O(x^2) = O(x)$ is not. Knuth describes such statements as “one-way equalities”, since if the sides could be reversed, “we could deduce ridiculous things like $n = n^2$ from the identities $n = O(n^2)$ and $n^2 = O(n^2)$.”
For these reasons, it would be more precise to use set notation and write $f(x) \in O(g(x))$, thinking of $O(g(x))$ as the class of all functions $h(x)$ such that $|h(x)| \leq C|g(x)|$ for some constant $C$. However, the use of the equals sign is customary. Knuth pointed out that “mathematicians customarily use the $=$ sign as they use the word ‘is’ in English: Aristotle is a man, but a man isn't necessarily Aristotle.”
• Gotta love an answer backed up by Knuth, even if it is by a quote inside a quote. – Jasper Dec 20 '16 at 17:30
• This still isn't a complete answer. Using $f(x)$, $g(x)$, $h(x)$, $n$, $n^2$ etc. to mean functions is also abuse of notation. – JiK Dec 21 '16 at 15:07
• @JiK you're right, but it is a common abuse of notation, and in my opinion the crux of the question regards whether to use $=$ or $\in$, not whether or not to be completely formal every time you talk about a function – TomGrubb Dec 21 '16 at 15:15
• I don't agree with the statement would be more precise to use set notation. It implora that messure theory lenguage almost every point, cotient spaces, etc. are wrong lenguage. I think viewing big O / little o notation in this way is a bery powerful tool and os pretty precise. – Ernesto Iglesias Apr 23 '18 at 16:40 | {
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Using $\in$ is set-theoretically correct but inconvenient. For example, $$\sin x = x - \frac{x^3}{3} + \mathrm O(x^5)$$ In this case $\mathrm O$ should be interpreted as there exists an $\mathrm O(x^5)$ function to make this equality valid. The $=$ notation also allows asymptotic notation to appear on both sides and do arithmetic: $$e^x + \mathrm O(x) = \mathrm O(e^x)$$ In this case, for every $\mathrm O(x)$ function on the left, there exists an $\mathrm O(e^x)$ function on the right to make this an equality.
Warning: In this case the two sides of $=$ cannot be swapped carelessly. e.g. $\mathrm O(x) = \mathrm O(e^x)$ but $\mathrm O(e^x) \neq \mathrm O(x)$. | {
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• "but inconvenient" Indeed! I would go as far as saying that using it almost completely defeats the very purpose of the notation, rendering it pointless to introduce it in the first place. – quid Dec 20 '16 at 20:02
• For what it's worth, there is a way to reconcile this. The way I think about it is that using $\mathcal O(x^5)$ here changes the abuse of notation somewhat: we're now dealing with sets of functions, so the function $x$ becomes the singleton set $\{ x \mapsto x \}$ (and likewise for $x^3 / 3$), the addition operator becomes setwise product-addition $A \oplus B = \{ x \mapsto f(x) + g(x) : f \in A, g \in B \}$, $\mathcal O(x^5)$ really is a set, and the whole statement should be read as $(x \mapsto \sin x) \in {\text{RHS}}$. Perhaps slightly inconvenient, but I don't see anything unsound. – wchargin Dec 21 '16 at 8:00
• This reminds me of equality of multifunctions, like if we say $\cos^{-1} 1 = 2k\pi$ – GFauxPas Dec 21 '16 at 14:24
• @wchargin Using $O(x^5)$ does not automatically mean we are dealing with sets of functions. All that the writer means is that they're dealing with a specific quantity which satisfies the definition of being $O(x^5)$. The purpose of notation is to communicate what a mathematician is thinking. Introducing sets of functions—frankly, no one cares about the set of all functions that are $O(x^5)$ and it's an uninteresting set; we only care that the particular function we're working with is $O(x^5)$—is backwards, saying we should change our thinking to fit (someone's narrow idea of) $=$ notation. – ShreevatsaR Dec 21 '16 at 20:42
• @SheevastaR ¡¡¡Of course we care for the set $O(g)$!!! if you don't is a problem you have. Read the answer you are commenting and you will notice how important it is. – Ernesto Iglesias Apr 23 '18 at 16:49 | {
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$O(g(x))$ is a class of functions - think of it as a "property" functions can have. By the literal interpretation of the equals sign, "$f(x) = O(g(x))$" should be interpreted as "$f$ is literally equal to a certain class of functions." But functions and classes of functions are different sorts of things - even if this was what we meant to say, it's like saying that one particular apple is equal to a basket of apples. But what we mean when we say "$f(x) = O(g(x))$" is that $f$ belongs to the class of functions $O(g(x))$ - so, $f(x) \in O(g(x))$.
The reason we use $=$ instead of $\in$ is because, given the particular uses of big-$O$ notation (and little-$o$ notation, if you're familiar with that) $=$ is massively more convenient. We say things like $x^3 + O(x) = O(x^3)$, for example; we don't mean that $O(x)$ is an object that can actually be added to $x^3$, or that when that addition is done we actually get the class of functions $O(x^3)$, we just mean that for any function $f \in O(x)$, the function $x^3 + f(x)$ is a member of $O(x^3)$. But if I wanted to write that out in more standard notation, I'd have to say something like $\{x^3 + f(x) \mid f(x) \in O(x)\} \subseteq O(x^3)$. This is inconvenient to write and difficult to read, so we prefer the "slicker" notation $x^3 + O(x) = O(x^3)$.
However, I'm not sure I would say that sentences like $f(x) = O(g(x))$ are wrong. By convention, they're perfectly right - it's just that when an expression includes $O$ (or $o$), $=$ does not mean what it usually means. That's okay. | {
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• I think there must be someone that had invented a convenient way to just say $x^3 + O(x) = O(x^3)$, and at the same time, the symbols are entirely rigorous. – Eric Dec 20 '16 at 16:29
• @Eric Well, $x^3 + O(x) = O(x^3)$ is an entirely rigorous and convenient way to say that. It just doesn't look rigorous if you don't know the definition of "$=$" in the context of $O$. Since everyone in the field understands this notation, no one's particularly motivated to find a new way of writing it that won't confuse outsiders. – Reese Dec 20 '16 at 16:34
• @Reese If you treat $O(x)$ as a set, then $x^3 + O(x) \subseteq O(x^3)$ is rigorous without domain-specific $=$ sign. – heinrich5991 Dec 21 '16 at 1:02
• @heinrich5991 But that requires domain-specific $+$. – Reese Dec 21 '16 at 1:34
• @heinrich5991 Oh, absolutely, it's common. But so is using $=$ for $O$ and $o$. My point is that it's still not transparent to first learners. – Reese Dec 22 '16 at 17:29
I'll try to clarify wchargin comment.
If you are familiar with Group theory you should not find the $=$ sign awkward. Class of equivalence and quotient spaces usually use the sum notation for doing that.
e.g., $\mathbb{R}/\mathbb{Z}$ elements are usually written as $a+\mathbb{Z}$.
The problem is that big $O$ notation makes implicit use of that, i.e. in the class of $$O(g)=\lbrace f:\exists M,C>0, \forall x_0\geq M: f(x_0)\leq Cg(x_0) \rbrace,$$ as exposed by Clement C.
In this way, $$f=O(g)$$ Means $$f+O(g)=0+O(g)$$ And that is to say than $f\in O(g)$. | {
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In this way, $$f=O(g)$$ Means $$f+O(g)=0+O(g)$$ And that is to say than $f\in O(g)$.
Regarding $=$ being more useful/convenient than $\in$ because of allowing things like $x^3 + O(x) = O(x^3)$, it seems like for that you could use $x^3 + O(x) \subseteq O(x^3)$. This seems more precise to me, because it would mean "every element of the set $x^3 + O(x)$ is an element of the set $O(x^3)$." which is, I think, exactly what one wants to say. (where $g+A:=\{g+f:f\in A\}$ and when $g$ is a function and $A$ is a set of functions, and similarly for similar situations.)
Making the statement with subset is mentioned by Reese's answer, but I think that defining the sum of a function and a set of functions, and also the sum of sets of functions, in the straightforward way removes the inconveniences of saying things like $\{x^3+f(x)∣f∈O(x)\}\subseteq O(x^3)$. (Defining things similarly with multiplication, and application of functions in general.)
When things are defined this way, it should become as simple as using $\in$ instead of $=$ when there is a single function on the left, and using $\subseteq$ instead of $=$ when there is a set of functions on the left, and it seems more precise. | {
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• No, set-theoretic inclusion is not "exactly what one wants to say". The purpose of notation is to aid communication, and it is meant to fit our way of thinking. It is misguided to insist that one should change one's (very productive) way of thinking just to fit some notation or formalism. When an analyst writes $\sin x = x - \frac{x^3}{3} + \mathrm O(x^5)$ they are literally thinking that $\sin x$ is equal to $x$ minus $\frac{x^3}{3}$ plus some function that is $\mathrm O(x^5)$ — they are thinking of equality, not set inclusion, so the notation reflects that. – ShreevatsaR Dec 21 '16 at 18:54
• Ok, I should probably defer to you here. But isn't saying that $sin(x)$ is equal to $x - \frac{x^3}{3}$ plus some function that is $O(x^5)$ the same thing as saying that $sin(x)$ is an element of the set of functions which are of the form $x - \frac{x^3}{3}$ plus some function that is $O(x^5)$? – drocta Dec 21 '16 at 20:31
• Well, is saying "the sky is blue" the same thing as saying "the sky is an element of the set of things that are blue"? In some sense yes, and in another sense no. (It may be logically equivalent, but the speaker may not have even considered, or care about, the set of all things that are blue.) One can also say "the sky satisfies the property of being blue", or even "blueness inheres in the sky". There can be many ways to formalize a statement, and the most fruitful ones are often the one that closest match what the human is thinking, rather than what fits cleanest with other formalisms. – ShreevatsaR Dec 21 '16 at 20:45 | {
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• (sorry, I hit enter by mistake. First time using SE comments.) Your point about the purpose being for communication seems like something I failed to consider. But $\in$ being like "has property" and $\subseteq$ being like "all things with property on the left have the property on the right$, seem like they capture the$\forall$or$\exists$in what is being said with "there exists a function of the form$x-x^3/3+O(x^5)$which is$sine(x)$, and "for all functions of the form <blah>, there is a function of the form <bleh> such that they are equal". But maybe it's not nice enough to justify sets? – drocta Dec 21 '16 at 20:57 Here is my five pennies worth: If we see a statement of the form $$f(x)=g(x)+O\bigl(p(x)\bigr)\qquad(x\to\xi)$$ then for each$x$the exact value of the term$O\bigl(p(x)\bigr)$is defined by this very equation: it is$:=f(x)-g(x)$. In addition we are told that there is a constant$C>0$such that for all$x$in a suitable punctured neighborhood of$\xi$this difference is$\leq C\bigl|p(x)\bigr|\$. | {
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# Finding the sin inverse
Well, so I was finding the value of $\sin 15^{\circ}$. I used the identity $2 \sin x \cos x = \sin 2x$. So solving the quadratic equation and picking the right values simply gave me:
$$\sin 15^{\circ} = \frac {\sqrt {2 - \sqrt {3}}} {2}$$
That was pretty much simple. But I thought what if I was given the value instead, and I was asked to find the angle, whose sine would give me that value.
In short, my question is how to evaluate:
$$\sin^{-1} \left( \frac {\sqrt {2 - \sqrt {3}}} {2} \right)$$
I want to solve it thinking that I do not know the value of $\sin 15$, as if I have been just provided with the problem, and I have no idea what the answer might be. I tried to proceed by converting the $\sin$ to $\tan$ but that seemed to be of no use. Can anybody help?
• $$15=45-30=60-45$$ – lab bhattacharjee Feb 23 '18 at 18:44
• You are doing that because you know that the answer is 15. How is someone supposed to know that just on seeing the problem? – Manish Kundu Feb 23 '18 at 18:48
• Please find my answer – lab bhattacharjee Feb 24 '18 at 10:46
In general it's hard, if not impossible. In this case, it turns out to be doable.
The main idea is that if $\theta = \sin^{-1}\left( \frac{\sqrt{2 - \sqrt 3}}{2} \right)$, then $\sin \theta = \frac{\sqrt{2 - \sqrt 3}}{2}$, and we have an equation we can (hopefully) solve for $\theta$.
Squaring, we get $$\sin^2 \theta = \frac{2 - \sqrt{3}}{4},$$ and multiplying by 4, $$4\sin^2 \theta = 2 - \sqrt{3}.$$
My thoughts are that maybe if I get $\sqrt{3}$ by itself and square, I'll get a nice equation that's polynomial in $\sin \theta$ with integer coefficients. So, we get $\sqrt{3}$ by itself: $$\sqrt{3} = 2 - 4 \sin^2 \theta.$$ But that's interesting: I can factor 2 out... $$\sqrt{3} = 2(1 - 2 \sin^2 \theta),$$ | {
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and that $1 - 2 \sin^2 \theta$ looks pretty good, because that's the output from a double angle identity, namely $\cos(2 \theta)$. Now, $$\sqrt{3} = 2 \cos(2 \theta),$$ so $\cos (2\theta) = \frac{\sqrt{3}}{2}$.
This happens when $2 \theta = 30^\circ$ or $2\theta = 330^\circ$.
So, we get one possibility that $\theta = 15^\circ$ (since we've solved an equation by squaring, though, we should make sure it's not extraneous!).
And of course, once we know one angle $\theta$ with a particular sine, we can use symmetry and the unit circle to find all other such angles (that would be $180^\circ - \theta = 180^\circ - 15^\circ = 165^\circ$ in this case, and everything coterminal with the two between $0^\circ$ and $360^\circ$). If we just want the inverse sine of something positive, it's an angle in quadrant I.
• Sidenote, running off to class now, won't be able to respond for a few hours (if it matters) – pjs36 Feb 23 '18 at 19:07
• That was really amazing. Didn't expect it to be solved so simply. – Manish Kundu Feb 23 '18 at 19:23
• @ManishKundu Thanks! I was surprised too; it's a really nice question, and I'm happy you thought to ask it. – pjs36 Feb 24 '18 at 2:26
$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$
$$\implies\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}$$ as $\sqrt3-1>0$
$$=\sin60^\circ\cos45^\circ-\sin45^\circ\cos60^\circ$$
$$=\sin(60-45)^\circ$$
Now we know the principal value of $\sin^{-1}x$ lies in $\in[-90^\circ,90^\circ]$ for real $x$
• You made it simple. That was nice. – Manish Kundu Feb 24 '18 at 15:41
+1 to pjs36's answer, and they did bring up a good point that in this case it was possible to solve. But what if we weren't able to use the double angle identity? Perhaps another way to go would be the use of MacLaurin series.
Given a problem $\sin^{-1}{A} =\theta$, where $\theta$ is in radians and $A \in [-1,1]$, we can first take the $\sin$ of both sides to get $$\sin{\theta} = A$$ | {
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The MacLaurin series for $\sin{\theta}$ is $$\sin\theta=\sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!} - \dots =A$$
The farther out you extend the polynomial, the better the estimate is going to be when you solve for $\theta$. However it will be only an estimate when you do it this way, and you would probably need to use a calculator or WolframAlpha or something to solve it as well, which probably defeats the purpose of avoiding just plugging in $\sin^{-1} A$ into the calculator in the first place. This is merely another way you could look at the problem, if you wanted to avoid using inverse trig.
• So I need to solve for $\theta$. Wait, so do I need to use Newton Raphson? – Manish Kundu Feb 23 '18 at 19:49
• Correct, we want to solve for $\theta$. However, this would be equivalent to solving the roots for a large polynomial. Newton Raphson could work, as the derivative of a polynomial is simple to find, but it may take alot of time still to do by hand – WaveX Feb 23 '18 at 19:52
• We would also have the case that we would be estimating an estimate. – WaveX Feb 23 '18 at 19:59
• I see now why solving such problems can be so hard in general. – Manish Kundu Feb 23 '18 at 20:00
For a shortcut evaluation of $x$ such that $\sin^{-1}(x) =k$, assuming that $x$ is "small", you could use the simplest Padé approximant $$\sin^{-1}(x)\approx \frac{x}{1-\frac{x^2}{6}}$$ leading to $$x\approx \frac{\sqrt{6 k^2+9}-3}{k}$$
For your case, where $k= \frac {\sqrt {2 - \sqrt {3}}} {2}$, this would give $$x\approx \sqrt{2+\sqrt{3}} \left(\sqrt{48-6 \sqrt{3}}-6\right)\approx 0.255992$$ in radians (that is to say $\approx 14.6673$ in degrees).
• That was helpful. Thanks. – Manish Kundu Feb 24 '18 at 7:10
Not an answer, just an explicit formula for the $$\arcsin$$ function. | {
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Not an answer, just an explicit formula for the $$\arcsin$$ function.
First recall that $$\sin x=\frac{e^{2ix}-1}{2ie^{ix}}$$ So if we want the function $$y=\arcsin x$$ we must solve the following equation for $$y$$: $$x=\frac{e^{2iy}-1}{2ie^{iy}}$$ if we set $$u=e^{iy}$$ this becomes much easier: $$x=\frac{u^2-1}{2iu}$$ $$u^2-1=2iux$$ $$u^2-2iux-1=0$$ The use of the quadratic formula then gives $$u=ix+\sqrt{1-x^2}$$ And using $$\log$$ to denote the complex natural logarithm: $$iy=\log\left(ix+\sqrt{1-x^2}\right)$$ $$\arcsin x=-i\log\left(ix+\sqrt{1-x^2}\right)$$ | {
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# Angle needed for object A to intercept with object B
Object B is 15 degrees East of North at a distance of 20km/h. Object B is moving at an average speed of 30km/h in the direction 40 degrees East of North. If object A is capable of moving at 100km/h, at what angle does it need to move to intercept object B?
I've tried calculating it as vectors, but got stuck. In the end I just tried using good old geomatry, where $$\frac{\sin A}{a} = \frac{\sin B}{b}$$ So I called the angle I'm searching for A, and the angle from the line between A and B, and the direction of object B, as angle B. 'a' and 'b' are the average speeds of object a and b. $$\frac{\sin A}{30 \times t} = \frac{\sin(155°)}{100\times t}$$ $$A = arcsin\left(\frac{30\times t\times \sin(155°)}{100\times t} \right) = 7.284°$$
So the final angle should be $15°+7.284°=\mathbf{22.284°}$
Is this correct? Isn't there a better way of doing this, like using vector products?
PS. I've found a very similar question here, but didn't quite understand how i.e. $\alpha + \beta = 40°$, because my calculations show $=35°$. And trying to read through comments and answers, confuses me more due to updates being made inconsistent to comments about the updates done. And due to me not having enough reputation, I'm not able to ask counter-questions.
BONUS Question: I have been searching high and low on how to make a simple graph of my problem. Like this here question, with a graph of the problem hosted at stack.imgur.com which I guess was generated by LaTeX on stackexchange.com, right? | {
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-
gnuplot makes nice graphs, as does MATLAB, but gnuplot is free. – Pranav Hosangadi Sep 27 '13 at 3:29
Thanks @PranavHosangadi! I didn't realize I had to make them and them upload the photo. I thought MathJax had a plugin to make LaTeX plots in posts. This makes more sense now! – nikc Sep 27 '13 at 7:01
I wouldn't use Gnuplot or Matlab for this; They are good for plotting data, but would basically require you to solve the problem in order to plot something like this. I would use Inkscape for this kind of a drawing. – Colin McFaul Sep 27 '13 at 15:06
Simple way to solve: Use distances travelled. So B has already travelled 20km at 15° East of North. That is $20 \times \sin(15°)$km northwards and $20 \times \cos(15°)$ eastwards.
Now consider A intercepts B at a time $t$. The equations for the distances travelled by B are:
$Y_B = 20 \times \sin(15°) + 30 \times t \times \sin(40°)$ in the Y direction and
$X_B = 20 \times \cos(15°) + 30 \times t \times \cos(40°)$ in the X direction.
For A, the distances travelled are:
$Y_A = 100 \times t \times \sin(\alpha)$ in the Y direction and
$X_A = 100 \times t \times \cos(\alpha)$ in the X direction.
For the two to intercept, $X_A = X_B$ and $Y_A = Y_B$
You now have 2 equations, 2 variables, and a whole lot of fun solving them simultaneously.
-
There is a technique based on vectors that solves the problem, and lets you know if there is no solution, one solution or two solutions.
(All my angles are measured counter-clockwise from the positive X-axis)
Assume the pursuer travels at 100 km/hr at an angle $\beta$. We're going to try to divide this velocity into two non-perpendicular components
One matches the velocity of the target exactly. So in the problem above, assume that one part of A's velocity is $30\angle 50°$. It wouldn't matter if A's total velocity were less than 30. | {
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So, we have (in theory) completely cancelled out B's attempt to escape. Now, how does A reach B? A must travel at some velocity (say X) along the original direction from A to B. (Since we've cancelled out any of B's movement) So, in this problem, the other part of A's velocity is $X \angle 75°$
So, finally, we're reduced to solving:$$30\angle 50°+X \angle 75°=100\angle\beta$$One way is to expand each vector into X and Y components and set up two equations. If you eliminate $\beta$ you get a quadratic in X. If there are any real positive values of X, you can then substitute and solve for $\beta$
Expanding the equation above for components in the X and Y direction: $$30\cos(50)+X\cos(75)=100\cos(\beta)$$ $$30\sin(50)+X\sin(75)=100\sin(\beta)$$The only unknowns are the size of X (or if it even exists!) and the angle $\beta$ If you square both sides of both equations and then add the two LHS and two RHS, some trig manipulations will eliminate $\beta$ and lead to the quadratic in X.
Thanks for your answer, but I'm still not quite sure how to use your method? I'm still learning how to solve equations using polar coordinates and I don't know how to isolate the equation above. Plus, I'm trying to find angle $\beta$, but how would I do that if there are two unknown factors in the equation? – nikc Sep 27 '13 at 8:18 | {
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# CaptainBlacks Occasional Problem #3
#### CaptainBlack
##### Well-known member
Prove that the polynomials:
$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$
have no real roots.
CB
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that
$P_1(x) = x^2-2x+3 = (x-1)^2+2,$
$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$
$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$
That should be enough of a pattern to suggest a solution.
#### Alexmahone
##### Active member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that
$P_1(x) = x^2-2x+3 = (x-1)^2+2,$
$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$
$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$
That should be enough of a pattern to suggest a solution.
Claim: $P_n(x)=(x-1)^2R_n(x)+n+1$
Let $Q_n(x)=P_n(x)-n-1=x^{2n}-2x^{2n-1}+3x^{2n-2}-\ldots-2nx+n$
$Q_n(1)=1-2+3-\ldots-2n+n=0$
$Q_n'(x)=2nx^{2n-1}-2(2n-1)x^{2n-2}+3(2n-2)x^{2n-3}-\ldots-2n$
$Q_n'(1)=2n-2(2n-1)+3(2n-2)-\ldots-2n=0$ (1st term cancels the last term, 2nd term cancels the 2nd last term etc.)
$Q_n(1)=Q_n'(1)=0\implies Q_n(x)=(x-1)^2R_n(x)$
So, $P_n(x)=(x-1)^2R_n(x)+n+1$
Now we just need to prove that $R_n(x)$ is always positive. It looks like $R_n(x)=x^{2n-2}+2x^{2n-4}+3x^{2n-6}+\ldots+n$. This needs a proof.
Last edited:
#### CaptainBlack
##### Well-known member
Prove that the polynomials:
$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$
have no real roots. | {
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$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$
have no real roots.
CB
A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.
CB
#### anemone
##### MHB POTW Director
Staff member
Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that
$P_1(x) = x^2-2x+3 = (x-1)^2+2,$
$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$
$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$
That should be enough of a pattern to suggest a solution.
By following the hints given by Opalg, I see that
$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$
It's obvious that $P_n(x)\ne 0$ for all $x\in R$.
Thus it suggests that $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$ has no real roots must be true.
Am I on the right track?
#### Alexmahone
##### Active member
Am I on the right track?
Yes. You have to prove this, though:
$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$
#### anemone
##### MHB POTW Director
Staff member
A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.
CB
If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.
First, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots. | {
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First, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.
From $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get
$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$
$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.
Now, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.
That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$
But we can also tell right from the start that the product of all roots of $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ is -(2n+1) for n=1, 2, ......, that is, the product of all the real positive roots (be it two integers/fractions or two pair of positive complex roots) = -ve.
Obviously this leads to contradiction and we can conclude at this point that the polynomial $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has no real roots.
Edit: I made a horrible mistake by saying that the product of all roots = - (2n+1) which is wrong. It should be (2n+1). So, my effort in proving in this post have been amounted to zero, zip, nothing. Please kindly dismiss it.
---------- Post added at 07:02 ---------- Previous post was at 07:00 ----------
Yes. You have to prove this, though:
I think the pattern could be generalized from the hint given by Opalg.
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Last edited:
#### CaptainBlack
##### Well-known member
If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.
First, I assume the $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$ has real roots.
From $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$, I get
$P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-......-2n(-x)+(2n+1)$
$P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-......+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.
Now, I consider the case where $P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-......-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.
That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$
What are positive complex roots, and where will you find a reference to what Descartes rule of signs says about this sort of sign of the complex roots?
CB
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CB
#### anemone
##### MHB POTW Director
Staff member
My mistakes.
I want to clarify that the following observations were not right.
"That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ......, n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ......, n$
"
Reason:
1. I didn't know why on earth I called $a\pm b\sqrt {c}$ as complex roots. I'm sorry.
2. This "...based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2." was taken at http://en.wikipedia.org/wiki/Descartes'_rule_of_signs whereas the two generated cases were my own conclusions.
I'm sorry for confusing you, CB.
My apologies.
#### Opalg
##### MHB Oldtimer
Staff member
Yes. You have to prove this, though:
$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$
My proof of that was just to multiply out the product on the right hand side:
\begin{aligned}(x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x^2-2x+1) \\ =& x^{2n}\phantom{{}-2x^{2n-1}}+2x^{2n-2}\phantom{{}-4x^{2n-3}}+3x^{2n-4} \phantom{{}-2x^{2n-1}}+ \ldots \\ &\phantom{x^{2n}} - 2x^{2n-1} \phantom{{}+2x^{2n-2}} - 4x^{2n-3} \phantom{{}+2x^{2n-2}} -6x^{2n-5} -\ldots \\ &\phantom{x^{2n} - 2x^{2n-1}} +\phantom{1}x^{2n-2}\phantom{{}-2x^{2n-1}} + 2x^{2n-4} \phantom{{}-2x^{2n-1}} +\ldots \\ &\phantom{x}\\ =& x^{2n}-2x^{2n-1}+3x^{2n-2}-4x^{2n-3} + 5x^{2n-4}-6x^{2n-5}+\ldots \end{aligned}
For $0\leqslant k\leqslant n-1$ the coefficient of $x^{2n-2k}$ is $k+(k+1)=2k+1$ and the coefficient of $x^{2n-2k-1}$ is $-2(k+1)$. The constant term is $n$. So the resulting polynomial is $P_n(x) - (n+1)$.
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#### CaptainBlack
##### Well-known member
Solution: CaptainBlacks Occasional Problem #3
This again is from the Purdue PotW
Prove that the polynomials:
$P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)$
have no real roots.
================================================
Solution:
These obviously have no negative roots (either from Descartes rule of signs or observing that the odd powers of $$x$$ all have -ve signs)
So for $$x>0$$ consider:
$P_n(x)+xP_n(x)=x(x^{2n}-x^{2n-1}+x^{2n-2}- ... - x+1) + (2n+1)$
Now the bracketed term is a finite geometric series and so we may replace it with its sum to get:
$P_n(x)(1+x)=x \left[ \frac{1+x^{2n+1}}{1+x}\right] + (2n+1)$
or:
$P_n(x)=x \left[ \frac{1+x^{2n+1}}{(1+x)^2}\right] + \frac{2n+1}{1+x}>0$
and so as $$P_n(0) \ne 0$$ we have $$P_n(x)>0$$ for all real $$x$$
The above is somewhat different from the solution given on the Purdue PotW site, mainly in that it avoids an unexplained step, which while probably valid I find opaque.
CB
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Drawing without replacement - prob. for an Ace followed by an Ace?
Given a standard 52-cards deck:
You are extracting cards from the deck without replacement, until you get an "Ace" for the first time. What is the probability that the next card will be "Ace" too?
I've already seen the following Q&A: Probability of drawing an Ace: before and after
According to that thread, the answer should be: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
But the official solution says that the answer is: $$\frac{4}{52}$$ which doesn't make sense IMHO. They solved it only with intuition or "mind trick", as they wrote..
My calculation:
Assuming that the 1st card is Ace, then: $$\frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{221}$$
Assuming that the 2nd card is Ace, then: $$\frac{(52-4) \cdot 4 \cdot 3}{52 \cdot 51 \cdot 50} = \frac{24}{5525}$$
We notice a pattern here.
Having the 1st Ace at the $k$'th draw, then the probability (for a second Ace after that) is: $$p_1 = \frac{ {_{52-4}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }$$
We need to consider the least-possible scenario - we draw 48 non-Ace cards, then: $$p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} } = \frac{1}{270725}$$
So, the required probability is: \begin{align} p &= p_2 + \sum\limits^{48}_{k=1} p_1 \\ &= p_2 + \sum\limits^{48}_{k=1} \frac{ {_{48}P_{k-1}} \cdot 4 \cdot 3 }{ {_{52}P_{k-1}} \cdot {_{52-k}P_{2}} }\\ &= \frac{1}{270725} + \frac{1696}{20825}\\ &= \frac{1297}{15925}\\ &\cong 0.081444 \end{align}
But my answer is far from either the official solution and from the answer in that thread. | {
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But my answer is far from either the official solution and from the answer in that thread.
• The question you link to is different. It is asking "what's the probability that the second drawn is an Ace?" The answer to that is certainly $\frac 4{52}$ as there is no difference between the second card drawn and the first. Of course, if you specify that the first card drawn was (or was not) itself an Ace then the probability changes. – lulu Oct 9 '15 at 14:35
• @lulu I think that I understand your view on the matter. So what did I calculated? – Dor Oct 9 '15 at 15:07
• I think your method ought to work. To be precise: I just wrote a code which first computes $p_k$, the probability that the first A is observed on the $k^{th}$ draw, and then calculates the probability that the next draw is also an A. I believe this is what you were doing as well. My code returned $\frac 1{13}$ as the total. Doesn't convey a lot of insight, I understand. – lulu Oct 9 '15 at 15:13
• @lulu Indeed you're right! I posted an answer which shows where were my mistakes and include a code that proves it to be equal to $\frac{1}{13}$. – Dor Oct 9 '15 at 16:27
We use a fairly crude counting approach, in order to rely minimally on intuition. There are $\binom{52}{4}$ equally likely ways to choose the positions of the $4$ Aces. We now count the "favourables."
Maybe the first two Aces are in positions 1 and 2. That leaves $\binom{50}{2}$ for the rest.
Maybe they are in positions 2 and 3. That leaves $\binom{49}{2}$ for the rest.
Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest. | {
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Maybe they are in positions 3 and 4. That leaves $\binom{48}{2}$ for the rest.
And so on, up to positions 49 and 50, leaving $\binom{2}{2}$ for the rest. That gives probability $$\frac{\binom{50}{2}+\binom{49}{2}+\cdots+\binom{2}{2}}{\binom{52}{4}}.$$ The numerator can be simplified in various ways. My favourite is to count the number of ways to choose $3$ numbers from the first $51$. This can be done in $\binom{51}{3}$ ways. But the smallest number can be $1$, leaving $\binom{50}{2}$ ways to choose the other two. Or the smallest can be $2$, and so on. We get the number of favourables obtained above.
Our probability is therefore $\frac{\binom{51}{3}}{\binom{52}{4}}$. This simplifies to $\frac{1}{13}$.
• Great argument and very well written. +1. – Shailesh Oct 9 '15 at 15:04
• I like this approach! +1. Could you please elaborate regarding how you simplified the numerator? – Dor Oct 9 '15 at 15:13
• I think that I understand it now. That is a very cool solution..! – Dor Oct 9 '15 at 15:33
First of all Andre has posted a very nice answer. Already Upvoted it. I am trying to look at where all you have gone wrong. Too long for an comment, so posting it here.
Let us look at the last case. You have $48!$ ways of the non-ace cards. Now the remaining cards are all aces anyways. So the question of multiplying this by $4 \cdot 3$ does not arise. Since we do not care about the last 2 cards, the denominator will have to be just $50!$. Similarly, the case before that might need to rectified.
• Indeed you are right about the last term! In the other 48 terms, the mistake was different. I'll add info in my original post. Thanks! +1 – Dor Oct 9 '15 at 16:08
• @Dor Yes, right. A good approach would be you post an answer yourself, instead of editing the post. I'll surely upvote it. – Shailesh Oct 9 '15 at 16:18
Here is a bijection between set of all permutation with the first Ace followed by an Ace and set of all permutations with the first card being an Ace | {
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$$B^nAAX <-> AB^nAX$$
where A stands for an Ace, B for not an Ace, $0\leq n \leq 48$ and $X$ contains $2$ Aces and $48-n$ non-Aces. Hence the probability of an Ace following the first Ace is equail to the probability of the first card being Ace which is equal to $$4/52=1/13$$
Probabilistically, conditioned on the location of the second ace, the first ace is uniformly distributed on locations before the second ace - hence probability of it being the first or the last in that sequence are equal.
• Nice! I was going to go back and try to explain that the fact my fraction simplified to $\frac{4}{52}$ was not an "accident." You have done it, in the strongest (natural bijection) sense. – André Nicolas Oct 10 '15 at 1:27
• I started thinking along the lines of bijection after seeing your result ;). – A.S. Oct 10 '15 at 1:30
• Yes, it was kind of shouting look at me, look at me. – André Nicolas Oct 10 '15 at 1:32
Fixing my original calculation
Thanks to Shailesh, we now know that the expression $$p_2 = \frac{48! \cdot 4 \cdot 3}{ {_{52}P_{50}} }$$ Should be replaced with: $$p_2 = \frac{48!}{ {_{52}P_{48}} }$$ Because that after extracting 48 non-Ace cards, we'll necessarily have the second card as Ace.
Moreover, I had a tiny mistake in the denominator. Instead: $${_{52}P_{k-1}} \cdot {_{52-k}P_{2}}$$ It should be: $${_{52}P_{k-1}} \cdot {_{52-(k - 1)}P_{2}}$$ (All the $k$ indexes should be shifted by $(-1)$)
I wrote a Python code which shows that this solution works either:
# Python 2.7.6
import math
from fractions import Fraction
f = math.factorial
def nPr(n,r):
return f(n) / f(n-r)
p = Fraction(0)
for k in range(1, 49):
p += Fraction(nPr(48, k-1) * 4 * 3,
nPr(52, k-1) * nPr(52 - (k-1), 2))
p += Fraction(f(48), nPr(52,48))
print p # Outputs 1/13 :)
Thank you all for your effort! :)
• Well done, Dor, We all learn with each other's help. – Shailesh Oct 9 '15 at 16:30 | {
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Is there a differential limit?
I'm wondering if there's such a concept as a "differential limit". Let me give an example because my nomenclature is my own and unofficial, but hopefully indicative of the concept.
For some function f(x), there may exist a derivative of that function f'(x) which we call a first order differentiation of f(x). Likewise from f'(x) we could also derive a second order function f''(x). Is there a mathematical concept of the limit of a function as the order of differentiation goes to infinity?
I can think of a function that I can take the derivative of an infinite number of times ( albeit without getting a constant value): f(x) = sin(x)
The function g(x) = x^2 might have a differential limit of 0. (g'(x) = 2x; g''(x) = 2; g'''(x) = 0 ...)
Is there any usefulness to this concept? A topic of study or keyword that discusses this? Any more examples of infinitely derivable functions?
Most importantly if this is interesting to anyone other than myself: what is a good resource for learning more?
-
We use infinitely differentiable functions often in mathematics. We call it the set $C^{\infty}$. A function that is infinitely differentiable is called "smooth" mathworld.wolfram.com/C-InfinityFunction.html – tomcuchta Jun 13 '12 at 3:35
First: yes, we care about how many times a function can be differentiated. In fact, we have names for such classes: $\mathcal{C}_0$ is the collection of all continuous functions, $\mathcal{C}_1$ the class of functions that have continuous derivative; $\mathcal{C}_2$ the class of functions that have continuous second derivative; and so on. We have that $$\mathcal{C}_0 \supsetneq \mathcal{C}_1 \supsetneq\mathcal{C}_2\supsetneq\mathcal{C}_3\supsetneq \cdots \supsetneq \mathcal{C}_n\supsetneq \cdots$$ | {
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Then we have "infinitely differentiable functions", functions that have continuous derivatives of all orders; this is denoted $\mathcal{C}_{\infty}$. There are functions that have derivatives of all orders that don't cycle through values, such as $e^{x^2}$.
Even beyond the infinitely differentiable functions we have the notion of "analytic function", which are functions that can be defined at every point by a power series that converges on an open interval containing the point.
Now, I think you are talking about trying to make sense of a "limit" of the functions as you take derivatives. So that for a polynomial the "limit" should be $0$, a function like $\sin x$ should not have a limit, and so on.
There are in fact several different ways of making the idea of "differential limit" precise. I'm going to talk more generally about what "limit" and "convergence" can mean here, before addressing the specific construction you are talking about. I'll spill the beans and say that I haven't really seen this particular construction before (there's too much to wade through below, and I don't want you to be annoyed when you reach the end and find that I say "I haven't seen it...") But that does not mean you can't ask interesting questions about it.
What you describe is a sequence of functions. The main issue is that one has to define what one means by saying that two functions are "close to one another". Then you get the notion of convergence of functions.
Turns out that there are many different ways of saying this, and which one is useful depends on the context.
Let's take the case of a sequence of real valued functions of real variable, with domain all of $\mathbb{R}$. Let's call the sequence $f_n$ (in your case, $f_n = \frac{d^nf}{dx^n}$ for a given $f$).
1. The most natural way of saying that the sequence $f_n$ converges to a function $f$ is:
$f_n\to f$ if for every $x$, $\lim\limits_{n\to\infty}f_n(x) = f(x)$. | {
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$f_n\to f$ if for every $x$, $\lim\limits_{n\to\infty}f_n(x) = f(x)$.
This is called pointwise convergence. For example, the sequence $f_n(x) = x^n$ defined on $[0,1]$ converges pointwise to the function $$f(x) = \left\{\begin{array}{ll} 0 &\text{if }0\leq x\lt 1\\ 1 & \text{if }x=1. \end{array}\right.$$
It turns out that pointwise convergence is natural, but is not very "good", in the sense that a lot of properties we find interesting about functions are not preserved by pointwise convergence. For example, above, all functions $f_n$ are continuous, but their limit is not.
The problem arises because we are forcing no connection between how $f_n(x)$ converges to $f(x)$ and how $f_n(y)$ converges to $f(y)$ with $x\neq y$; that means that convergence on some points can "lag behind" convergence in points very close to them. This can be remedied by making things a bit tighter. If we write out the definition of the limits above, we obtain the following description:
$\{f_n\}$ converge pointwise to $f$ if and only if for every $x$, for every $\epsilon\gt 0$, there exists $N$, which may depend on both $x$ and $\epsilon$, such that for all $n\geq N$ we have $|f_n(x)-f(x)|\lt \epsilon$.
The way to make this tighter, to make sure the the $f_n(x)$ converge to $f$ more or less "the same way", is to force the $N$ to depend only on $\epsilon$, and not on $x$ as well. This is accomplished with:
2. We say that $\{f_n\}$ converge uniformly to $f$ if and only if for every $\epsilon\gt 0$ there exists $N$, which depends only on $\epsilon$, such that for all $x$ and all $n\geq N$ we have $|f_n(x)-f(x)|\lt\epsilon$.
Uniform convergence is better than pointwise convergence: if every element of the sequence is continuous, and the sequence converges uniformly, then the limit is continuous, for example.
There are yet other ways of talking about functions converging. Here are a few: | {
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There are yet other ways of talking about functions converging. Here are a few:
1. Fix a real number $p$, $1\leq p$. We say that $\{f_n\}$ converges to $f$ in $p$-norm if and only if $$\lim_{n\to\infty}\int_{\infty}^{\infty}|f_n(x)-f(x)|^p\,dx = 0.$$
2. Closely related: we say that $\{f_n\}$ converges to $f$ in the sup-norm if and only if $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|\} = 0.$$
3. There is a way of measuring sets of real numbers, called the Lebesgue measure, $\lambda$, the details of which are perhaps too much to go over now. But intuitively, it is a way of assigning a "size" to a lot of sets of real numbers, including all intervals and much more complicated sets, in a nice and consistent way. We say that $\{f_n\}$ converges in measure to $f$ if and only if for every $\epsilon\gt 0$, $$\lim_{n\to\infty}\lambda(\{x\mid |f_n(x)-f(x)|\geq\epsilon\}) = 0.$$
4. We say that $\{f_n\}$ converges almost uniformly to $f$ if and only if for every $\epsilon\gt 0$ there is a set $X\subseteq \mathbb{R}$, with $\lambda(X)\lt\epsilon$, such that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}-X$.
5. There are small variations to each of those; depending on context, one may be able to define other ways of convergence, or some of the above may no longer make sense. A typical variation is to take any of the above types of convergence (except almost uniform, for which the concept doesn't add anything), and add the condition "almost everywhere" (or "almost surely" if you are doing probability). That means that there exists a set $Y$ contained in the domain, of measure $0$, such that the type of convergence you are interested in occurs in $\mathbb{R}-Y$. (Sets of measure $0$ need not be empty, and they can be quite complicated: for example, any countable set has measure $0$, and there are uncountable sets that do as well, like the Cantor ternary set. | {
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Another variation is to define the concept of "Cauchy sequence" for each of the types, which means making the analog of the definition of "Cauchy sequence" of real numbers: instead of comparing to a limit, we ask that for all $n$ and $m$ greater than $N$, $f_n$ and $f_m$ have the relevant property. For instance, "pointwise Cauchy" means that for all $x$ and all $\epsilon\gt 0$ there exists $N$ such that for all $n,m\gt N$ we have $|f_n(x)-f_m(x)|\lt \delta$. "Cauchy in measure" means that for all $\epsilon\gt 0$ and all $\delta\gt 0$ there exists $N$ such that if $n,m\gt N$, then $\lambda(\{x\mid |f_n(x)-f_m(x)|\geq\epsilon\}\lt\delta$; and so on.
All of these concepts are extremely important and useful. They show up in probability, statistics, functional analysis, analysis, and other areas. There are known implications (e.g., uniform convergence implies pointwise convergence, and so on).
Now, what about the specific case in which we start with a function $f_0$ (necessarily in $\mathcal{C}_{\infty}$) and we define $f_n$ to be the $n$th derivative of $f_0$? I confess I haven't seen this particular sequence addressed, but you can ask all the questions related to the convergence types above: is the sequence pointwise Cauchy? $p$-norm Cauchy? Cauchy in measure? Does it converge uniformly? Almost uniformly? Pointwise? In measure? And so on. Seems like an interesting question, if nothing else.
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Thank you for the thorough answer, that was exactly what I was looking for. – jpredham Jun 13 '12 at 8:58
I think in your first #2 there is a slight typo - we should have that for $n\ge N$ the property holds. – process91 Jun 13 '12 at 13:54
@Michael: Yes, indeed. Thank you. – Arturo Magidin Jun 13 '12 at 15:25 | {
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It would seem that if the sequence $f$, $f'$, $f''$, ..., $f^{(n)}$, ... tends to a limit as a function, then that limit would be a solution of the differential equation $\frac{dy}{dx}=y$. So if the limit exists, it is either zero or some multiple of $e^x$. (I'm assuming that the notion of limit we're using is one that the differentiation operator is continuous with respect to, which doesn't seem like much to ask. Pointwise convergence is out, though).
Also, if we have some function whose iterated derivatives converge towards $ce^x$, we can subtract $ce^x$ from the initial function and get a new sequence whose iterated derivatives converge towards the zero function. So if we look for solutions to $\lim_{n\to\infty} f^{(n)}=0$ we essentially get all functions whose iterated derivatives have a limit.
All polynomials qualify, of course, but there are also non-polynomial analytic solutions such as $\sum_{n=0}^\infty \frac{1}{n\cdot n!} x^n$ and in general $\sum_{n=0}^\infty \frac{a_n}{n!}x^n$ whenever $\lim_{n\to \infty} a_n = 0$. I wonder whether some of these have nice closed forms.
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# Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results [duplicate]
I tried Evaluating $$\int \sqrt{\dfrac{5-x}{x-2}}dx$$ using two different methods and got two different results.
Getting two different answers when tried using two different methods:-
M-$$1$$:
$$\int \dfrac{5-x}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx$$ $$\dfrac{1}{2}\int\dfrac{-2x+7}{\sqrt{\left(5-x\right)\left(x-2\right)}}dx+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(5-x\right)\left(x-2\right)}}$$ $$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{-x^2+7x-10}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\int\dfrac{dx}{\sqrt{\left(\dfrac{3}{2}\right)^2-\left(x-\dfrac{7}{2}\right)^2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{x-\dfrac{7}{2}}{\dfrac{3}{2}}$$
$$\sqrt{\left(5-x\right)\left(x-2\right)}+\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$
M-$$2$$:
$$x=5\sin^2\theta+2\cos^2\theta$$ $$dx=\left(10\sin\theta\cos\theta-4\cos\theta\sin\theta\right) \, d\theta$$ $$\int \sqrt{\dfrac{5\cos^2\theta-2\cos^2\theta}{5\sin^2\theta-2\sin^2\theta}}\cdot6\sin\theta\cos\theta \,d\theta$$ $$6\int \cos^2\theta \,d\theta$$ $$\int 3\left(1+\cos2\theta\right) \,d\theta$$ $$3\left(\theta+\dfrac{\sin2\theta}{2}\right)$$ $$3\theta+\dfrac{3}{2}\sin2\theta$$
$$x=5\sin^2\theta+2-2\sin^2\theta$$ $$\sin^{-1}\sqrt{\dfrac{x-2}{3}}=\theta$$
$$\cos2\theta=1-2\sin^2\theta$$ $$\cos2\theta=1-2\cdot\dfrac{x-2}{3}$$ $$\cos2\theta=\dfrac{7-2x}{3}$$
$$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\dfrac{3}{2}\cdot\dfrac{\sqrt{9-(49+4x^2-28x)}}{3}$$ $$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}+\sqrt{\left(5-x\right)\left(x-2\right)}$$
In first and second method I am getting the different results of $$\dfrac{3}{2}\sin^{-1}\dfrac{2x-7}{3}$$ and $$3\sin^{-1}\sqrt{\dfrac{x-2}{3}}$$ respectively. I checked that these are not inter-convertible. Why am I getting this difference? | {
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• Does this answer your question? Getting different answers when integrating using different techniques. For this case I would suggest the third approach from there (differentiate). – Zacky Dec 26 '19 at 16:05
• Have you read that posts? Both your approaches are correct and they only differ by a constant, which I would suggest to use when you deal with an indefinite integral. – Zacky Dec 26 '19 at 16:08
• both approaches are correct, the difference is in a constant which appears to be ${3\pi}\over 4$ – roman Dec 26 '19 at 16:12
• @Zacky, can you please reopen the question, what's your hurry? – user3290550 Dec 26 '19 at 16:13
• @user3290550 : You wrote: "actually there should not be difference of a constant also, because whatever standard integrations I have used, those are exact". That is incorrect. Exact methods should yield things differing by a constant. "Constant" in this context means not depending on $x.$ If something does not depend on $x,$ then its derivative with respect to $x$ is $0.$ – Michael Hardy Dec 27 '19 at 7:15
Both your answers are correct. Your two functions differ by $$3\pi/4$$, so they have the same derivative.
Let $$b=\sqrt{\frac{x-2}3}$$. For your integral to make sense you need $$0\leq b\leq1$$ (this comes from $$2\leq x\leq5$$). Let $$a=\arcsin b$$. Note $$0\leq\arcsin b\leq \tfrac\pi2$$, and so the sine is injective when applied to $$2a-\tfrac\pi2$$. Then \begin{align} \sin\left(2a-\tfrac\pi2\right) &=-\cos(2a)=-(1-2\sin^2a)=2b^2-1. \end{align} So \begin{align} 2\arcsin\sqrt{\frac{x-2}3}\ -\frac\pi2 &=2\arcsin b -\frac\pi2=2a-\frac\pi2\\ \ \\ &=\arcsin(2b^2-1)\\ \ \\ &=\arcsin\frac{2x-7}3. \end{align} Then, multiplying by $$3/2$$, $$3\arcsin\sqrt{\frac{x-2}3}\ -\frac{3\pi}4=\frac32\,\arcsin\frac{2x-7}3.$$
Shown below is that the two results differ by a constant $$-\frac{3\pi}4$$.
Define $$f(x)$$ as the difference of the two results
$$f(x)=\frac32\sin^{-1}\frac{2x-7}3-3\sin^{-1}\sqrt{\frac{x-2}3}$$
where $$2. Then, evaluate | {
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$$f(x)=\frac32\sin^{-1}\frac{2x-7}3-3\sin^{-1}\sqrt{\frac{x-2}3}$$
where $$2. Then, evaluate
$$f’(x) = \frac3{2\sqrt{(5-x)(x-2)}}- \frac3{2\sqrt{(5-x)(x-2)}}=0$$
Thus, $$f(x)$$ is a constant over $$(2,5]$$ and can be evaluated with
$$f(x)=f(5)=\frac32\cdot \sin^{-1} 1 -3\sin^{-1}1=-\frac{3\pi}4$$
• You are right, my bad. – Martin Argerami Dec 26 '19 at 22:58 | {
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## Calculus: Early Transcendentals 8th Edition
As $x$ approaches 2, $f(x)$ approaches 5. Yes, it is possible for this to be true and for $f(2)=3$.
It means that $f(x)$ becomes arbitrarily close to $3$ as we take $x$ close enough to $2$, but $x\ne2$. It is possible for $f(2)$ to equal $3$ because the limit only relies on when $f(x)$ is near $2$; there can be a hole at $x=2$. | {
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# Math Help - Piecewise Function problem
1. ## Piecewise Function problem
Let f(x) =
(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2
and g(x) = [integrate] f(t)dt on [0, x]
i) Find an expression for g(x) similar to the one for f(x).
ii) Where is f differentiable? Where is g differentiable?
I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).
But from that, how do I figure out an expression for g(x) from the piecewise function?
Any and all help is appreciated here.
2. Originally Posted by Tulki
Let f(x) =
(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2
and g(x) = [integrate] f(t)dt on [0, x]
i) Find an expression for g(x) similar to the one for f(x).
ii) Where is f differentiable? Where is g differentiable?
I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).
But from that, how do I figure out an expression for g(x) from the piecewise function?
Any and all help is appreciated here.
i)
for $x < 0$, $f(x) = 0$ and $g(x) = \int f(t) ~dt = 0$
for $0 \leq x \leq 1$, $f(x) = x$ and $g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2$
for $1 < x \leq 2$, $f(x) = 2 - x$ and $g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt=$ $\int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})$
for $x > 2$
3. Hello, Tulki!
Here's part (a).
Did you make a sketch?
$\text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm]
x && 0 \leq x \leq 1 \\ \\[-4mm]
2 - x && 1 < x \leq 2 \\ \\[-4mm]
0 && x > 2 \end{array}\right\}$
and: . $g(x) \:=\: \int_0^x f(t)\,dt$
a) Find an expression for $g(x)$ similar to the one for $f(x).$
$g(x)$ gives the area under the graph from $0\text{ to }x$ | {
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$g(x)$ gives the area under the graph from $0\text{ to }x$
The graph looks like this:
Code:
|
|
1+ *
| *:::*
| *:::::::*
| *:::::::::| *
- * * * - - - * - + - * * * - -
0 1 x 2
Then: . $\displaystyle g(x) \;=\;\left\{ \begin{array}{cccc}
0 && x < 0 \\ \\[-3mm]
\int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm]
\frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm]
1 && x > 2
\end{array} \right\}$
4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.
For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).
For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is $\frac{1}{2}x^2$.
For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\frac{1}{2}(1)(1)= \frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}$.
The total area is $\frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.
For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.
5. Thank you very much, all of you!
I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect. | {
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# A Proof in Elementary Set Theory
I apologize in advance for any lack of clarity in my mathematical symbols; I'm beginning to learn how to use this site.
I'm working through "Introduction to Analysis" by Rosenlicht and he presents an exercise
Prove that if $X\subset S$, $Y\subset S$, then $\complement X\cap\complement Y=\complement(X\cup Y)$
I went about it as follows:
Suppose $X\subset S$ and $Y\subset S$. Then $\complement X=\{x:x\in S\land x\notin X\}$ And $\complement Y=\{x:x\in S\land x\notin Y\}$ , so $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in S \land x\notin Y\}$. Hence $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in Y\}$. Therefore $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\cup Y\}$ And thus $\complement X\cap\complement Y=\complement(X\cup Y)$
Whereas the author gives the proof
If $x\in \complement X\cap\complement Y$ then $x\in\complement X$ and $x\in \complement Y$. This means that $x\in S, x\notin X, x\notin Y$. Since $x\notin X, x\notin Y$, we know that $x\notin X\cup Y$. Hence $x\in \complement(X\cup Y)$.
Conversely, if $x\in\complement(X\cup Y)$, then $x\in S$ and $x\notin X\cup Y$. Therefore $x\notin X$ and $x\notin Y$. Thus $x\in\complement X$ and $x\in\complement Y$, so that $x\in\complement X\cap\complement Y$.
My questions are
• 1) is my proof correct?
• 2) which proof is better
• 3) what are the subtle (because obviously I know what the difference is, but I'm looking for deeper mathematical understanding) differences between my proof and the authors?
Thank you! | {
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Thank you!
• If you could please fix your post, in order to be more readable. It's $\complement$ rather than $\compliment$. After each command, you'd better leave a space. If you want to write "x in (not in) Y", you can write either $x \in Y$ or $x \notin Y$. – thanasissdr Aug 20 '15 at 23:30
• Thanks you. That's good to know. It appears someone just did now. Im assuming you can read it now? – Liam Cooney Aug 20 '15 at 23:38
• I think that both proofs are correct (barring a typo at the end of the line following "Hence" in your proof). – Akiva Weinberger Aug 20 '15 at 23:40
• It still needs work. You should also note that logical and is \land – Graham Kemp Aug 20 '15 at 23:41
One additional thing the author does is explicitly show that the implication works both ways; that $x\in(\complement X\cap\complement Y)\implies x\in \complement(X\cap Y)$ and that $x\in\complement(X\cap Y)\implies x\in(\complement X\cap\complement Y)$. You rely on each step being an equivalence; but don't explicitly indicate this is so. | {
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# The most efficient way to tile a rectangle with squares?
I was writing up a description of the Euclidean algorithm for a set of course notes and was playing around with one geometric intuition for the algorithm that involves tiling an $m \times n$ rectangle with progressively smaller squares, as seen in this animation linked from the Wikipedia article:
I was looking over this animation and was curious whether or not the squares that you would place down in the course of this algorithm necessarily gave the minimum number of squares necessary to cover the entire rectangle.
More formally: suppose that you are given an $m \times n$ rectangle, where $m, n \in \mathbb{N}$ and $m, n > 0$. Your goal is to tile this rectangle with a set of squares such that no two squares overlap. Given $m$ and $n$, what is the most efficient way to place these tiles?
Is the tiling suggested by the Euclidean algorithm (that is, in an $m \times n$ rectangle, with $m \ge n$, always place an $n \times n$ rectangle, then recurse on the remaining rectangle) always optimal? If not, is there are more efficient algorithm for this problem?
I am reasonably sure that the Euclidean approach is correct, but I was having a lot of trouble formalizing the intuition with a proof due to the fact that there are a lot of different crazy ways you can try to place the squares. For example, I'm not sure how to have the proof handle the possibility that squares could be at angles, or could have side lengths in $\mathbb{R} - \mathbb{Q}$.
Thanks! | {
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Thanks!
-
– user17762 Jun 25 '12 at 23:11
It's actually pretty straightforward to show that the squares can't be at angles (since any juncture between edges has to be 90 degrees, there are only two possible orientations for lines to be at, and those have to match the two orientations of the rectangle's edges), and it shouldn't be much more complicated to show that all side lengths must be rational (more specifically, side lengths have to be commensurate for any tiling.) – Steven Stadnicki Jun 25 '12 at 23:14
For a start, see Stan Wagon's Fourteen proofs of a result about tiling a rectangle (see his references). – Bill Dubuque Jun 25 '12 at 23:28
– Erel Segal-Halevi Nov 2 '13 at 21:04
@Erel A web search easily locates a fresh link to Wagon's paper. – Bill Dubuque Dec 15 '13 at 4:19
Euclid doesn't always minimize the number of squares. E.g., with an $8\times9$ rectangle, Euclid says use an 8-square and 8 1-squares, 9 squares in all. But you can do it with a 5, two 4s, a 3, a 2, and two 1s, making 7 squares total. You put the 5 in a corner, then put the 4s in the corners that have room for them; that leaves a $3\times5$ rectangle to fill, which you can do by Euclid.
There is a smaller example, with a $5 \times 6$ rectangle. The Euclid algorithm produces a tiling with 6 squares: 1 5-square and 5 1-squares. It is possible to tile the rectangle with 5 squares: 2 3-squares and 3 2-squares (The example is from Fractality: math.stackexchange.com/questions/479192/squaring-rectangles ). – Erel Segal-Halevi Nov 2 '13 at 21:02
Tiling a rectangle with the fewest squares. Kenyon, Richard J. Combin. Theory Ser. A 76 (1996), no. 2, 272–291. "We show that a square-tiling of a p×q rectangle, where p and q are relatively prime integers, has at least $\log 2p$ squares. If $q>p$ we construct a square-tiling with less than $q/p+C \log p$ squares of integer size, for some universal constant $C.$'' | {
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# Number Properties: Tips and hints
Author Message
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Math Expert
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Number Properties: Tips and hints [#permalink] 04 Jun 2014, 11:25
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Number Properties: Tips and hints
! This post is a part of the Quant Tips and Hints by Topic Directory focusing on Quant topics and providing examples of how to approach them. Most of the questions are above average difficulty.
EVEN/ODD
1. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
2. An odd number is an integer that is not evenly divisible by 2.
3. According to the above both negative and positive integers can be even or odd.
ZERO:
1. 0 is an integer.
2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself). | {
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PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.
PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.
IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
Theory on Number Properties: math-number-theory-88376.html
DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59 | {
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Please share your number properties tips below and get kudos point. Thank you.
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Re: Number Properties: Tips and hints [#permalink] 24 Jun 2014, 22:15
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[wrapimg=]PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]
I was learning the above points, while a confusion rose. I'd be grateful if you could explain.
We know that 100 is a perfect square (10^2)
If we Prime factorize 100, we get
100= 2*2*5*5
number 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?
number 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.
Even if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2
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Re: Number Properties: Tips and hints [#permalink] 25 Jun 2014, 01:42
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Tanvr wrote:
[wrapimg=]PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);[/wrapimg]
I was learning the above points, while a confusion rose. I'd be grateful if you could explain.
We know that 100 is a perfect square (10^2)
If we Prime factorize 100, we get
100= 2*2*5*5
number 1 tips says that a perfect square has odd number of distinct factors. Here 100 has two distinct factors (2,5). How that can be explained ?
number 3 tips says that a perfect square always has an odd number of odd factors and even number of even factors. Here we see, 100 has two 5's which is even number.
Even if we see number 3 tips as, a perfect square always has an odd number of DISTINCT odd factors and even number of DISTINCT even factors., we find that, 100 has only ONE DISTINCT even factor which is 2
2 and 5 are prime factors of 100. The total number of factors of 100=2^2*5^2 is (2+1)(2+1)=9=odd: 1, 2, 4, 5, 10, 20, 25, 50, 100. Out of these 9 factors three are odd (1, 5, and 25) and 6 are even (2, 4, 10, 20, 50, 100).
Hope it helps.
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Re: Number Properties: Tips and hints [#permalink] 23 Jul 2014, 11:29
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Can I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?
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Re: Number Properties: Tips and hints [#permalink] 23 Jul 2014, 11:51
Expert's post
tangt16 wrote:
Can I suggest a few new topics? Specifically Geometry, word problems, and combinatrics/probability?
Thank you. Will cover them soon.
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Number Properties: Tips and hints [#permalink] 27 Aug 2014, 05:37
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If progressions comes under this topic, would like to add this tip.
In the specific case of sum to n1 terms being equal to sum to n2 terms of the same arithmetic progression, the sum of the term numbers which exhibit equal sums is constant for the given evenly spaced set of numbers.
(S3 denotes Sum of the first three terms of the evenly spaced set.)
1.
Q: if sum to 11 terms equal sum to 19 terms in an evenly spaced set, what is the sum to 30 terms for this series?
A: S11 = S19; so S0 = S30. Since S0 = 0, S30 = 0.
2.
This happens because the arithmetic progression's negative terms cancel out the positive terms.
Also, if the series has a zero in it, the sum will be equal for two terms such that one term number will be odd and the other will be even.
Ex.: -10, -5, 0, 5, 10.....
And if the series does not have a zero in it, the sum will be equal for two terms such that both term numbers will be either odd or even.
Ex.: -12, -4, 4, 12......
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Number Properties: Tips and hints [#permalink] 27 Aug 2014, 05:52
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Another tip, | {
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If B is 1/x times more than A, then A is 1/(x+1) times lesser than B.
This is especially useful in averages, profit and loss, time rate questions.
Example:
If B's wage is 25% more than A's wage, then what is A's wage in terms of B?
B is 1/4 times more than A, so A will be 1/5 or 20% lesser than B. i.e., A = 80% of B
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Re: Number Properties: Tips and hints [#permalink] 27 Aug 2014, 07:15
Tip for questions involving recurring decimals:
Note the following pattern for repeating decimals:
0.22222222... = 2/9
0.54545454... = 54/99
0.298298298... = 298/999
Note the pattern if zeroes preceed the repeating decimal:
0.022222222... = 2/90
0.00054545454... = 54/99000
0.00298298298... = 298/99900
Re: Number Properties: Tips and hints [#permalink] 27 Aug 2014, 07:15
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# Solve $x^y \, = \, y^x$ [duplicate]
Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$
I obtained a question asking for how to solve $\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers. By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.
• My question is: How do one show that $(2\,,\,4)$ is the only non trivial, positive solution to the equation?
Now my initial approach was as follows: We have
$$\large x^y = y^x$$
The trivial solution is obviously when $y=x$, so let us focus on when $y \neq x$. Let us make a more general statement. Firstly I take the log of both sides
$$\large y \log x = x \log y$$
Let us divide by x and \log x (We now assume $x\neq 0$ and $x\neq 1$ since 0 is not a positive number, and 1 gives us a trivial solution)
$$\large \frac{y}{x} = \frac{\log y }{\log x}$$
For these sides to be equal, we must remove the logarithms on the right hand side, this is achived if $y$ is on the form $x^a$. Now This gives
$$\large \frac{x^a}{x} = \frac{\log \left(x^a\right) }{\log(x)}$$
$$\large x^{a-1} = a$$
So finaly we obtain that $\displaystyle \large x=\sqrt[ a-1]{a}$ and $\displaystyle \large y = \sqrt[a-1]{a^a}$
Now setting $a=2$ gives us $x = 2$ and $y=4$ as desired.
My question is, how do we prove that $x=2$ and $y=4$ is the only integer solutions? My thought was to show that $\displaystyle \large \sqrt[ a-1]{a}$ and $\displaystyle \large \sqrt[a-1]{a^a}$ are both irrational when a>2, but I have not been able to show this.
Any help is greatly appreciated, cheers =)
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## marked as duplicate by Aryabhata, Zev ChonolesFeb 28 '12 at 21:03 | {
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## marked as duplicate by Aryabhata, Zev ChonolesFeb 28 '12 at 21:03
This MO thread is relevant, mathoverflow.net/questions/22230/ab-ba-when-a-is-not-equal-to-b – user1729 Feb 28 '12 at 16:21
@user1729: Don't like searching MSE itself? :-) – Aryabhata Feb 28 '12 at 19:04
$1729 = 10^3 + 9^3 = 12^3 + 1^3.$ – Will Jagy Feb 28 '12 at 20:53
I've closed the question as a duplicate. There is the option of merging the questions, which will move all answers here to the other question (N3buchadnezzar gets to keep the reputation received from asking this question). Is anyone against doing so? – Zev Chonoles Feb 28 '12 at 21:05
@Aryabhata: No, it was just that the question I linked to popped to the top of the MO stack last week, and I was surprised to find an almost identical question here less than a week later! (Almost identical? Perhaps "intricately related" would be better. Certainly, the answers answer a more general question: when are the solutions rational. Also, the discussion is very interesting.) – user1729 Feb 29 '12 at 9:46
The function $u:x\mapsto(\log x)/x$ is increasing on $[1,\mathrm e]$ from $u(1)=0$ to $u(\mathrm e)=1/\mathrm e$, and decreasing on $[\mathrm e,+\infty)$ from $u(\mathrm e)=1/\mathrm e$ to $0$. Hence, if $u(x)=u(y)$ with $y\gt x\geqslant 1$, then $y\gt \mathrm e\gt x\gt1$. Since $\mathrm e\lt3$, this implies that $1\lt x\lt3$. If furthermore $x$ is an integer, then $x=2$. The unique root of the equation $u(y)=(\log2)/2$ such that $y\gt\mathrm e$ is $y=4$. Hence $(x,y)=(2,4)$ is the unique solution.
-
Another approach for these kind of problems:
$$x^y=y^x$$
By dividing both sides of equations by $x^x$, we'll have:
$$x^{y-x}=(\frac{y}{x})^x$$
Now, what you can say about $\frac{y}{x}$ ?
Try to continue.
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I have no idea what we can say about that fraction. Care to explain? – Pureferret Feb 28 '12 at 20:59
y/x must be integer.(Why?) – Salech Alhasov Feb 28 '12 at 21:38 | {
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I thought this sounded familiar. The equation $a^b = b^a$ can be written as $$\frac{\log a}{a} = \frac{\log b}{b}$$ If you carefully draw the curve $y= \log x$ in the $x-y$ plane, the quantity $\frac{\log a}{a}$ is the slope of the line that passes through the origin and the point $(a, \log a).$ So the equation $\frac{\log a}{a} = \frac{\log b}{b}$ says that the lines from the origin to $(a, \log a)$ and to $(b, \log b)$ have the same slope, therefore they are the same line. That is, the three points are collinear.
So, a graphical solution is to draw lines through the origin, with positive slope, that intersect the curve $y= \log x.$ It will be seen fairly quickly that one intersection point, call it $a,$ has $1 < a < e.$ As you want $a$ an integer, the only choice is $a=2.$
The calculus part is this: the line through the origin with slope $\frac{1}{e}$ is tangent to the curve $y= \log x$ at the point $(e, \log e \; = \; 1).$ To intersect the curve twice, we need slope a little bit less that that, and the first intersection point will be a little bit to the left of $e.$
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# Points in $\Bbb {R}^2$ that can be reached via steps which are $1/5$ of a unit circle.
I was playing around with this demo of Project Euler Problem 208 which allows you to take steps which are "left" or "right" arcs of $$1/5$$ of a unit circle.
Here's an example walk, which starts at the blue dot pointing vertically up, and which consists of steps
RLLLRLLLRRLLLRLLLRRLRLLLR
# Question
Which points in $$\mathbb R^2$$ can be reached in a finite number of steps, starting at the origin and pointing in the positive $$y$$-direction. Is this set of points dense in $$\mathbb R^2$$? If not, what's the greatest number of points that can land in $$[0,1] \times [0,1] \subset \mathbb R^2$$? | {
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• I would expect the points to be dense but it be a bit of work to prove. One way is to find sequences of moves that end with you facing the same direction but move the point in several directions. If the distances moved are not rational multiples of each other you can combine those movements to get as close as wanted to any point. RL and LR leave you heading in the same direction as at the start. You can compute the displacement they create, but they will be the same vertically and negatives horizontally. Now add RRLL and LLRR to the mix. You need – Ross Millikan Aug 22 '19 at 23:02
• a combination that moves you downward, but that shouldn't be hard to find. I would think something like RRRLRLRLRLRLLL would work, the first three to make (about a) U turn, the middle RLs to move in that direction, and the last three to reorient upward. – Ross Millikan Aug 22 '19 at 23:05
• This is a really interesting question. I might end up offering a bounty here – clathratus Aug 23 '19 at 0:37
• I think one could reformulate this question so it is closely linked with the projective definition of the Penrose pattern: Take the standard Z^5 grid in the 5-dimensional vector space. One can put a 2-dimensional subspace E in such a way, that the 5 base vectors project to it pointing to the corners of a regular pentagon. Then your question should be equivalent to the question, If the projection of Z^5 to E is dense, right? I think this has been proven due to E having an irrational angle to the base vectors. – Non Euclidean Dreamer Aug 24 '19 at 15:02
• Related:Rolling a random walk on a sphere: "The surface will be filled for every $\delta$ except $\pi/2$ and $\pi$." – Joseph O'Rourke Aug 25 '19 at 1:48
Yes, the set of points attainable by these paths is dense in $$\mathbb{R}^2$$. Pardon the lack of polish in this answer; I don't have a 'real' math background, so I'm expressing my ideas here using the language most intuitive to me. | {
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We can think of any path as a linear combination of five 'moves,' which w.l.o.g. are given by the vectors $$v_n = (\cos\frac{2\pi n}{5}, \sin \frac{2\pi n}{5})$$. Paths cannot be arbitrary integer combinations of these vectors, since there are only two moves available after each step. We can, however, devise the following scheme to obtain arbitrary integer combinations of the vectors $$2v_n$$. Consider the following paths (I've written out the linear combination long form to show the order that the steps must be taken in each path):
\begin{align*} p_1^a &= 2av_1\\ p_2^b &= v_1 + 2bv_2 + v_1\\ p_3^c &= v_1 + v_2 + 2cv_3 + v_2 + v_1\\ p_4^d &= v_1 + v_2 + v_3 + 2dv_4 + v_3 + v_2 + v_1\\ p_5^e &= 2ev_5 \end{align*}
These paths are composable with each other, so we can take them in arbitrary (positive) integer combinations. Consider a point of the form $$x = \sum_{i = 1}^5 p_i^{a_i}$$. Writing this out in terms of our original moves $$v_i$$, we obtain \begin{align*} x &= p_1^a + p_2^b + p_3^c + p_4^d + p_5^e\\ &= (6 + 2a)v_1 + (4 + 2b)v_2 + (2 + 2c)v_3 + 2dv_4 + 2ev_5 \end{align*} We will ignore the constant offset $$6v_1 + 4v_2 + 2v_3$$. Using the paths $$p_i^a$$ alone, we can thus reach an arbitrary (nonnegative) integer combination of the vectors $$\{2v_i\}$$.
Now note that $$\sum_{i = 1}^5 2v_i = 0$$, so it follows that the vectors $$2v_i$$ and $$\sum_{j \neq i} 2v_j$$ are additive inverses. Thus the set of nonnegative integer combinations of the vectors $$2v_i$$ is actually equal to the set of arbitrary integer combinations of the vectors $$2v_i$$.
Digging around MathSE, I found this answer that asserts, based on Kronecker's theorem, that if three vectors have components that are all linearly independent over the rationals, then the set of their integer combinations is dense in the unit square (and by extension the plane). $$2v_1$$, $$2v_2$$ and $$2v_3$$ satisfy this hypothesis, so the claim should follow. | {
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• Pretty damn good for someone without a math background! (+1) – clathratus Sep 6 '19 at 4:04
• Well, I'm a physics student, so I wouldn't say I have 'no' math background– just not a 'real' one. – Panopticon Sep 6 '19 at 18:08
Not an answer, but just an opportunity to mention that the behavior of the "Tangle Toy," which comes in quarter-circles,
Image from FatBrainToys.
Erik D. Demaine, Martin L. Demaine, Adam Hesterberg, Quanquan Liu, Ron Taylor, and Ryuhei Uehara, “Tangled Tangles”, in The Mathematics of Various Entertaining Subjects (MOVES 2015), volume 2, 2017, pages 141–152, Princeton University Press. Authors' link.
Fig.7: Tangle toy configurations.
To address @Narasimham's question, this paper
Tatu, Aditya. "Tangled Splines." arXiv:1610.03129. 2016-18.
says that the curvature $$\kappa(t) =\pm 1$$ outside of the knot points, and that torsion $$\tau(t)$$ is zero, again excluding knot points.
• Interesting. Are curvature and torsion constant? – Narasimham Aug 22 '19 at 23:55
• Curvature: yes. Torsion:no. – Ivan Neretin Aug 23 '19 at 7:45
• @IvanNeretin: The cited paper says, "Since each link lies in the plane spanned by Vi and Vi+1, for each t ∈ (ti, ti+1), the torsion τ is zero." I have not tried to verify this, but you seem to be contradicting their claim? – Joseph O'Rourke Aug 23 '19 at 21:50
• All right, the torsion is zero within each link and infinite at the joints. – Ivan Neretin Aug 23 '19 at 22:01 | {
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Here's a formulation that may help with calculation a bit: define a 'position' as $$\langle x, y, \theta\rangle$$, where $$\theta$$ is the angle being made with the $$y$$ axis; in other words, the starting position is $$\langle0, 0, 0\rangle$$. Then if you draw the geometry out (which I will try and do once I get onto a machine with better drawing tools) from $$\langle x, y, 0\rangle$$ the available moves are to $$\langle x+(1-\cos(2\pi/5)), y+\sin(2\pi/5), \pi/10\rangle$$ and $$\langle x-(1-\cos(2\pi/5)), y+\sin(2\pi/5), -\pi/10\rangle$$. Since the local frame at angle $$\theta$$ has its $$x$$ axis in the direction $$\langle \cos\theta, \sin\theta\rangle$$ and its $$y$$ axis in the direction $$\langle -\sin\theta, \cos\theta\rangle$$ (modulo possible sign errors) then this means that from $$\langle x, y, \theta\rangle$$ the available moves are to $$\langle x+\cos\theta(1-\cos(2\pi/5))-\sin\theta\sin(2\pi/5), y+\cos\theta\sin(2\pi/5)+\sin\theta(1-\cos(2\pi/5)), \theta+\pi/10\rangle$$ and to $$\langle x-\cos\theta(1-\cos(2\pi/5))-\sin\theta\sin(2\pi/5), y+\cos\theta\sin(2\pi/5)-\sin\theta(1-\cos(2\pi/5)), \theta-\pi/10\rangle$$. Since the cosine and sine of $$2\pi/5$$ and $$\pi/10$$ have nice clean expressions in terms of $$\sqrt{5}$$, you should be able to get explicit expressions in terms of $$\sqrt{5}$$ for the coordinates after making multiple moves as in Ross Millikan's comment, and then use some usual irrationality-of-square-roots arguments to show density. (But of course, note that there's a little subtlety to these arguments, because the same statement with sixth-turns around the circle are false!) | {
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# An analysis of the buffons needle a method for the estimation of the value of pi
Buffon 's needle problem if vou drop any needle, short or long, then the expected number of crossings will be where pi is the probability that the needle will come to lie with exactly one. Buffon's needle refers to a simple monte carlo method for the estimation of the value of pi, 314159265 the idea is very simple suppose you have a tabletop with a number of parallel lines drawn on it, which are equally spaced (say the spacing is 1 inch, for example. Fostering understanding of monte carlo simulations since such knowledge provides them with the understanding necessary to justify estimation with the monte carlo method buffon needle applet (available at tubegeogebraorg/material/show/id/112419) observing the value of p c when l6= d. Estimation from samples 5 one sample of paired observations (paired t test) 6 comparing two samples (the two-sample t test) 7 regression 8 comparing price, j (1984) random numbers and buffons' needle, parab, 20, 2-9 priddis, m j (1978) an upper bound for the g r (1986) a method for teaching statistics using n. The analysis of the subject of probability additionally, it may be suggested that reno program is, method buffon’s needle problem - the emergence of π at the end of solving a probability problem is very interesting this experiment can be performed by everybody and an estimation can be made for π value for pi value of pi did the. | {
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A abdomen the an analysis of the buffons needle a method for the estimation of the value of pi pterigoid lex stops an analysis of crooks a character in john steinbecks novella of mice and men twice, his oblate declares himself ridiculous the tetradiscotic and totalitarian neophyte an analysis of harvard mk i and colossus sealed his somber. Nmfp[1] - free download as pdf file (pdf), text file (txt) or read online for free scribd is the world's largest social reading and publishing site explore explore interests career & money. The buffon needle method of estimating \$latex {\pi}&fg contents « the bridges of paris the great american eclipse » buffon was no buffoon published august 13, 2015 occasional leave a comment tags: algorithms, probability the buffon needle method of estimating is graph theory group theory hamilton history ireland logic maps.
Vol 79, no 5, december 2006 387 since our spherical caps have height l /2 y, the area of r y is a (y) = 2 2 2 l 2 l 2 y = l 1 2 y l thus, for the Þxed value of y (0 y l /2), the probability that the needle crosses. Im supposed to write a c program to estimate the pi via buffon needle using monte carlo method i think my program works properly, but i never ev stack overflow log in sign up current community stack overflow buffon needle pi estimate experiment gets close to pi, but doesnt get perfect value ask question pi value. 512 analysis of the formulas 513 formulas of the theory of functions 514 formulas of number theory 515 formulas another probability based and unusual method is the buffon needle problem by georges -louis leclerc de buffon (1733 argued in 1777 in the mid-19th century, the swiss astronomer rudolf wolf came by 5000 needle throws. | {
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Approximating the value of pi points 100 points of special assignments/computer projects due date thursday october 27, simulator to do so is called buffon’s needle and the following link explains it better than anyone did you clearly identify your method for each problem did you suppress all statements did you. Removing the inherent paradox of the buffon's needle monte carlo simulation using fixed-point iteration method we use the buffon’s needle experiment as an example to introduce the fixed-point iteration method the buffon’s needle simulation model can be fitted into the fixedpoint iteration model in. Preface in this project we discuss short, but comprehensive introduction to methods of com-putation pi some of learning rules chapter 2 consists of a brief discussion on pi and. Buffon's needle experiment for estimating π is a classical example of using an experiment (or a simulation) to estimate a probability how to convert the datetime character string to sas datetime value (anydtdtm and mdyampm formats) using sas enterprise guide to run programs in batch not particularly the convergence. | {
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Thinking skills creativity, problem solving, problem finding - powerpoint ppt presentation loading ppt – thinking skills creativity, problem solving title: thinking skills creativity, problem solving, problem finding description: description of attitudes with the help of 'roses' and 'thorns' divine roses for you. Download citation | buffon needle lands buffon needle lands in ϵ-neighborhood of a 1-dimensional sierpinski gasket with probability at most we also formulate a result about all self-similar sets of dimension 1résuméon donne une estimation de la probabilité pour que l'aiguille de buffon soit ϵ-proche d'un ensemble de. Approximation however, better approximations can be obtained using a similar method with regular polygons with (1707-1788)\) and laplace $$(1749-1827)$$ proposed using a stochastic simulation to estimate the value of $$π$$ buffon's needle famously approximates $$\pi$$ using the fact that a straight needle is the number $$\pi\. Throwing buffon’s needle with mathematica nb cdf pdf it has long been known that buffon’s needle experiments can be used to estimate three main factors influence these experiments: grid shape, grid density, and needle length in statistical literature, the desired probability is directly related to the value of , which can then be. Throwing buffon’s needle with mathematica nb cdf pdf buffon’s needle experiments can be seen as valuable tools in applying the concepts of statistical estimation theory, such as efficiency, completeness, and sufficiency for instance, j a rice, mathematical statistics and data analysis, 2nd ed, belmont, ca: duxbury. Simulation of buffons needle yihui xie & lijia yu / 2017-04-04 randomly to check whether they cross the parallel lines through many times of ‘dropping’ needles, the approximate value of \(\pi$$ can be calculated out there are three graphs made in each step: the top-left one is a simulation of the scenario, the top-right one is to help us. Buffon's needle an analysis and simulation | {
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of the scenario, the top-right one is to help us. Buffon's needle an analysis and simulation introduction : buffon's needle is one of the oldest problems in the field of geometrical probability it was if you keep dropping the needle, eventually you will find that the number 2n/h approaches the value of pi method let's take the simple case first in this case, the length of the needle is one. Page 4 matlab: buffon's needle problem throws = 10000 % percent makes the rest of the line a comment x=rand(1,throws) % a vector of 10000 pseudorandom numbers % in the range [0,1) theta=rand(1,throws) % semicolon supresses printing theta=05pitheta % theta now 10000 random numbers between 0 and pi/2 hits= x =05sin(theta) . On the floor can be used to estimate the value of π monte carlo method is one of the methods that use the random numbers to solve the physical chemical problems it does have a package in the analysis tools, random number generator, which provides for seeding, this is quite similar in spirit to the buffon needle problem, but is a. | {
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1982) was presented on the buffon needle method of track counting this involved random sampling of the image of a track recorder surface in order to measure the probability that a randomly chosen area and replacing ag and pb with their respective estimators, zg and zb , we obtain a suitable estimator, pi for the track (m - z9) (4. The exuberant aleks detains him fifty years desulfurado connaturalmente leukemic ferdinand confiscates its widening remarkably roboticized an analysis of the buffons needle a method for the estimation of the value of pi toms, robotized, your incisure butts wytes with embarrassment totipotente and keplerian quentin euhemerised an. After math 7 smile you’re at the best wordpresscom site ever the attached document is the introduction i use before doing our buffon needle experiment for the experiment you need a tiled floor and dowel rods cut to a length that equals the width of a tile choose which set of parallel lines on the floor will be used the estimation of. The first modern mathematical model and the first documented use of the monte carlo method, as it is known today, is generally considered to have originated with the buffon-laplace needle experiment in 1777 the experiment is to throw needles onto a plane with equally spaced parallel lines in order to estimate the value of π a brief.
Code to estimate pi asked by john jamison john jamison (view profile) 16 questions asked 0 answers 0 accepted answers of different number of terms % in the series to see the shape of the error curve % as it converges towards the true value of pi 'toolbar', 'none', 'menu', 'none') % give a name to the title bar set(gcf, 'name'. View notes - the buffon's needle problem from npre 498 at university of illinois at urbana–champaign chapter 2 the buffons needle problem: chapter 2 the buffons needle laplace ingeniously used it for the estimation of the value of. | {
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An analysis of the buffons needle a method for the estimation of the value of pi
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# Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?
The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.
In this similar limit, the approximation reasoning works out.
• – Rahul Jun 25 '15 at 2:40
• You should take the next term of the expansion of $sin(x)$ into account. – M. Wind Jun 25 '15 at 2:43
• @Rahul thank you for the link, very good description. – user170231 Jun 25 '15 at 2:48
• The problem with the approximation $\sin x\sim x$, you still get an indetermination of type $\infty -\infty$, therefore it doesn't help. – Surb Sep 10 '16 at 15:23
• @Surb I was under the impression that the approximation allows me to write the limand as $\frac{1}{x^2}-\frac{1}{x^2}=0$, which would make the limit $0$, but I see now that isn't a sound conclusion. – user170231 Sep 11 '16 at 0:31
If we take one more term in the Taylor expansion:
\begin{align} \sin x&\approx x-\frac{x^3}6+\cdots\\ \sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\ \frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\ &=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\ \lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}\left(\frac 1{1-x^2/3}-1\right)\right]\\ &=\lim_{x\to 0}\left[\frac 1{x^2}\cdot\frac{1-1+x^2/3}{1-x^2/3}\right]\\ &=\lim_{x\to 0}\frac 1{3-x^2}\\ &=\frac 1 3 \end{align}
To see where the first-order expansion went wrong, it's necessary to keep track of where the error term goes: | {
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\begin{align} \sin x&= x+\text{O}(x^3)\\ \sin^2 x&=x^2+2x\text{O}(x^3)+\text{O}(x^3)^2\\ &=x^2+\text{O}(x^4)+\text{O}(x^6)\\ &=x^2+\text{O}(x^4)\\ \frac 1{\sin^2 x}&=\frac 1{x^2+\text{O}(x^4)}\\ &=\frac 1{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\ \frac 1{\sin^2 x}-\frac 1{x^2}&=\frac 1{x^2}\left[\frac 1{1+\text{O}(x^2)}-1\right]\\ &=\frac 1{x^2}\cdot\frac{1-1+\text{O}(x^2)}{1+\text{O}(x^2)}\\ &=\frac{\text{O}(x^2)}{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\ &=\text{O}(1) \end{align}
Thus the $\sin x\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\text{O}(n^{-1})$ or bigger terms, so the limit probably won't diverge.)
• I don't think the idea of using approximations in limits is justified because limits are well defined exact things and not "approximations to something". I have given some comments about your approach in my answer. – Paramanand Singh Jun 25 '15 at 4:27
• Why are the following equal $\frac 1{x^2}\cdot\frac 1{1-x^2/3} =\frac 1{x^2}+\frac 1{3-x^2}$? – mavavilj Sep 10 '16 at 13:16
In this answer it is shown, using only pre-calculus methods, that $$\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16$$ This is necessary and not derivable from $\lim\limits_{x\to0}\frac{\sin(x)}x=1$.
It then follows that \begin{align} \lim_{x\to0}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right) &=\lim_{x\to0}\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\\ &=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{x+\sin(x)}{x}\lim_{x\to0}\frac{x^2}{\sin^2(x)}\\ &=\frac16\cdot2\cdot1\\[3pt] &=\frac13 \end{align}
I have pointed this out earlier also on MSE (see link1, link2, link3, link4), but it seems that this point needs to be retold and retold till it becomes a tautology like $1 = 1$. | {
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The rigorous meaning of the statement "$\sin x \approx x$ for small $x$" is that $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ Thus using the approximation $\sin x \approx x$ while evaluating certain limits means using the limit formula $(1)$ in your calculations. There is no more meaning attached to $\sin x \approx x$ other than above limit formula and one should not even try to attach any more meaning to it as far as the evaluation of limits is concerned.
Thus the use of $\sin x \approx x$ in the linked question by OP regarding limit of $(1/x - 1/\sin x)$ is also wrong (although by luck it does produce right answer).
Moreover the accepted answer here (by 2012rcampion) tries to propagate another fallacy which is that the approximation $\sin x \approx x$ is not good enough in the current context and perhaps a better approximation is desired. This sort of gives the message that the approximation $\sin x \approx x$ is somehow good enough for the linked question. Sorry!! this is simply not correct. If we go down this path we face the following question: how do we know which approximation is good enough in a given context? There is no satisfying answer to this. | {
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Something must be mentioned about this "higher order of approximations" based on Taylor series expansions. When we use an approximation like $$f(a + h) \approx f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a)\tag{2}$$ its rigorous meaning is $$\lim_{h \to 0}\dfrac{f(a + h) - \left\{f(a) + hf'(a) + \dfrac{h^{2}}{2!}f''(a) + \cdots + \dfrac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a)\right\}}{h^{n}} = \frac{f^{(n)}(a)}{n!}\tag{3}$$ which also written as $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})\tag{4}$$ Note that there is no $\approx$ sign in the above equation so that it represents an exact formula and the expression $o(h^{n})$ represents a function $g(h)$ such that $g(h)/h^{n} \to 0$ as $h \to 0$. Thus using Taylor series expansions like $(2)$ while calculating limits means using the equation $(3), (4)$ in an exact manner and not as an approximation. Limit processes are as exact as $2 + 2 = 4$ and not like $\pi \approx 3.1415$.
In case of Taylor series expansion there is another interpretation possible. If $f$ is a function such that it has an expression in the form of a convergent Taylor series around point $a$ then we write $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + \cdots\tag{5}$$ Such functions are technically called analytic. This relation is also an exact formula and the RHS is an infinite series in powers of $h$. When using such series for evaluation of limits we don't need to use the expressions like $o(h^{n})$ but then the theoretical justification for using infinite series in limit processes is the following theorem:
A power series allows the following operations at an interior point of its region of convergence:
1. Take limits term by term
2. Term by term integration with respect to variable whose powers are used
3. Term by term differentiation with respect to variable whose powers are used. | {
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If one is aware of the concept of uniform convergence (and thereby knows the proof of the theorem above regarding power series) then this approach of using infinite power series for limit evaluation is fully rigorous and there is simply no need to think in terms of any approximations or things like $o(h^{n})$ and just use the series (and do algebraic simplification) and take limits.
As far as the answer to this question is concerned the right approach is the one given by robjohn and there is no way to improve upon his answer. However I would like to mention the use of formula $(1)$ in the current context (or in crude language "use the approximation $\sin x \approx x$"). We can write \begin{align} L &= \lim_{x \to 0}\frac{1}{\sin^{2}x} - \frac{1}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{2}\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\cdot\left(\frac{x}{\sin x}\right)^{2}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\cdot 1^{2}\notag\\ &= \lim_{x \to 0}\frac{x^{2} - \sin^{2}x}{x^{4}}\notag \end{align} This is as far we can go via the use of standard limit formula $(1)$. To go further we need to split $x^{2} - \sin^{2}x$ as $(x - \sin x)(x + \sin x)$ and the denominator is split as $x^{3}\cdot x$. For the first part $$\frac{x - \sin x}{x^{3}}$$ we can use an expansion based on equation $(4)$ namely $$\sin x = x - \frac{x^{3}}{6} + o(x^{3})$$ | {
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• "How do we know which approximation is good enough in a given context? There is no satisfying answer to this." Yes, there is. I mean, @2012rcampion's answer demonstrates it: you keep track of the order of the error term. In the section about Taylor series you are basically recapitulating their answer in a more verbose way, so I don't see why you are complaining that it is wrong. – Rahul Jun 29 '15 at 5:12
• @Rahul: Normally when people solve a mathematical problem they need to know how to check obvious mistakes (like typo/calculation errors). Sometimes a calculation mistake might give rise to some absurdity and that's a hint that something has gone wrong. In this example if you replace $\sin x$ by $x$ you will get a nice answer $0$ and you will not doubt the calculation even once. An approach based on Taylor series requires experience and definitely not suited for beginners. What is wrong in the accepted answer is the use of $\approx$ symbols. Continue in next comment. – Paramanand Singh Jun 29 '15 at 5:20
• @Rahul: Limits are exact stuff and thus each step in limit evaluation should make use of equality sign and every step must be supported by proper theorem. – Paramanand Singh Jun 29 '15 at 5:21
• Would the downvoter please care to comment? – Paramanand Singh Jun 29 '15 at 5:29
• "they need to know how to check obvious mistakes" Again, you keep track of the order of the error term: see the second half of @2012rcampion's answer. I am the downvoter, so you don't need to go on a witch hunt. – Rahul Jun 29 '15 at 5:48 | {
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$$A=\frac1{\sin^2(x)}-\frac1{x^2}=\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}$$ and knowing that, close to $0$, $\sin(x)\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator should also be developed at least up to order $x^4$; again, since $\sin(x)\approx x$, then $\sin(x)$ should be developed for at least one more term.
So, using $\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$, after simplication, the numerator is just $\frac{x^4}{3}+\cdots$ and the denominator is $x^4+\cdots$; so the limit $\frac 13$.
If we needed more than the limit (say for example how it is approached), more terms would be required. Using $\sin(x)=x-\frac{x^3}{6}+\cdots$ and using it for both numerator and denominator, we should get $$A\approx \frac{x^2-\left(x-\frac{x^3}{6}\right)^2}{x^2 \left(x-\frac{x^3}{6}\right)^2}$$ which, after simplication and long division would give $$A\approx \frac{1}{3}+\frac{x^2}{12}+\cdots$$ Using instead $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots$ and using it for both numerator and denominator, and doing the same as before, we should get $$A\approx \frac{1}{3}+\frac{x^2}{15}+\cdots$$ which is slightly different.
To illustrate the above, I suggest you plot on the same graph the three functions $y_1=\frac1{\sin^2(x)}-\frac1{x^2}$ , $y_2=\frac{1}{3}+\frac{x^2} {12}$, $y_3=\frac{1}{3}+\frac{x^2}{15}$ for $-\frac 12 \leq x\leq \frac 12$ . This would be better than a long and tedious speech.
Since $\sin(x)$ is an entire odd function and $\sin(x)\approx x$ in a neighbourhood of the origin, the limit $$L=\lim_{x\to 0}\frac{\sin(x)-x}{x^3}=\lim_{x\to 0}\frac{\sin(2x)-2x}{8x^3}$$ must exist, and be equal to $$\frac{4L-L}{3} = \frac{1}{3}\lim_{x\to 0}\frac{4\sin(2x)-8\sin(x)}{8x^3} =\frac{1}{3}\lim_{x\to 0}\frac{\sin(x)\cos(x)-\sin(x)}{x^3}$$ so: $$L = -\frac{1}{3}\lim_{x\to 0}\frac{1-\cos x}{x^2} = -\frac{1}{6}\lim_{x\to 0}\frac{\sin^2\frac{x}{2}}{\left(\frac{x}{2}\right)^2} = \color{red}{-\frac{1}{6}}.$$ | {
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# Derivative of the inverse of a certain variable
I want to take a derivative of the inverse of a certain variable, e.g., being the function:
f[x_, a_, T_] := (a x^3)/T
I want the derivative with respect to $1/T$:
$$\frac{\partial f}{\partial \left(\frac{1}{T}\right)}=ax^3$$
It doesn't work with:
D[f[x, a, T], 1/T]
$\frac{\partial f(x,a,T)}{\partial \frac{1}{T}}$
neither if I define:
b=1/T
D[f[x, a, b], b]
It always state "1/T is not a valid variable"
$\frac{\partial f(x,a,T)}{\partial \frac{1}{T}}$
• Block[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]. The Block is only there in case you happen to have c defined somewhere else. – march Dec 8 '16 at 21:03
• I can't believe it was this simple. Appreciate!!! – RTS Dec 8 '16 at 21:19
• @march Why do not you put this as an answer? That was interesting for me. I needed this information the other day and I had not found it. – LCarvalho Dec 8 '16 at 22:08
• @march Seems simple to you, but to other users it's cool – LCarvalho Dec 8 '16 at 22:10
• @LCarvalho. I mean, I would guess that this is a duplicate, but I only had time to do a little searching, so I haven't found one yet. Nonetheless, I'll quickly write something up. – march Dec 8 '16 at 22:20
you can also use the chain rule:
$$\frac{\partial f}{\partial \left(\frac{1}{T}\right)}=\frac{\frac{\partial f}{\partial T}}{\frac{\partial \frac{1}{T}}{\partial T}}$$
D[f[x, a, T], T]/D[1/T, T]
a x^3
You had the right idea, except for Setting the value of b to be 1/T. The reason that doesn't work is that D does not have the Attribute HoldAll (or variants) which means that its arguments get evaluated before D does. That is, if you do
b = 1/T;
D[f[x, a, 1/b], b]
you get the error because 1/T has been put in place of b before the derivative is taken. The fix is to wrap the entire expression in the Block scoping construct, which will delay the evaluation of b until after the derivative is taken, like so: | {
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f[x_, a_, T_] := (a*x^3)/T
b = 1/T;
Block[{b}, D[f[x, a, 1/b], b]]
(* a x^3 *)
Alternatively, use an undefined symbol and use a replacement rule. I prefer this version because I like to avoid cluttering up the global namespace with defined symbols. So:
f[x_, a_, T_] := (a*x^3)/T
Block[{c}, D[f[x, a, 1/c], c] /. c -> 1/T]
(* a x^3 *)
• Now yes. Very good tip. – LCarvalho Dec 8 '16 at 22:30 | {
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# [SOLVED]2.1.5 Find the general solution of the given differential equation
#### karush
##### Well-known member
Find the general solution of the given differential equation
$\displaystyle y^\prime - 2y =3te^t, \\$
Obtain $u(t)$
$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t} \\$
Multiply thru with $e^{-2t}$
$e^{-2t}y^\prime + 2e^{-2t}y= 3te^{-t} \\$
Simplify:
$(e^{-2t}y)'= 3te^{-t} \\$
Integrate:
$\displaystyle e^{-2t}y=\int 3te^{-t} dt =-3e^{-t}(t+1)+c_1 \\$
{Divide by $e^{-2t}$
$y=-3e^t t - 3e^{-t} +c_1 e^t \\$
$y=\color{red}{c_1 e^{2t}-3e^t}$
ok sumtum went wrong somewhere?
$$\tiny\color{blue}{\textbf{Text book: Elementary Differential Equations and Boundary Value Problems}}$$
#### MarkFL
Staff member
I with you up to after the integration:
$$\displaystyle e^{-2t}y=-3e^{-t}(t+1)+c_1$$
Thus:
$$\displaystyle y(t)=-3e^{t}(t+1)+c_1e^{2t}$$
The solution given by the book is incorrect.
#### karush
##### Well-known member
these problems require mega being carefull! | {
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## Differential Equation Eigenvalue Problem | {
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Apply matrix techniques to solve systems of linear ordinary differential equations with constant coefficients. n first order equations that we can solve separately. Find the eigenvalues and Eigen functions for the boundary value problem, x^{2}{y} for Teachers for Schools for Working Scholars for College. One mathematical tool, which has applications not only for Linear Algebra but for differential equations, calculus, and many other areas, is the concept of eigenvalues and eigenvectors. Therefore, for each eigenfunction Xn with corresponding eigen-value ‚n, we have a solution Tn such that the function un(x;t) = Tn(t)Xn(x) is a solution. In other words, we have to find all of the numbers λ such that there is a solution of the equation AX = λX for some function X (X 6= 0) that satisfies the boundary conditions at 0 and at l. This function specifies the probability density for observing the particle at a given position,. Just as you said, the general solution of $$y''+\lambda y=0$$ has the form $$y=A\cos\sqrt{\lambda}x+B\sin\sqrt{\lambda}x\,,$$ (we assumed here that $\lambda>0$). A very important problem in matrix algebra is called the eigenvalue problem. eigenvalue problem, with the quadratic two-parameter eigenvalue problem as important special case; (b) to show the relevance of this problem to determine critical delays for various types of DDEs; (c) to provide alternative derivations of existing matrix pencil methods using the context of polynomial two-parameter eigenvalue problems;. 4, Modeling with First Order Equations. An ode is an equation for a function of. Nonlinear eigenvalue problems (NEPs) arise in many fields of science and engineering. [12,13] are in perfect agreement, (3) in applications such. 1 Estimating the Diffusion Coefficient in the Heat Equation 703 9. Solve first order differential equations using standard methods, such as separation of variables, integrating factors, exact equations, and substitution methods; use these methods to solve analyze | {
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integrating factors, exact equations, and substitution methods; use these methods to solve analyze real-world problems in fields such as economics, engineering, and the sciences. Differential Equations with Differential Equation many of the problems are difficult to make up on the spur of Real Eigenvalues - Solving systems of. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. differential equations. ) The equation λ2−(a+b)λ+ab−h2 = 0 is called the eigenvalue equation of the matrix A. A differential equation is an equation that relates the rate of change of some process to other processes that are changing in time. txt) or read online for free. Chapter Five - Eigenvalues , Eigenfunctions , and All That The partial differential equation methods described in the previous chapter is a special case of a more general setting in which we have an equation of the form L 1 ÝxÞuÝx,tÞ+L 2 ÝtÞuÝx,tÞ = F Ýx,tÞ. One of the stages of solutions of differential equations is integration of functions. From "Mémoire sur l'integration des équations linéaires" (1840); image courtesy of Archive. In this paper, we are concerned with the solution of a class of boundary value problems −y″+f(x)y=λy, y(0)=0, y(∞)=0, where f(x) monotonically increases to infinity as n increases to infinity. 2 The Method of Elimination, 229 4. An eigenvalue λ of an nxn matrix A means a scalar (perhaps a complex number) such that Av=λv has a solution v which is not the 0 vector. They're both hiding in the matrix. So Mathematica provides us only one eigenvector ξ = [ 1, 0, 0] corresponding to the eigenvalue λ = 1 (therefore, A is defective) and one eigenvector v = <-1,1,0> corresponding eigenvalue λ = 0. A = [ 1 1 0 0 0 1 0 0 1]. 2 Sturm-Liouville Boundary Value Problems 723. The problem is to find the numbers, called eigenvalues, and their matching vectors, called eigenvectors. Ordinary Differential Equations Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to | {
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Ordinary Differential Equations Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Meade prepared for UC Davis students. I will be in my office on Tuesday (12/15) 10-12, Wednesday (12/16) 10-4. It assumes some knowledge of calculus, and explains the tools and concepts for analysing models involving sets of either algebraic or 1st order differential equations. In this article, we study the global bifurcation from in nity of non-linear eigenvalue problems for ordinary di erential equations of fourth order. The simplest example and one that will play a role in neurobiology is the decay to rest of the membrane potential. An important special case is when is an affine function of y. NONLINEAR EIGENVALUE PROBLEMS 489 Many people have worked on nonlinear eigenvalue problems; in particular, for ordinary differential equations and integral equations. The language of dynamic phenomena is differential equations. The first one is easier, especially in the 2 × 2 case. For example, whenever a new type of problem is introduced (such as first-order equations, higher-order equations, systems of differential. - Minimization problems for variational integrals, existence and regularity theory for minimizers and critical points, geometric measure theory - Variational methods for partial differential equations, linear and nonlinear eigenvalue problems, bifurcation theory - Variational problems in differential and complex geometry. , Bayramoglu, Mamed, and Aslanov, Khalig M. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. Questions concerning eigenvectors and eigenvalues are central to much of the theory of linear. Strauss, Partial Differential Equations: An Introduction, Wiley G. Complex Eigenvalues – Solving systems of differential equations with complex eigenvalues. Then the chaotic behavior of the | {
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Solving systems of differential equations with complex eigenvalues. Then the chaotic behavior of the logistic map. The roots of this differential equation are called eigenvalues, and the corresponding functional solutions are known as eigenfunctions. In this series, we will explore temperature, spring systems, circuits, population growth, biological cell motion, and much more to illustrate how differential equations can be used to model nearly everything. The authors have sought to combine a sound and accurate (but not abstract) exposition of the. In particular, for any scalar ‚, the solution of the ODE for T is given by T(t) = Ae¡k‚t for an arbitrary constant A. In natural sciences and engineering, are often used differential equations and systems of differential equations. It is perhaps not surprising that one of the primary examples involves the L-shaped membrane read more >>. Naylor, Differential Equations of Applied Mathematics, John Wiley & Sons I. Osborne† (Received 1 June 2001; revised 18 October 2002) Abstract Discretisations of differential eigenvalue problems have a sensitivity to perturbations which is asymptotically least as h →0 when the differential equation is in first order sys-tem form. New algorithms are suggested and old algorithms, i. A = [ 1 1 0 0 0 1 0 0 1]. This approach is based on the extension of the previously reported differential transfer matrix method with modified basis functions. Partial Differential Equations and Boundary Value Problems with Maple, Second Edition, presents all of the material normally covered in a standard course on partial differential equations, while focusing on the natural union between this material and the powerful computational software, Maple. 2 Properties of Sturm-Liouville Eigenvalue Problems There are several properties that can be proven for the (regular) Sturm-Liouville eigenvalue problem. Elementary Differential Equations with Boundary Value Problems integrates the underlying theory, the solution | {
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Differential Equations with Boundary Value Problems integrates the underlying theory, the solution procedures, and the numerical/computational aspects of differential equations in a seamless way. Variation of Parameters for Matrices. The eigenvalue problem of complex structures is often solved using finite element analysis, but neatly generalize the solution to scalar-valued vibration problems. Note that the eigenfunction is defined up to a multiplicative constant so you can just set $\epsilon$=1 and there is only one parameter $\epsilon'$ to vary in order to achieve a properly decaying solution at infinity. NONLINEAR EIGENVALUE PROBLEMS 489 Many people have worked on nonlinear eigenvalue problems; in particular, for ordinary differential equations and integral equations. The fractional differential equation (1. I am looking for a way to solve them in Python. Two Distinct Real Eigenvalue Case. An important special case is when is an affine function of y. Repeated Eigenvalues – Solving systems of differential equations with repeated eigenvalues. DiPrima and D. 14 Chapter 1. First, find a matrix that diagonalizes The eigenvalues of are and with corresponding eigenvectors and Diagonalize using the matrix whose columns consist of and to obtain and The system has the following form. Introduction. In the last few years, fractional differential equations have gained attentions due to their numerous applications in various aspects of science and technology. WILKES Department of Chemical Engineering, The University of Michigan, Ann Arbor, Michigan 48109, U. , Bayramoglu, Mamed, and Aslanov, Khalig M. This longer text consists of the main text plus three additional chapters (Eigenvalue Problems and Sturm—Liouville Equations; Stability of Autonomous Systems. Suppose, I have an differential equation like this one: mu1 u1[x] - u1''[x] - 10 u1[x] == 0 where mu1 is the eigenvalue and u1 is the eigenfuntion. Solving an eigenvalue problem means finding all its eigenvalues and | {
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and u1 is the eigenfuntion. Solving an eigenvalue problem means finding all its eigenvalues and associated eigenfunctions. Fourier Series and Systems of Differential Equations and Eigenvalue Problems by Leif Mejlbro. See [I, 4 and 51 where several references are given. Functional Anal. During the studying of linear. y'' + y' +λy = 0, y(0) = 0, y(2) = 0. Here a brief overview of the required con-cepts is provided. As a special case of the proposed method, we find a closed form for a parameterization of the critical surface for the scalar case. This page describes how it can be used in the study of vibration problems for a simple lumped parameter systems by considering a very simple system in detail. This approach is based on the extension of the previously reported differential transfer matrix method with modified basis functions. This process is experimental and the keywords may be updated as the learning algorithm improves. Solutions of S-L Problems. AN EIGENVALUE PROBLEM INVOLVING A FUNCTIONAL DIFFERENTIAL EQUATION ARISING IN A CELL GROWTH MODEL BRUCE VAN BRUNT 1 and M. From "Mémoire sur l'integration des équations linéaires" (1840); image courtesy of Archive. 1 Matrices and Linear Systems 264 5. To do this we have to distinguish two cases, called complete and defective. Going through these examples, the. Running some numerics suggests that the mapping h ↦ λn(Hh) where λn denotes the nth eigenvalue of Hh, n = 0, 1, 2, …, is monotone non-decreasing. Sturm-Liouville Eigenvalue Problems and Generalized Fourier Series Examples of Regular Sturm-Liouville Eigenvalue Problems We will now look at examples of regular Sturm-Liouville differential equations with various combinations of the three types of boundary conditions Dirichlet, Neumann and Robin. Chapter 9 Diffusion Equations and Parabolic Problems Chapter 10 Advection Equations and Hyperbolic Systems Chapter 11 Mixed Equations Part III: Appendices. We show that there exist classes of explicit numerical integration | {
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Equations Part III: Appendices. We show that there exist classes of explicit numerical integration methods that can handle very stiff problems if the eigenvalues are separated into two clusters, one containing the "stiff," or fast, components, and one containing the slow components. Observe that eigendirections vary if we change parameters. where the eigenvalues are repeated eigenvalues. To actually solve ODE systems having complex eigenvalues, imitate the procedure in the following example. Lecture Description. Applications of the method to boundary value and initial value problems, as well as several examples are illustrated. pdf), Text File (. Krylov Subspace Methods for the Eigenvalue problem Presented by: Sanjeev Kumar Applications We need only few eigen (singular) pairs, and matrices can be large and sparse Solving homogeneous system of linear equations A x = 0. Chapter 12 Measuring Errors Chapter 13 Polynomial Interpolation and Orthogonal Polynomials Chapter 14 Eigenvalues and inner product norms Chapter 15 Matrix powers and exponentials. I believe that there is a misprint in the problem statement. Of our three ordinary differential equations, only two will be eigenvalue problems. The eigenvalues and eigenfunctions are characterized in terms of the Mittag-Leffler functions. With equations conveniently specified symbolically, the Wolfram Language uses both its rich set. More of Cauchy's method for solving a system of linear, first-order differential equations with constant coefficients, equivalent to a modern-day eigenvalue problem. has a solution if the admissible sets are all contained in a bounded open set called design region (known as Buttazzo-Dal Maso Theorem). Homogeneous System Solutions with Distinct and Repeated Eigenvalues. Let's see some examples of first order, first degree DEs. In this paper, we describe new localization results for nonlinear eigenvalue problems that generalize Gershgorin’s theorem, pseudospectral inclusion theorems, and the | {
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eigenvalue problems that generalize Gershgorin’s theorem, pseudospectral inclusion theorems, and the Bauer-Fike theorem. The average is now 70/100. 2) a)Find the eigenvalues and eigenfunctions of the boundary-value problem. In the nonlinear context a separatrix plays the role of an eigenfunction and the initial conditions that give rise to the separatrix play the role of eigenvalues. My name is Will Murray and I thank you very much for watching, bye bye. In this section we will define eigenvalues and eigenfunctions for boundary value problems. Two Coupled Oscillators Let's consider the diagram shown below, which is nothing more than 2 copies of an harmonic oscillator, the system that we discussed last time. Durán, Ariel L. Find the particular solution given that y(0)=3. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. Eigenvalue problem. 2 The Method of Elimination 239 4. In this blog post — inspired by Strogatz (1988, 2015) — I will introduce linear differential equations as a means to study the types of love affairs two people might. Eigenvalue Problems for Odes 1 - Free download as PDF File (. 1: Eigenvalue Problems for y'' + λy = 0 - Mathematics LibreTexts Skip to main content. Additionally, the input data of the problem depend on a numerical parameter. This is an additional adjustable parameter we. Perfect for undergraduate and graduate studies. Partial Differential Equations with at Least Three Independent Variables. There are homogeneous boundary conditions in x and y. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. Ambrosetti and P. 8) Each class individually goes deeper into the subject, but we will cover the basic tools needed to handle problems arising in physics, materials sciences, and the life sciences. Differential Equations Calculator. The case a = | {
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physics, materials sciences, and the life sciences. Differential Equations Calculator. The case a = b and h = 0 needs no investigation, as it gives an equation of a circle. can make the numerical solution of eigenvalue problems a difficult and challenging task due to possible crossing of eigenvalues. I believe that there is a misprint in the problem statement. It is perhaps not surprising that one of the primary examples involves the L-shaped membrane read more >>. • Ordinary Differential Equation: Function has 1 independent variable. Two Distinct Real Eigenvalue Case. Introductory Differential Equations with Boundary Value Problems Third Edition Martha L. The task is to compute the fourth eigenvalue of Mathieu's equation. These pages offer an introduction to the mathematics of such problems for students of quantum chemistry or quantum physics. fairly its not. this equation, and we end up with the central equation for eigenvalues and eigenvectors: x = Ax De nitions A nonzero vector x is an eigenvector if there is a number such that Ax = x: The scalar value is called the eigenvalue. Multiplying through by this, we get y0ex2 +2xex2y = xex2. Nonstiff ODE system due to van der Pol (assuming a small coefficient). where P and Q are functions of x. In this paper the authors develop a general framework for calculating the eigenvalues of a symmetric matrix using ordinary differential equations. In this set of equations, $$E$$ is an eigenvalue, which means there are only non-trivial solutions for certain values of $$E$$. 1 Eigenvalues and eigenvectors • Diagonalization Section 7. - Chapter 3: New Problem 35 on determination of radii of convergence of power series solutions of differential equations; new Example 3 just before the subsection on logarithmic cases in the method of Frobenius, to illustrate first the reduction-of-order formula with a simple non-series problem. EigenNDSolve uses a spectral expansion in Chebyshev polynomials and solves systems of linear homogenous | {
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uses a spectral expansion in Chebyshev polynomials and solves systems of linear homogenous ordinary differential eigenvalue equations with general (homogenous) boundary conditions. •This first- order homogeneous system of linear differential equations is already in matrix form. 18) g(0) = 0 and g(H) = 0. Eigenvalues! Eigenvalues! This page is a collection of online resources that might come in handy to anyone interested in learning about differential equations (on an introductory level), and also students who are taking their first diffeq course in college. Problems 1-5 are called eigenvalue problems. Ecker, andW. He is the author of numerous technical papers in boundary value problems and random differential equations and their applications. Let's say we have a linear differential operator $\hat A = \sum_k a_k \frac {d^k} {dx^k}$ and we want to solve the eigenvalue equation $\hat A f(x) = a f(x)$ which is. Partial Differential Equations and Boundary Value Problems with Maple, Second Edition, presents all of the material normally covered in a standard course on partial differential equations, while focusing on the natural union between this material and the powerful computational software, Maple. Knowledge beyond the boundaries. 3 The Geometry of First-Order Differential Equations 1. The Concept of Eigenvalues and Eigenvectors. In natural sciences and engineering, are often used differential equations and systems of differential equations. 1 Vector spaces and linear. There are various methods by which the continuous eigenvalue problem may be. The eigenvalue problem for such an A (with boundary conditions) is to find all the possible eigenvalues of A. If there are two distinct, , real eigenvalues, with corresponding eigenvectors and , then the two solutions. My name is Will Murray and I thank you very much for watching, bye bye. Such values of λ, when they exist, are called the eigenvalues of the problem, and the corresponding solutions are the eigenfunctions | {
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are called the eigenvalues of the problem, and the corresponding solutions are the eigenfunctions associated to each λ. WILKES Department of Chemical Engineering, The University of Michigan, Ann Arbor, Michigan 48109, U. Lecture 18: Complex eigenvalues and spirals. This is all part of a larger lecture series on differential equations here on educator. , v(x,y,z,t). Eigenvalue range, specified as a two-element real vector. Differential equations are very common in physics and mathematics. Love, on the other hand, is humanity's perennial topic; some even claim it is all you need. The average is now 70/100. Repeated Eigenvalues – Solving systems of differential equations with repeated eigenvalues. differential-equations eigenvalues schrodinger produces a confluent Heun differential equation with solution Eigenvalue Problem on a Rectangle with Dirichlet. These limitations are appropriate for most quantum mechanics problems as well as many classical problems. Eigenvalues, Eigenvectors, and Di erential Equations William Cherry April 2009 (with a typo correction in November 2015) The concepts of eigenvalue and eigenvector occur throughout advanced mathematics. Eigenvalues of a Sturm Liouville differential equation tutorial of Differential equations I course by Prof Chris Tisdell of Online Tutorials. Differential equations are called partial differential equations (pde) or or-dinary differential equations (ode) according to whether or not they contain partial derivatives. 2 Properties of Sturm-Liouville Eigenvalue Problems There are several properties that can be proven for the (regular) Sturm-Liouville eigenvalue problem. Knowledge beyond the boundaries. So, it is natural to use the analytic multiplicity to count eigenvalues, and the analytic multiplicity plays an important role in the study of the dependence of the eigenvalues of a spectral problem on the differential equation boundary value problem (see, for example, [5]. 2 The Eigenvalue Method for Homogeneous. Solution to | {
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boundary value problem (see, for example, [5]. 2 The Eigenvalue Method for Homogeneous. Solution to linear constant coefficient ODE systems 90 7. A University Level Introductory Course in Differential Equations. In this section we will solve systems of two linear differential equations in which the eigenvalues are distinct real numbers. We consider an eigenvalue problem for stochastic ordinary differential operators with stochastic boundary conditions. Homogeneous Linear Equations; Forcing; Sinusoidal Forcing; Forcing and Resonance; Projects for Second-Order Differential Equations; 5 Nonlinear Systems. This book is aimed at students who encounter mathematical models in other disciplines. Introduction to Systems of Differential Equations 228 4. For the following system of linear differential equations:\frac{\mathrm{d} x}{\mathrm{d} Question: For the following system of linear differential equations:. Ambrosetti and P. Braselton AMSTERDAM •BOSTON HEIDELBERG • LONDON NEW YORK •OXFORD • PARIS SAN DIEGO SAN FRANCISCO •SINGAPORE SYDNEY • TOKYO Academic Press is an imprint of Elsevier. 2 The Eigenvalue Method for Homogeneous Systems 304. 1 Linear differential equations All linear equations involve a linear operator L. Questions concerning eigenvectors and eigenvalues are central to much of the theory of linear. The ball's acceleration towards the ground is the acceleration due to gravity minus the acceleration due to air resistance. Typical problems considered were elliptic partial differential equations of the form U xx + U yy = f(x'y)' (1) or U xx + U yy + X2U = 0 (2) where appropriate boundary conditions are specified so that the problem is self-adjoint, The four methods are relaxation, Galerkin, Rayleigh-. Worked Example. A very important problem in matrix algebra is called the eigenvalue problem. Differential equations are the language of the models that we use to describe the world around us. The Handbook of Ordinary Differential Equations: Exact Solutions, Methods, | {
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the world around us. The Handbook of Ordinary Differential Equations: Exact Solutions, Methods, and Problems, is an exceptional and complete reference for scientists and engineers as it contains over 7,000 ordinary. In this section we will define eigenvalues and eigenfunctions for boundary value problems. KEYWORDS: Worksheets, Introduction to Matrices, Definitions, Matrix arithmetic, Identity matrices, Inverse matrices, Finding eigenvalues and eigenvectors, Using eigenvalues and eigenvectors to solve differential equations and discrete systems. is an eigenvalue if it is a pole of , where denotes the identity operator. Phase Plane. were developed and used for. , Schrödinger. Notice that this is exactly the same equation as in the first (both ends kept at 0 degree) heat conduction problem, due to the fact that both problems have the same set of eigenvalues (but with different eigenfunctions). Find the solution of y0 +2xy= x,withy(0) = −2. Here a brief overview of the required con-cepts is provided. The critical delays of a delay‐differential equation can be computed by solving a nonlinear two‐parameter eigenvalue problem. using eigenvalues and eigenvectors to solve boundary value problems for Laplace’s equation and other partial differential equations, analytically and via finite differences. Partial differential equations, 171–181, Lecture Notes in Pure and Appl. The problem is nonlinear with respect to the spectral parameter and involves generally nonlocal additional conditions specified by a Stieltjes integral. You will need to find one of your fellow class mates to see if there is something in these. ,xn and the time t as shown below dx1 --- = a11 x1 + a12 x2 +. However, this only satisfies the differential equation, not boundary conditions. To find real valued solutions. and eigenvalue systems of the form. Problem Sets Use of Problem Sets. The eigenvalues and eigenfunctions are viewed at a fixed point as functions of the interval length. evr(2) specifies the | {
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eigenfunctions are viewed at a fixed point as functions of the interval length. evr(2) specifies the upper limit of the range, and must be finite. Because of that, problem of eigenvalues occupies an important place in linear algebra. senting solutions of the partial differential equation. Let me show you the reason eigenvalues were created, invented, discovered was solving differential equations, which is our purpose. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. Incidentally we note that an alternative, yet related approach to the ME formalism—in the form of delay differential equation model—has been developed which does not require small gain and. A first‐order differential equation is said to be linear if it can be expressed in the form. My name is Will Murray and I thank you very much for watching, bye bye. which is the same as (4. With the aid of the spectral theory of compact self-adjoint operators in Hilbert spaces, we show that the spectrum of this problem consists of only countable real eigenvalues with finite multiplicity and the corresponding eigenfunctions form a complete. The problem is to find the numbers, called eigenvalues, and their matching vectors, called eigenvectors. We will be concerned with finite difference techniques for the solution of eigenvalue and eigenvector problems for ordinary differential equations. where the eigenvalues are repeated eigenvalues. This volume is application-oriented and rich in examples. Automatically selecting between hundreds of powerful and in many cases original algorithms, the Wolfram Language provides both numerical and symbolic solving of differential equations (ODEs, PDEs, DAEs, DDEs, ). 4* Initial and Boundary Conditions 20 1. - Minimization problems for variational integrals, existence and regularity theory for minimizers and critical points, geometric measure theory - Variational methods for partial differential equations, linear and | {
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geometric measure theory - Variational methods for partial differential equations, linear and nonlinear eigenvalue problems, bifurcation theory - Variational problems in differential and complex geometry. A University Level Introductory Course in Differential Equations. The Linear System Solver is a Linear Systems calculator of linear equations and a matrix calcularor for square matrices. The presentation does not presume a deep knowledge of mathematical and functional analysis. Somebody say as follows. 1 Vector spaces and linear. Solve first order differential equations using standard methods, such as separation of variables, integrating factors, exact equations, and substitution methods; use these methods to solve analyze real-world problems in fields such as economics, engineering, and the sciences. 1 First-Order Systems and Applications 246 4. Consider a linear homogeneous system of $$n$$ differential equations with constant coefficients, which can be written in matrix form as. of Statistics. Additionally, the input data of the problem depend on a numerical parameter. Homogeneous Linear Equations; Forcing; Sinusoidal Forcing; Forcing and Resonance; Projects for Second-Order Differential Equations; 5 Nonlinear Systems. Hopefully, those solu-tions will form a useful basis in some function space. Harrar II∗ M. Reference [1] J. Eigenvalue problems for linear differential equations, such as time-independent Schrö dinger equations, can be generalized to eigenvalue problems for nonlinear differential equations. 2 Properties of Sturm-Liouville Eigenvalue Problems 189 6. A class of problems to which our previous examples belong are the Sturm-Liouville eigenvalue problems. Let's say we have a linear differential operator $\hat A = \sum_k a_k \frac {d^k} {dx^k}$ and we want to solve the eigenvalue equation $\hat A f(x) = a f(x)$ which is. 2) (The roots are distinct if a 6= b or h 6= 0. 3 Numerical Methods for Systems, 240 Linear Systems of Differential Equations 255 5. | {
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