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Recall, you can add constants together into a single constant (as was done with C).
Also, you can define a $C_1$ and $C_2$ - one to each side. You can also show the constant as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest.
Example: Consider the separable equation:
$$y' = x y$$
After separation, we can integrate both sides as:
$$\int \dfrac{1}{y}~dy = \int x~dx$$
Approach 1: Single constant (we could have also put $C$ on the LHS - try this)
$$\ln y = \dfrac{x^2}{2} + C$$
We take the exponential of each side and have:
$$y = e^{x^2/2 + C} = e^{x^2/2}~e^C = w~e^{x^2/2}$$
Note: $w = e^C$, which is just some constant (totally arbitrary).
Approach 2: Constant on each side
$$\ln y + c_1 = \dfrac{x^2}{2} + c_2$$
Taking exponential of both sides:
$$e^{\ln y + c_1} = e^{x^2/2 + c_2}$$
The RHS is as above and the LHS, we have ($q$ is just any arbitrary constant):
$$e^{\ln y}~e^{c_1} = y~q$$
Now we write:
$$q y = c_2~e^{x^2/2}$$
Dividing constants and just calling it $w$, we have:
$$y(x) = w~e^{x^2/2}$$
In both approaches, we just get some constant $w$.
Now, if you were provided with an initial condition like $y(0) = 2$, you would plug in $x=0$ and see that $w = 2$. | {
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# Maximise population coverage subject to budget constraint
Let $t_i =$
$1$ if transmitter i is to be constructed
and $0$ otherwise,
$c_j =$
$1$ if community j is covered
and $0$ otherwise.
Obj func:
Max
$$z = [10, 15, ..., 10] \cdot c$$
s.t.
1. the budget constraint
$$[3.6, 2.3, ..., 3.10] \cdot t \le 15$$
1. If community $j$ is covered, it is done so by at least one constructed transmitter $i$:
eg if $c_1$ is covered, then $t_1$ or $t_3$ is constructed:
$$c_1 \to (t_1 \bigvee t_3)$$
$$\iff \neg c_1 \bigvee (t_1 \bigvee t_3)$$
$$\iff 1 - c_1 + t_1 + t_3 \ge 1$$
$$\iff c_1 \le t_1 + t_3$$
Similarly, we have:
$$c_2 \le t_1 + t_2$$
$$\vdots$$
$$c_{15} \le t_7$$
1. If a transmitter $i$ is constructed, at least one community $j$ is covered:
eg if $t_1$ is constructed, then $c_1$ and $c_2$ are covered:
$$t_1 \to (c_1 \bigwedge c_2)$$
$$\iff \neg t_1 \bigvee (c_1 \bigwedge c_2)$$
$$\iff (\neg t_1 \bigvee c_1) \bigwedge (\neg t_1 \bigvee c_2)$$
$$\iff 1 - t_1 + c_1 \ge 1 \ \text{and} \ 1 - t_1 + c_2 \ge 1$$
$$\iff c_1 \ge t_1 \ \text{and} \ c_2 \ge t_1$$
Similarly, we have:
$$c_2, c_3, c_5 \ge t_2$$
$$c_1, c_7, c_9, c_{10} \ge t_3$$
$$\vdots$$
$$c_{12}, c_{13}, c_{14}, c_{15} \ge t_7$$
Is that right?
From Chapter 3 here.
• Note if you choose transmitters $1, 2$, then there is overlap of communities, and your function will double count the population of community $2$. – Macavity Mar 15 '16 at 5:05
• @Macavity Oh thanks. What to do then? $z'=z-y \cdot x$ where $y_i$ corrects for double or triple counting? – BCLC Mar 15 '16 at 5:10
• @Macavity I think Kuifje found away around the double count. Do you agree? – BCLC Mar 17 '16 at 17:54
Here is a way to get the constraints right. Define $x_i$ exactly as you have done. Define $p_i$ to be an indicator $1/0$ depending on whether community $i$ has been covered or not. | {
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The budget constraint remains as you have set, and the objective function is now of form $\max z = 10p_1+15p_2 + \cdots + 10p_{15}$
Now how do we ensure that $p_i$ is set correctly, for any given choice of $\{x_i\}$? One way is to note that the objective function gives positive weight to $p_i$, so define a constraint for each population based on transmitters which could cover it - e.g. for the first one : $p_1 \le x_1+x_3$, for the second $p_2 \le x_1 + x_2$ etc. This will force the population indicator to turn $0$ if none of the relevant transmitters are selected.
• Thanks Macavity. Edited. How is it? – BCLC Mar 15 '16 at 6:08
• Your $p_i$ needs to be replaced by $c_i$ in the objective function and $x_i$ has no role to play in the objective. – Macavity Mar 15 '16 at 6:10
• Thanks. We answered in class, and we sort of got different answers. Prof is still going to think about it. How do you find sol? – BCLC Mar 15 '16 at 11:31
• Using an $M$ is also a good way, though not needed here. Try out the Integer Programs and see how it works. – Macavity Mar 15 '16 at 11:43
• @BCLC Yes, your try 4 seems what I meant, and it works. – Macavity Mar 20 '16 at 18:31
This is a classical set covering problem.
Let $y_i$ be a binary variable that equals $1$ if and only if transmitter $i=1,\cdots,7$ is built, and $x_j$ another binary that equals $1$ if community $j$ is covered. Let $p_j$ be the population of community $j=1,\cdots,15$.
You want to maximize the total coverage: $$\mbox{Maximize }Z=\sum_{j=1 }^{15}p_jx_j$$ subject to budget constraints: $$\sum_{i=1}^7c_iy_i\le 15 \\$$ To link the variables, proceed like Macavity suggests: $$x_j\le \sum_{i\;|\;i \;covers\; j}y_i\quad \forall j=1,\cdots,15\\ y_i,x_j\in \{0,1\}\quad \forall i=1,\cdots,7\;\forall j=1,\cdots,15$$ | {
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• Sorry I was editing while you asked the question. Variables $x_j$ will make sure there is no double counting, and are activated as soon as one of the variables $y_i\; |\;i$ covers $j$ is. – Kuifje Mar 17 '16 at 18:03
• No apologies needed. I'm the one asking a favour. It's really 'Try 2' and not 'Try 1' ? My prof seems to disagree with Macavity – BCLC Mar 17 '16 at 18:14
• What do you mean by 'Try 1' and 'Try 2'? I think Macavity is right. He basically explained the nature of my second constraint. What specific point does your professor disagree with? – Kuifje Mar 17 '16 at 18:25
• I disagree with "Try1". Try 1 (first equation) implies that if communities $1$ and $2$ are covered, than transmitter $1$ is built. This is not necessarily true, as they can be covered by transmitters $3$ and $2$ for example. It is the other way around: if community $1$ is covered, than at least one transmitter among those that cover the community must be built. – Kuifje Mar 18 '16 at 0:34
• yes indeed, for the same reasons. – Kuifje Mar 18 '16 at 14:30 | {
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# Why does factoring eliminate a hole in the limit?
$$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)$$
I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much. | {
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• Very laudable that you want to understand instead of "just memorize". Aug 7, 2013 at 19:21
• The expression $\frac{(x+5)(x-5)}{x-5}=x+5$, just so long as $x\neq 5$, since $\frac{(x+5)(x-5)}{x-5}$ has the bad fortune of not being defined at $x=5$. Aug 7, 2013 at 19:29
• I don't understand what your question is. Are you asking why the function $\frac{x^2-25}{x-5}$ is undefined at $x=5$? or why we're allowed to cancel $(x-5)$ from the top and bottom? or why the limits before and after cancellation are equal? Aug 7, 2013 at 20:09
• @BlueRaja-DannyPflughoeft yes, all of those questions I'm asking :) Aug 7, 2013 at 22:35
• To answer the question - I've only been told the methodology of expanding the x2−25 into (x−5)(x+5). It goes like this: (in reverse) (x-5)(x+5) = x(x+5)-5(x+5) = x^2+5x-5x-25 = x^2-25 So, working above the other way takes x^2-25 back to (x-5)(x+5) Aug 8, 2013 at 22:52
First, and by definition, when dealing with
$$\lim_{x\to x_0}f(x)$$
we must assume $\,f\,$ is defined in some neighborhood of $\,x_0\,$ except , perhaps, on $\,x_0\,$ itself, and from here that in the process of taking the limit we have the right and the duty to assume $\,x\,$ approaches $\,x_0\,$ in any possible way but it is never equal to it.
Thus, and since in our case we always have $\,x\ne x_0=5\,$ during the limit process , we can algebraically cancel for the whole process.
$$\frac{x^2-25}{x-5}=\frac{(x+5)\color{red}{(x-5)}}{\color{red}{x-5}}=x+5\xrightarrow[x\to 5]{}10$$
The above process shows that the original function behaves exactly as the straight line $\,y=x+5\,$ except at the point $\,x=5\,$ , where there exists "a hole", as you mention. | {
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• said hole can, of course be filled in by plugging it with the limit ... Aug 13, 2013 at 22:33
• Some definitions of the limit do not require that we consider points that are not equal to $x_0$. How do you handle these cases then? The definition goes as $\lim f(x) = L$ provided for each $\epsilon\gt 0$ there is some $\delta \gt 0$ such that $\vert x - x_0\vert \lt \delta\ldots$ and so on. Sep 9, 2016 at 15:50
• @AlexOrtiz The epsilon delta definition of the limit for general metric spaces has $0 < d(x,c) < \delta$. Note the first inequality, which implies $x \neq c$. Nov 21, 2020 at 0:19
This image of mine seems apropos:
In the case of $\lim_{x\to 5} \frac{x^2-25}{x-5}$, the message here is: Away from $x=5$, the function $\frac{x^2-25}{x-5}$ is completely identical to $x+5$; thus, what we expect to find as we approach $x=5$ is the value $5+5$. This anticipated value is what a limit computes.
The fact that the original function isn't defined at $x=5$ is immaterial. Walley World may be closed for repairs when you arrive, but that doesn't mean you and your dysfunctional family didn't spend an entire cross-country road trip anticipating all the fun you'd have there. | {
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• Extremely helpful answer as always, Blue. You're incredible at understanding what people are really asking and providing strong intuition. Thank you for your contribution to this site! Jun 1, 2015 at 15:55
• @Blue Your answer made me think something: Can we always find another function which - away from a certain value - is completely identical to the former function? I mean, what is done in this example is the following: We have an algebraic expression and we can find another algebraic expression that have that property. Jul 6, 2017 at 1:21
• BTW: I read your image as the jingles in this video. Jul 6, 2017 at 1:21
• Super answer...the craziest I ever read in here Jul 13, 2019 at 22:17
• @Blue: Shameless plug: A downloadable poster version of the "Journey/Destination" image is available at my Etsy shop, and a t-shirt is available from Spring. (I'll delete this comment if there are objections.)
– Blue
Oct 12, 2021 at 10:00
Let's consider a simpler example first. Consider the function $f(x) = \frac{2x}x$. This says you take some number $x$, multiply by 2, then divide by the original number $x$. Obviously the answer is always 2, right? Except that when $x$ is zero, the division is forbidden and there is no answer at all. But for every $x$ except 0, we have $\frac{2x}x = 2$. In particular, for values of $x$ close to, but not equal to 0, we have $\frac{2x}x = 2$.
The function $\frac{x^2-25}{x-5}$ is similar, just a little more complicated. Calculating $x^2-25$ always gives you the same as $(x-5)(x+5)$. That is, if you take $x$, square it, and subtract 25, you always get the same number as if you take $x$, add 5 and subtract 5, and then multiply the two results. So we can replace $x^2-25$ with $(x+5)(x-5)$ because they always give the same number regardless of what you start with; they are two ways of getting to the same place. And then we see that $$\frac{x^2-25}{x-5} = \frac{(x+5)(x-5)}{x-5} = x+5$$ | {
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except that if $x-5$ happens to be zero (that is, if $x=5$) the division by zero is forbidden and we get nothing at all. But for any other $x$ the result of $\frac{x^2-25}{x-5}$ is always exactly equal to $x+5$. In particular, for values of $x$ close to, but not equal to 5, we have $\frac{x^2-25}{x-5} = x+5$.
The limit $$\lim_{x\to 5} \ldots$$ asks what happens to some function when $x$ close to, but not exactly equal to 5. And while this function is undefined for $x=5$, because to calculate it you would have to divide by zero, it is perfectly well-behaved for other values of $x$, and in particular for values of $x$ close to 5. For values of $x$ close to 5 it is equal to $x+5$, and so for values of $x$ close to 5 it is close to 10. And that is exactly what the limit is calculating.
I think that what confuses you is the difference between "solving the algebraic expression", and "finding the limit". Given:
$$f_1=\frac{x^2-25}{x-5} \quad f_2 = (x+5)$$
Then, $f_1$ and $f_2$ are most definitely NOT the same function. This is because they have different domains: 5 is not a member of the domain of $f_1$, but it is in the domain of $f_2$.
However, when we go from: $$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} \quad to \quad \lim _{x\rightarrow 5}\frac{(x-5)(x+5)}{x-5} \quad to \quad \lim_{x\rightarrow 5} (x+5)$$
We are not saying that the expressions inside the limits are equal; maybe they are, maybe they are not. What we are saying that they have the same limit. Totally different statement.
Above, the transformation of the second expression to the third one allows us to find a different function for which a) we know that the limit is the same, and b) we know how to trivially calculate that limit.
The big question, then: what transformations can I make to the function $f_1$ so that the limit stays the same? I think this is usually poorly explained in introductory courses -- a lot of hand-waving going on. | {
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Obviously you can do any algebraic manipulation that leaves $f_1$ unchanged. You can also make any manipulation that removes and/or introduces discontinuities (points for which the function does not exist), as long as the new function stays continuous for an arbitrarily small neighborhood around $a$ (except possibly at $a$ itself). Your example is a case of such a transformation.
Here I'm myself cheating because I'm not defining 'continuity' for you. I'm sorry; please use an intuitition of what continuous means ("no holes, no jumps"), until you are presented with a formal one.
More complex transformations exist, but they have to be justified individually. You'll get to them eventually.
• you clarified a lot, thanks Aug 8, 2013 at 20:08
• Sixth line from the end - "You" will be "Your".
– MrAP
Jul 5, 2017 at 19:18
• @MrAP, thanks. You should feel free to post an edit directly Jul 6, 2017 at 1:04
Here's the basic idea: You're given a rational function, $f\colon \Bbb R \setminus \{5\}\to \Bbb R$, which is continuous everywhere in its domain.
You want to find the limit of that function at $5$.
One way to do that is to construct a continuous extension of that function, $g\colon \Bbb R \to \Bbb R$, such that $g(x) = f(x)$ whenever $x$ is in the domain of $f$. Then $$\lim_{x\to 5}f(x) = \lim_{x\to 5} g(x) = g(5).$$
In this case, factoring and cancelling accomplishes that objective.
• i really like this idea of a "continuous extension", what is the formal name of this rule/theorem? I'd like to read more about it. Thanks for the answer @dfeuer Aug 8, 2013 at 20:07
• @user4150: Try this google search: "removable discontinuity" "continuous extension" Aug 8, 2013 at 21:24
• @dfeuer Typo? It should be $$\lim_{x\to 5}f(x) = \lim_{x\to 5} g(x) = g(\color{royalblue}5).$$ Oct 1, 2014 at 8:06
• @Hakim, right you are. Fixed. Oct 1, 2014 at 9:55
The chill pill you are looking for is | {
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The chill pill you are looking for is
If $f_1$ = $f_2$ except at $a$ then $\lim _{x{\rightarrow}a}f_1(x)=\lim_{x{\rightarrow}a}f_{2}(x)$
• Yes! You made my day! Apr 14, 2016 at 16:42
• Assuming both function to be continuous at $x=a$. May 26, 2017 at 11:45
Intuitively, we can start the other way round, by simply considering $\lim_{x\rightarrow 5} (x+5)$ which we're all agreed we understand. Now consider, independently,
$$\frac{x-5}{x-5}$$
this is obviously 1 everywhere, except where it's undefined at $x = 5$. So, what would you expect to be the effect of multiplying the two? It's just multiplying by 1, except that it also introduces a hole at $x = 5$
To understand this more broadly, it is convenient to check L'Hôpital's rule, which basically boils down to this:
Given a function $f(x)=\frac{g(x)}{h(x)}$, where for some $x=x_0$ both $g(x_0)=0$ and $h(x_0)=0$, the actual value $f(x_0)$ can be obtained as*
$$\lim_{x\to x_0}f(x) = \lim_{x\to x_0}\frac{g'(x)}{h'(x)}$$
(where the prime denotes derivation by $x$) Note that if $g'(x_0)=0=h'(x_0)$, you can re-apply the same rule again and again until either numerator or denominator are not 0. In your example, we get
$$\lim_{x\to 5}\frac{x^2-25}{x-5} = \lim_{x\to 5}\frac{2x}{1}=10=\lim_{x\to5}\,(x+5)$$
To get an even deeper understanding, the rule's proof might be an interesting read.
*: This is only one case, you can also have $|g(x_0)|=|h(x_0)|=\infty$, but not $g(x_0)=0$ and $f(x_0)=\pm\infty$ | {
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• I don't think it's appropriate to bring up L'Hôpital in a question like this. To justify the rule, the reader needs to be familiar with derivatives, differentiability, the sandwich theorem and Cauchy's mean value theorem just to get started. The OP is asking for understanding, not higher-level magic hand-waving. L'Hôpital's rule is a great practical tool, but it's also wielded too often as a substitute for thinking. Aug 8, 2013 at 11:53
• @EuroMicelli Woah there, I didn't intend to hand-wave magically, I just wanted to provide a different answer that, if one is interested enough, can not only explain the removable singularities in polynomial fractions but also in more complicated cases. Though I agree one shouldn't blindly apply Hôpital to just anything if there is another maybe more elegant way Aug 8, 2013 at 13:04
• Also you don't obtain the $f(x_0)$ value - you get $f(x_0)$ value provided it is defined and $f$ is continous. The OP function is not defined at $5$ but it's limit is. Aug 9, 2013 at 7:05
• @MaciejPiechotka True, stricly speaking the example function is equivalent to $$f(x) = \begin{cases} x+5 & x\neq 5 \\ \text{undefined} & x = 5\end{cases}$$ but the singularity at $x=5$ is removable as in $$f(x)= \begin{cases}\tfrac{x^2-25}{x-5} & x\neq 5 \\ 10 & x=5 \end{cases} \Rightarrow f(x) \equiv x+5$$ But you're right that one should always keep removed singularities in mind as they may have severe influence to e.g. applications in Physics (though I can't tell one ad hoc) Aug 9, 2013 at 7:10
One of definitions of $\lim_{x \to A} f(x) = B$ is:
$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - A\right| < \delta}\left|f(x) - B\right| < \varepsilon$$
The intuition is that we can achieve arbitrary 'precision' (put in bounds on y axis) provided we get close enough (so we get the bounds on x axis). However the definition does not say anything about the value at the point $f(A)$ which can be undefined or have arbitrary value. | {
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One of method of proving the limit is to find the directly $\delta(\varepsilon)$. Hence we have following formula (well defined as $x\neq 5$):
$$\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \epsilon$$
As $x\neq 5$ (in such case $\left|x - 5\right| = 0$) we can factor the expression out
$$\forall_{0 < \left|x - 5\right| < \delta} \left|x + 5 - 10\right| < \varepsilon$$ $$\forall_{0 < \left|x - 5\right| < \delta} \left|x - 5 \right| < \varepsilon$$
Taking $\delta(\varepsilon) = \varepsilon$ we find that:
$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \varepsilon$$
The key thing is that we don't care about value at the limit.
Look at this example:
$$\lim _{\text{thing} \to zero}(\dfrac{\text{Thing}}{\text{Thing}}) \cdot \text{Another thing}$$
Here our $\text{'Thing'}$ is tending to zero, but not zero, it is something real, call it a real $\text{Thing}$, which can be cancelled merrily.
• What do you have against using letters as variables? Mar 8, 2014 at 18:54
The hole doesn't always disappear but when it does it is because one expression is equivalent to another except right at that hole. For example $$\frac{x^2 - 25}{x - 5}= \frac{(x+5)(x-5)}{x-5}= (x + 5)$$ everywhere except at x = 5. However, if we are very close to 5 and we had infinite place arithmetic (as we do theoretically) the calculation of the first expression $\frac{x^2 - 25} {x - 5}$ would be very close to the last expression $(x+5)$, that is the limit as x approaches 5 of the first expression is the same limit as the limit as x approaches 5 of last expression. Since that latter is a 'nice number', we can define the value at 5 for the first expression and remove the "everywhere except at $x = 5$" by just putting 10 as the value and not calculating it. | {
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The trick is to think of limit a bit differently: A limit is the value a function would have at a point were it continuous at that point, with nothing else being different.
You can see this by comparing the epsilon-delta definitions of the two concepts. We say $$f$$ is continuous at $$c$$, if $$f$$ is defined at $$c$$ and for every $$\epsilon > 0$$ there is a $$\delta > 0$$ such that
$$|f(x) – f(c)| < \epsilon$$ whenever $$0 < |x – c| < \delta$$.
Likewise, $$f$$ has limit $$L$$ at $$c$$ if for every $$\epsilon > 0$$ there is a $$\delta > 0$$ such that
$$|f(x) – L| < \epsilon$$ whenever $$0 < |x – c| < \delta$$.
Thus, if we have a function with isolated gaps and there is a different function $$f^{*}$$ such that
1. $$f^{*}(x) = f(x)$$ wherever $$f(x)$$ is defined and continuous,
2. $$f^{*}$$ is defined and continuous everywhere, i.e. has domain $$\mathbb{R}$$ (versus the original function having domain $$\mathbb{R}$$ minus a set of isolated points),
then we must have that the value this $$f^{*}$$ takes at the points where $$f$$ was not defined or continuous must be the limiting values at those points.
And that's what you find by cancellation: by rewriting the expression describing the function in a form that involves only addition and multiplication with no division, you have created a formula that describes a function that must agree where the division part is defined, but since no division is involved, is defined everywhere. Moreover, since addition and multiplication (and powers, though you can consider them repeated multiplication) are continuous, then it follows that this function must be continuous as well. Hence the above are met, and the holes fill in. | {
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The idea is that if two functions $f$ and $g$ differ only at $a$ but are identical otherwise, we have $\displaystyle \lim_{x \to a} f(x) = \lim_{x \to a} g(x)$. In this case $f(x) = \dfrac {x^2-25}{x-5}$ and $g(x) = x+5$ are identical everywhere except $x=5$, so $\displaystyle \lim_{x \to 5} \dfrac {x^2-25}{x-5} = \lim_{x \to 5} x+5 = 10.$
In limits the value of x is always tending and not exact
x→a means the value of x tends to a. Similarly x→5 means tending to 5. So basically the value of x is not 5, but a little more or less than 5.
Hence you cannot take the denominator 0 in any case as the denominator will be, again, tending to zero. That is why we try to factorize the numerator first in order to check if there are any terms common both on the numerator and the denominator.
$$L:=\lim_{x\to x_0}f(x)$$ is the value that would make $$f$$ continuous at $$x_0$$.
If you known another function $$g$$ which is continuous and coincides with $$f$$ around $$x_0$$, then perforce
$$L=g(x_0).$$ | {
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# HP Forums
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
.
InvH:
« .577215664902 - InvPsi 1. -
»
InvPsi:
« DUPDUP DUP InvP InvP DUP InvP InvP DUPDUP InvP InvP DUP InvP DUP EXP .5 + Psi OVER - - EXP .5 +
»
InvP:
« DUP EXP .5 + Psi OVER - - EXP .5 + Psi OVER - -
»
(*) On the HP 49G, replace Psi with 0 PSI.
Examples:
4.012007 InvH --> 30.523595226 (in less than half a second); this is a very belated solution to one of the famous Valentin's challenges (#3 here).
.577215664902 InvH --> 0.46163214497 ; InvH(Euler-Mascheroni constant) = xmin (local minimum of the continuous factorial function).
4 pi 2 / - 3 ENTER 2 LN * - InvH --> 0.24999999999 ; one of the special values for fractional arguments examples here.
1.5 InvH --> 2. ; H2
1 Psi --> -0.577215664902 InvPsi --> 1.00000000009
2 Psi --> 0.422784335098 InvPsi --> 2.
Background:
Let H(x) be the continuous function associated with Harmonic Numbers. Then,
$H(x)=\gamma +\psi \left ( x+1 \right )$
$\psi \left ( x+1 \right )=H(x)-\gamma$
$x+1 =\psi^{-1} \left ( H(x)-\gamma \right )$
$x+1 =\psi^{-1} \left ( H(x)-\gamma \right )$
$x =\psi^{-1} \left ( H(x)-\gamma \right )-1$
or
$H^{-1}(x) =\psi^{-1} \left ( x-\gamma \right )-1$
That is, in order to obtain the inverse of H(x) we only need the Inverse Digamma Function and the Euler-Mascheroni constant. No problem with the constant, but the Inverse Digamma might be a problem since Digamma is not easily invertible.
A rough approximation is
$\psi^{-1} \left ( x\right )=e^{x}+\frac{1}{2}$
The equivalent HP 50g program is
P1:
« EXP .5 +
»
However, this is good only for x >= 10, not good enough for our purposes:
10 P1 --> 22026.9657948 Psi --> 10.0000000001.
But,
9 P1 --> 8103.58392758 Psi --> 9.00000000064
8 P1 --> 2981.45798704 Psi --> 8.00000000469
7 P1 --> 1097.13315843 Psi --> 7.00000003465
So, let's try to improve the accuracy a bit:
P2:
« DUP P1 Psi OVER - - P1
» | {
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So, let's try to improve the accuracy a bit:
P2:
« DUP P1 Psi OVER - - P1
»
10 P2 --> 22026.9657926 Psi --> 9.99999999999
9 P2 --> 8103.58392239 Psi --> 8.99999999999
8 P2 --> 2981.45797306 Psi --> 8.
7 P2 --> 1097.13312043 Psi --> 7.
6 P2 --> 403.928690211 Psi --> 6.
5 P2 --> 148.912878357 Psi --> 5.00000000001
But,
4 P2 --> 55.0973869316 Psi --> 4.00000000039
3 P2 --> 20.5834634677 Psi --> 3.00000002131
Proceeding likewise, we get
P3:
« DUP P2 Psi OVER - - P2
»
This is good for x as low as 2, but not for x = 1:
2 P3 --> 7.88342863120 Psi --> 2.
1 P3 --> 3.20317150637 Psi --> 1.0000000139
A couple more steps suffice for x around -0.6 and greater, which is good enough for our purposes. That's what the InvPsi program above does, albeit in an inelegant way. Also, this is just an intuitive and somewhat inneficient approach. Better methods suggestions are welcome.
Edited to fix a couple of typos.
Edited again to include a printout of my current directory:
InvPsi will accept arguments greater or equal -1. About 0.05 s, 0.10 sor 0.50 s, depending on the arguments.
Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.
John
(04-27-2016 04:57 PM)John Keith Wrote: [ -> ]Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.
If the arguments are always plain harmonic numbers, that is, the ones obtained from the discrete function, then the inverse function can be implemented simply as
InvHn:
« .577215664902 - EXP IP »
Examples:
137 ENTER 60 / InvHn --> 5. ; 137/60 = H(5)
10 Hx --> 2.92896825397 ; H(10)
InvHn --> 10.
2E10 --> 24.2962137754 ; H(2x10^10)
InvHn --> 20000000000.
where Hx is the program for the continuous function:
Hx:
« 1. + Psi .577215664902 + »
Regards, | {
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where Hx is the program for the continuous function:
Hx:
« 1. + Psi .577215664902 + »
Regards,
Gerson.
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | {
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Why do small angle approximations have to be in radians?
Why do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.
• @EmilioPisanty perhaps you're correct, but anonymous disparagement is hardly the best way to greet a new user. Anyway, user28435, try Googling one-line answers before asking in future (this must have been asked hundreds of times on Math SE).
– Meow
Aug 19, 2013 at 21:36
• Don't think of sine as being defined in terms of angles - rather, arcs length of a circle. (The approximations come from the fact that a circle is approximately a straight line, especially near the point of tangency.) Oct 21, 2014 at 10:55
The reason why we use radians for $\sin x$ (or other trigonometric functions) in calculus is explained here and here in Wikipedia.
Having known that, notice that small angle approximation is just the Taylor expansion of $\sin x$ and $\cos x$ around $x=0$:
$$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\tag{1}$$ $$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\tag{2}$$
If you scale $x$ by some constant $\omega$, then you must replace $x$ with $\omega x$ in $(1)$ and $(2)$. So, working in degrees $($ $\omega=\frac{\pi}{180}$ $)$, the approximation will become: $$\sin \theta\approx \frac{\pi}{180}\theta\tag{\theta in degrees}$$
There are several correct ways to answer this that illuminate different aspects of what is going on (and I wouldn't be surprised if the answer is present on this site somewhere EDIT: indeed the other responses to this question are different ways of looking at it). Here is a geometrical answer.
The basic formula is
$$l = r \theta$$ | {
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The basic formula is
$$l = r \theta$$
where $l$ is the arc length of a segment of a circle with radius $r$ subtending an angle $\theta$. For this formula to be true, $\theta$ needs to be in radians. (Just try it out, is the arc length of the whole circle equal to $360 r$ or $2\pi r$?) In fact, that formula is really how you define what you mean by radians.
Now consider a straight line segment connecting the two endpoints of the arc subtended by $\theta$. Call it's length $a$. You can show by fiddling with triangles that $$a=2r\sin\left(\frac{\theta}{2}\right)$$
In the limit that the angle is small (so only a small piece of the circle is subtended), you should be able to convince yourself that $a\approx l$. This is the core of the small angle approximation.
Using the two relationships above we have
$$r\theta \approx 2r \sin\left(\frac{\theta}{2}\right)$$ or $$\sin\left(\frac{\theta}{2}\right)\approx\frac{\theta}{2}$$ You can see that using radians was crucial here because it allowed us to use $l=r\theta$.
A 'small angle' is equally small whatever system you use to measure it. Thus if an angle is, say, much smaller than 0.1 rad, it will be much smaller than the equivalent in degrees. More typically, saying 'small angle approximation' typically means $\theta\ll1$, where $\theta$ is in radians; this can be rephrased in degrees as $\theta\ll 57^\circ$.
(Switching uses between radians and degrees becomes much simpler if one formally identifies the degree symbol $^\circ$ with the number $\pi/180$, which is what you get from the equation $180^\circ=\pi$. If you're differentiating with respect to the number in degrees, then, you get an ugly constant, as you should: $\frac{d}{dx}\sin(x^\circ)={}^\circ \cos(x^\circ)$.)
In real life, though, you wouldn't usually say 'this angle should be small' without saying what it should be smaller than. If the latter is in degrees then the former should also be in degrees. | {
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That said, though: always work in radians! Physicists tend to use degrees quite often, but there is always the underlying understanding that the angle itself is a quantity in radians and that degrees are just convenient units. Trigonometric functions, in particular, always take their arguments in radians, so that all the math will work well. Always differentiate in radians, always work analytically in radians. And at the end you can plug in the degrees.
It's because this relationship $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$ (i.e. $\sin(x) \approx x$) only holds if $x$ is in radians, as, using L'Hoptial's rule, $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\rightarrow0}\frac{d}{dx}\sin(x)=\begin{cases} 1 & \text{ if } x\text{ is in radians}\\ \frac{\pi}{180} & \text{ if } x \text{ is in degrees} \end{cases}.$$ That is, $\frac{d}{dx}(\sin(x))=\cos(x)$ only if $x$ is in radians. The reason can be divined from this proof without words by Stephen Gubkin:
• Sure but WHY does this only hold in radians. That's the bit I can't understand. Everyone says it holds in radians but no one explains why lol. Thanks anyway. Aug 19, 2013 at 21:03
• @user28435: because in radians the sin of an angle basically equals the angle itself, at small enough angles, so it very much simplifies the algebra if you don't have to keep converting degrees.
– Mike Dunlavey
Aug 20, 2013 at 0:12
You need to clarify what you are really asking: in physics the small angle approx. is typically used to approximate a non-linear differential equation by a linear one - which is much easier to solve - by allowing us to approximate the sin(x) function by x. The resulting equation is only a good one for small angles. I.e., its a physics thing, not a math thing. | {
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because the sine of an angle is dimensionless [i.e. just a number], the angle has to be also dimensionless [i.e. the radian is dimensionless] ... units on both sides of an equation have to match | {
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# Determining the remainder of a small integer n raised to a high power when divided by m
I recently took a test that asked me to determine the value of $3^{82} \mod 5$.
I was unable to figure out how to do it on the test, but when I got home I noticed that there is a pattern to the remainders of $3^n$,
• $3^{1} \mod 5 = 3$
• $3^{2} \mod 5 = 4$
• $3^{3} \mod 5 = 2$
• $3^{4} \mod 5 = 1$
• $3^{5} \mod 5 = 3$
• $3^{6} \mod 5 = 4$
• $3^{7} \mod 5 = 2$
• $3^{8} \mod 5 = 1$
• $3^{9} \mod 5 = 3$
Is the best strategy to follow this pattern up to the $82^{\text{nd}}$ remainder (obviously in an intelligent way) or is there some much more obvious trick to this question that I missed?
-
As you have observed $$3^4 \equiv 1 \pmod 5$$ This means that $$3^{4k} \equiv 1 \pmod 5$$ Hence, $$3^{80} \equiv 1 \pmod 5 \implies 3^{82} \equiv 3^2 \pmod{5} = 4 \pmod 5$$ In general, $$3^{4k + r} \equiv 3^r \pmod{5}$$
Thank you for your answer. It's not entirely clear to me how $3^4 \mod 1 \equiv 1$ tells us that $3^{4k} \mod 1 \equiv 1$. Could you explain this step? – crazedgremlin Nov 7 '12 at 1:07
@crazedgremlin $\rm\: \color{#0A0}{3^4}\equiv\color{#C00} 1\:\Rightarrow\: 3^{4k} = (\color{#0A0}{3^4})^k\equiv \color{#C00}1^k\equiv 1\:$ by the Congruence Product Rule. It is essential to learn the arithmetic of congruences, as opposed to less flexible mod operations. – Bill Dubuque Nov 7 '12 at 1:23 | {
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# Thread: Limits / Series question
1. ## Limits / Series question
hi
Let
P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
n=2,3,4,......
Find Limit P when N tends to infinity.
2. Originally Posted by champrock
hi
Let
P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
n=2,3,4,......
Find Limit P when N tends to infinity.
Factorise $\frac{n^3-1}{n^3+1}$ as $\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$, and notice that $n^2+n+1 = (n+1)^2 - (n+1) + 1$. You will then find that you have a telescoping product.
3. what should i do with n^2 - n + 1 ?
I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
4. You easily undestand if you write the sequence of terms...
$P= \prod_{n=2}^{\infty} \frac{(n-1)\cdot (n^{2} + n + 1)}{(n+1)\cdot (n^{2} - n + 1)}= \frac {1\cdot 7}{3\cdot 3}\cdot \frac {2\cdot 13}{4\cdot 7}\cdot \frac {3\cdot 21}{5\cdot 13}\cdot \frac {4\cdot 31}{6\cdot 21}\cdot \dots$
... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]
Finally is...
$P=\frac{2}{3}$
Very nice!...
Kind regards
$\chi$ $\sigma$
5. Originally Posted by champrock
what should i do with n^2 - n + 1 ?
I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
Use the hint I gave before: $n^2+n+1 = (n+1)^2 - (n+1) + 1$. That tells you that the term $n^2+n+1$ in the numerator of each fraction cancels with the term $(n+1)^2 - (n+1) + 1$ in the denominator of the next one.
6. I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3
7. The term of order n is...
$a_{n}= \frac{(n-1)\cdot (n^{2} +n + 1)}{(n+1)\cdot (n^{2} - n + 1)}$
For $n=3$ is...
$a_{3}= \frac{2\cdot 13 }{4\cdot 7}$
Kind regards
$\chi$ $\sigma$ | {
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Log InorSign Up. The form of the covariance matrix σ in the unrotated system follows from equation (14) using R. The standard form of the equation of an ellipse with center (h, k) and major and minor axes of lengths 2a and 2b, respectively, where 0 < b < a, is. They both have shape (eccentricity) and size (major axis). The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (∘, ∘), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. | bartleby. A graph of the two equations is presented here. As stated, using the definition for center of an ellipse as the intersection of its axes of symmetry, your equation for an ellipse is centered at $(h,k)$, but it is not rotated, i. It has co-vertices at (5 ± 3, –1), or (8, –1) and (2, –1). Once we have those we can sketch in the ellipse. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel. How It Works. The parametric equations of an ellipse are and. |C|y the equation is transformed into σx02 + τy02 = −F. Here a > b > 0. This equation defines an ellipse centered at the origin. the graph is an ellipse if AC > 0, and in Section 5. You may ignore the Mathematica commands and concentrate on the text and figures. Ellipse drawing tool. Constructing (Plotting) a Rotated Ellipse. The equation x 2 – xy + y 2 = 3 re presents a "rotated ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. For a given chord or triangle base, the. The length of the major axis is 2a, and the length of the minor axis is 2b. +- hat j b. Complex Growth. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. Let x^2+3y^2=3 be the equation of an ellipse in the x-y plane. You may ignore the Mathematica commands and concentrate on the text and figures. Vertical Major Axis Example. They both have shape (eccentricity) and | {
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on the text and figures. Vertical Major Axis Example. They both have shape (eccentricity) and size (major axis). Since B = − 6 3 ≠ 0, the equation satisfies the condition to be a rotated ellipse. {\displaystyle Ax^ {2}+Bxy+Cy^ {2}+Dx+Ey+F=0}. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. Write the equation of the circle in standard form given the endpoints of the diameter: (-12, 10) and (-18, 12). They are located at (h±c,k) or (h,k±c). Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. HELP?! Rotate the axes to eliminate the xy-term in the equation. centerof the ellipse gets rotated about point pand the new ellipseat the new center gets rotated about the new centerby angle a. We have step-by-step solutions for your textbooks written by Bartleby experts!. Here we plot it ContourPlotA9 x2-4 x y + 6 y2 − 5, 8x,-1, 1<, 8y,-1, 1<, Axes fi True, Frame fi False,. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x¿y¿-term x¿ 2 4 + y¿ 1 = 1. I wish to plot an ellipse by scanline finding the values for y for each value of x. H(x, y) = A x² + B xy + C y² + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. The standard form of the equation of an ellipse with center (h, k) and major and minor axes of lengths 2a and 2b, respectively, where 0 < b < a, is. Since (− 6 3) 2 − 4 ⋅ 7 ⋅ 13 = − 346 < 0 and A ≠ C since 7 ≠ 13, the equation satisfies the conditions to be an ellipse. 18 < : 1 1 1 9 = ; (10) 3. By using this website, you | {
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satisfies the conditions to be an ellipse. 18 < : 1 1 1 9 = ; (10) 3. By using this website, you agree to our Cookie Policy. Activity 4: Determining the general equation of an ellipse/ Determining the foci and vertices of an ellipse. I want to plot an Ellipse. Rotate the hyperbola : In the above graph, the preimage is in blue and the image (rotated) is. When you rotate the ellipse about y = 5, the "tire" above will be coming-out and going-in through z-direction. EINSTEIN, ALBERT and MICHELE BESSO. Here are formulas for finding these points. I know that i can draw it using ellipse equation, then rotate it, compute points and connect with lines. If an ellipse is rotated about one of its principal axes, a spheroid is the result. If the rotation is small the resulting ellipse is very nearly round, but if the rotation is large the ellipse becomes very flattened (or very elongated, depending upon how you look at the effect), and if the circle is rotated until it is edge-on to our line of sight the "ellipse" becomes just a straight line segment. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. In the interesting first case the set Q is an ellipse. k is y-koordinate of the center of the ellipse. If the center of the ellipse is at point (h, k) and the major and minor axes have lengths of 2a and 2b respectively, the standard equation is. HELP?! Rotate the axes to eliminate the xy-term in the equation. Sketch the graph of Solution. When rotated inside a square of side length 2 having corners at ), the envelope of the Reuleaux triangle is a region of the square with rounded corners. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. (1) Ellipse (2) Rotated Ellipse (3) Ellipse Representing Covariance. The following 12 points are on this ellipse: The ellipse is symmetric about the lines y. Write the equation of the circle in standard form given the endpoints | {
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symmetric about the lines y. Write the equation of the circle in standard form given the endpoints of the diameter: (-12, 10) and (-18, 12). Here we plot it ContourPlotA9 x2-4 x y + 6 y2 − 5, 8x,-1, 1<, 8y,-1, 1<, Axes fi True, Frame fi False,. Given the equation of a conic, identify the type of conic. What are the applications of Ellipse in real life? The ellipse has a close reference with football when it is rotated on its major axis. Equation 3 thus becomes: rz n S(y, cos & - x, sin &)2 - N Eq. If and are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. | bartleby. The equation of an ellipse is a generalized case of the equation of a circle. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. Let us consider the figure (a) to derive the equation of an ellipse. minimal expansion. lationship between two images is pure rotation, i. If the rotation is small the resulting ellipse is very nearly round, but if the rotation is large the ellipse becomes very flattened (or very elongated, depending upon how you look at the effect), and if the circle is rotated until it is edge-on to our line of sight the "ellipse" becomes just a straight line segment. Let the coordinates of F 1 and F 2 be (-c, 0) and (c, 0) respectively as shown. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x¿y¿-term x¿ 2 4 + y¿ 1 = 1. h is x-koordinate of the center of the ellipse. (4) into the following canonical form (x 0 x 0) 2 a2 + (y0 y 0) 2 b2 = 1; (5) in which (x 0 0;y 0) is the center of the ellipse in the rotated coordinate system, and aand bare the lengths of the semi-axes. Expanding the binomial squares and collecting like terms gives. The equation b 2 = a 2 – c 2 gives me 16 | {
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the binomial squares and collecting like terms gives. The equation b 2 = a 2 – c 2 gives me 16 – 9 = 7 = b 2. {\displaystyle B} is zero then the conic is not rotated and sits on the x- and y- axes. ; If and are nonzero and have opposite signs. Rotating an Ellipse. We shall now study the Cartesian representation of the hyperbola and the ellipse. If you enter a value, the higher the value, the greater the eccentricity of the ellipse. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. Here a > b > 0. Thanks, Michal Bozon. For a non-rotated conic: A. Complex Growth. However, when you graph the ellipse using the parametric equations, simply allow t to range from 0 to 2π radians to find the (x, y) coordinates for each value of t. If the Circle option is selected, the width and height of the drawn shape is kept the same. We have step-by-step solutions for your textbooks written by Bartleby experts!. | bartleby. − <: Ellipse − >: Hyperbola; Firstly, if B is not zero then the graph represents a rotated conic. When a>b, we have a prolate spheroid, that is, an ellipse rotated around its major axis; when a K > -1 = ellipse, -1 = parabolic, and K < -1 is hyperbolic; R is the radius of curvature. General Equations of Conics The graph of Ax 2 Cy is one of the 2 Dx Ey F 0 following : 1. However now i want the ellipse to rotate to the slopes so that the top of the ellipse faces in the direction of the line segments normal. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. Ellipse configuration panel. 5 (a) with the foci on the x-axis. When we add an x y term, we are rotating the conic about the origin. I have the verticles for the major axis: d1(0,0. In this section, we will discuss the equation of a conic section which is rotated by. Because A = 7, and C = 13, you have (for 0 θ < π/2) Therefore, the equation in | {
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which is rotated by. Because A = 7, and C = 13, you have (for 0 θ < π/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. Reversing translation : 137(X−10)² − 210(X−10)(Y+20)+137(Y+20)² = 968 This is equation of rotated ellipse relative to original axes. If an ellipse is rotated about one of its principal axes, a spheroid is the result. Moment of inertia is defined with respect to a specific rotation axis. Several exam. (4) into the following canonical form (x 0 x 0) 2 a2 + (y0 y 0) 2 b2 = 1; (5) in which (x 0 0;y 0) is the center of the ellipse in the rotated coordinate system, and aand bare the lengths of the semi-axes. In terms of the geometric look of E, there are three possible scenarios for E: E = ∅, E = p 1 p 2 ¯, the line segment with end-points p 1 and p 2, or E is an ellipse. It follows that 0 £e< 1 and p> 0, so that anellipse in polar coordinates with one focus at the origin and the other onthe positive x-axis is given by. The major axis in a vertical ellipse is represented by x = h; the minor axis is represented by y = v. Notes College Algebra teaches you how to find the equation of an ellipse given a graph. You may ignore the Mathematica commands and concentrate on the text and figures. A “standard ellipsoid” has a circular midsection. {\displaystyle Ax^ {2}+Bxy+Cy^ {2}+Dx+Ey+F=0}. So I'm trying to find the intersections of the equations $$\ {x^2\over 1^2} + {y^2\over 2^2} = 1$$ $$5x^2 - 6xy + 5y^2 = 8$$ Both of the equations represent an ellipse, with the first ellipse being a vertical ellipse and the second ellipse being first one rotated 315 degrees counterclockwise. The center is at (h, k). The equation may be cast in a more general form since the standard deviation is to be calculated as the axis is rotated about the point of average location. {\displaystyle B} is non-zero, then the conic is rotated about the axes, with the rotation centred on the origin. If a > b, the ellipse is stretched further in | {
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the axes, with the rotation centred on the origin. If a > b, the ellipse is stretched further in the horizontal direction, and if b > a, the ellipse is stretched further in the vertical direction. The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (∘, ∘), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. First, notice that the equation of the parabola y = x^2 can be parametrized by x = t, y = t^2, as t goes from -infinity to infinity; or, as a column vector, [x] = [t] [y] = [t^2]. Thus, for the equation to represent an ellipse that is not a circle, the coefficients must simultaneously satisfy the discriminant condition B 2 − 4 A C < 0 B^2 - 4AC< 0 B 2 − 4 A C < 0 and also A ≠ C. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). The form of the covariance matrix σ in the unrotated system follows from equation (14) using R. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. A parametric form for (ii) is x=5. Find the points where the ellipse crosses the x-axis, and show that the tangent lines at these points are parallel. Let be the angle of rotation. Let’s see what happens when. a is the ellipse axis which is parallell to the x-axis when rotation is zero. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). The equation x^2 - xy + y^2 = 3 represents a "rotated ellipse," that is, an ellipse whose axes are not parallel to the coordinate axes. The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (∘, ∘), and rotation angle (the angle from the positive horizontal | {
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axis , center coordinates (∘, ∘), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. Equation of ellipse from its focus, directrix, and eccentricity Last Updated: 20-12-2018 Given focus(x, y), directrix(ax + by + c) and eccentricity e of an ellipse, the task is to find the equation of ellipse using its focus, directrix, and eccentricity. Let us consider the figure (a) to derive the equation of an ellipse. Squashed Circles and Gardeners. If we deform this circle according to the general two dimensional linear transformation, the eq below can be derived:. HELP?! Rotate the axes to eliminate the xy-term in the equation. An ellipse is a flattened circle. k is y-koordinate of the center of the ellipse. Two fixed points inside the ellipse, F1 and F2 are called the foci. You may ignore the Mathematica commands and concentrate on the text and figures. The parametric equations of an ellipse are and. Points p 1 and p 2 are called foci of the ellipse; the line segments connecting a point of the ellipse to the foci are the focal radii belonging to that point. The general transformation is Y = RX with inverse X = RTY. Below is the C++ representation of the above problem. By using a transformation (rotation) of the coordinate system, we are able to diagonalize equation (12). Step 2 : Trigonometric identity : Substitute and in above equation. I'm looking for a Cartesian equation for a rotated ellipse. Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. Our mission is to provide a free, world-class education to anyone, anywhere. , x(t), y(t), z(t)) or a more canonical form (e. I have the verticles for the major axis: d1(0,0. Determine the foci and vertices for the ellipse with general equation 2x^2+y^2+8x-8y-48. If $$\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\displaystyle D<0$$, a parabola for $$\displaystyle D = 0$$, and a hyperbola for $$\displaystyle D>0$$. This | {
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D<0$$, a parabola for $$\displaystyle D = 0$$, and a hyperbola for $$\displaystyle D>0$$. This can always be converted to VBA Code. Entering 0 defines a circular ellipse. Find the points where the ellipse crosses the x-axis, and show that the tangent lines at these points are parallel. The equation of a line through the point and cutting the axis at an angle is. In particular that the shape made by rotation around the x-axis can stand on the top, if it is made from wood. The velocity equation for a hyperbolic trajectory has either + , or it is the same with the convention that in that case a is negative. The blue ellipse shows the original plot. The super ellipse belongs to the Lamé curves. For a surface obtained by rotating a curve around an axis, we can take a polygonal approximation to the curve, as in the last section, and rotate it around the same axis. The ellipse points are P = C+ x 0U 0 + x 1U 1 (1) where x 0 e 0 2 + x 1 e 1 2 = 1 (2) If e 0 = e 1, then the ellipse is a circle with center C and radius e 0. This equation of an elliptic cylinder is a generalization of the equation of the ordinary, circular cylinder (a = b). Several exam. A circle if A = C 2. The image of the disk will be an ellipse with these directions as the major and minor axes. Thus, the graph of this equation is either a parabola, ellipse, or hyperbola with axes parallel to the x and y-axes (there is also the possibility that there is no graph or the graph is a “degenerate” conic: a point, a line, or a pair of lines). The general equation's coefficients can be obtained from known semi-major axis , semi-minor axis , center coordinates (∘, ∘), and rotation angle (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:. Rotated Parabolas and Ellipse. General equations as a function of λ X, λ Z, and θ d λ’= λ’ Z +λ’ X-λ’ Z-λ’ X cos(2θ d) 2 2 γ λ’ Z-λ’ X sin(2θ d) 2 tan θ d = tan θ S X S Z α = θ d - θ (internal rotation) λ’ = 1 λ λ X = quadratic elongation | {
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d) 2 tan θ d = tan θ S X S Z α = θ d - θ (internal rotation) λ’ = 1 λ λ X = quadratic elongation parallel to X axis of finite strain ellipse λ Z = quadratic elongation parallel to Z axis of finite. All Forums. Erase the previous Ellipse by drawing the Ellipse at same point using black color. k is y-koordinate of the center of the ellipse. : Activity 2 - Using the Graph-Rotation Theorem. attempt to list the major conventions and the common equations of an ellipse in these conventions. Number of decimal places for input variable: (Note: Input value of 0 means input variable will be integer. Rewrite the equation in the general form, Identify the values of and from the general form. the graph is an ellipse if AC > 0, and in Section 5. is along the ellipse’s major axis, the correlation matrix is σ′ = σ′2 1 0 0 σ′2 2. I am trying to find an algorithm to derive the 4 angles, from the centre of a rotated ellipse to its extremities. Here is a simple calculator to solve ellipse equation and calculate the elliptical co-ordinates such as center, foci, vertices, eccentricity and area and axis lengths such as Major, Semi Major and Minor, Semi Minor axis lengths from the given ellipse expression. | bartleby. If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. He provided me with some equations that I combined with a neat ellipse display program written by Wayne Landsman for the NASA Goddard IDL program library to come up with a "center of mass" ellipse fitting program, named Fit_Ellipse. 06274*x^2 - y^2 + 1192. The blue ellipse shows the original plot. Ellipse definition, a plane curve such that the sums of the distances of each point in its periphery from two fixed points, the foci, are equal. with the axis. Do the intersection points of two rotated parabolas lie on a rotated ellipse? 1. tive number has a square root. Combine multiple words with dashes(-), and seperate tags with spaces. The equation x 2 – xy + y 2 = 3 re | {
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multiple words with dashes(-), and seperate tags with spaces. The equation x 2 – xy + y 2 = 3 re presents a "rotated ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. 1-c/a cos( q) Usually, we let e= c/aand let p= b2/a, where eis called the eccentricityof the ellipse and pis called theparameter. The moment of inertia of any extended object is built up from that basic definition. If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. At first blush, these are really strange exponents. I have the verticles for the major axis: d1(0,0. The chord perpendicular to the major axis at the center is the minor axis. R = distance between axis and rotation mass (ft, m) The moment of all other moments of inertia of an object are calculated from the the sum of the moments. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). com Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse. Since (− 6 3) 2 − 4 ⋅ 7 ⋅ 13 = − 346 < 0 and A ≠ C since 7 ≠ 13, the equation satisfies the conditions to be an ellipse. In the interesting first case the set Q is an ellipse. The center of this ellipse is the origin since (0, 0) is the midpoint of the major axis. The equation x^2 - xy + y^2 = 3 represents a "rotated ellipse," that is, an ellipse whose axes are not parallel to the coordinate axes. Disk method. The parametric equation of a parabola with directrix x = −a and focus (a,0) is x = at2, y = 2at. Express the equation in the standard form of a conic section. the graph is an ellipse if AC > 0, and in Section 5. This ellipse is called the distortion ellipse. The transformed ellipse is de-scribed by the equation a0x2. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. g, with major axis aligned with X-axis, minor axis aligned with Y-axis … but | {
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group ingannatore. g, with major axis aligned with X-axis, minor axis aligned with Y-axis … but that’s certainly not an “arbitrary” ellipse. +- hat j b. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. The amount of correlation can be interpreted by how thin the ellipse is. ) Then my equation is: Write an equation for the ellipse with vertices (4, 0) and (–2, 0). How It Works. Our mission is to provide a free, world-class education to anyone, anywhere. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x − h and y by y − k in the equation of the curve. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. An ellipse represents the intersection of a plane surface and an ellipsoid. , the 3D analog to the 2D form ((X*X)/a)+((Y*Y)/b)=1). Moment of inertia is defined with respect to a specific rotation axis. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse. This gives a surface composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. E, qua, euros' Sale, per forza di guerra. Differential Equations (10) Discrete Mathematics (4) Discrete Random Variable (5) Disk Washer Cylindrical Shell Integration (2) Division Tricks (1) Domain and Range of a Function (1) Double Integrals (3) Eigenvalues and Eigenvectors (1) Ellipse (1) Empirical and Molecular Formula (2) Enthalpy Change (2) Expected Value Variance Standard. The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. In the rotated the major axis of the ellipse lies along the We can write the equation of the ellipse in this rotated as Observe that there is no in the equation. When rotated inside a square of | {
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ellipse in this rotated as Observe that there is no in the equation. When rotated inside a square of side length 2 having corners at ), the envelope of the Reuleaux triangle is a region of the square with rounded corners. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. Solution The equation of an ellipse usually appears when the plot of the from MECHANICAL MAE351 at Korea Advanced Institute of Science and Technology. Constructing (Plotting) a Rotated Ellipse. The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 3. When the center of the ellipse is at the origin and the foci are on the x-axis or y-axis, then the equation of the ellipse is the simplest. The longer axis, a, is called the semi-major axis and the shorter, b, is called the semi-minor axis. A circle if A = C 2. Rotating Ellipse. Equations When placed like this on an x-y graph, the equation for an ellipse is: x 2 a 2 + y 2 b 2 = 1. First I want to look at the case when , and. This can always be converted to VBA Code. x¿y¿-system x¿-axis. I should also mention, the ellipse is to be drawn on linear or log-log coordinates. If the x- and y-axes are rotated through an angle, say θ,. Let be the angle of rotation. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. (Since I wasn't asked for the length of the minor axis or the location of the co-vertices, I don't need the value of b itself. The equations of tangent and normal to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ at the point $$\left( {{x_1},{y_1}} \right)$$ are \frac. If $a>b$, the ellipse is stretched further in the horizontal direction, and if $b>a$, the ellipse is stretched further in the vertical | {
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further in the horizontal direction, and if $b>a$, the ellipse is stretched further in the vertical direction. Find dy dx. Suppose an ellipse is described by the equation ax2 + bxy + cy2 + dx + ey + f = 0 with a;b;c;d;e;f 2 F. 4 degrees and 90. asked • 04/02/15 find the equation of the image of the ellipse x^2/4 + y^2/9 when rotated through pi/4 about origin. Below is a list of parametric equations starting from that of a general ellipse and modifying it step by step into a prediction ellipse, showing how different parts contribute at each step. I want to plot an Ellipse. Torna cartesian equation of rotated ellipse ogni moment group ingannatore. Moment of inertia is defined with respect to a specific rotation axis. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse. com Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse. xcos a − ysin a 2 2 5 + xsin. I am not very sure if my solution is correct but I'd rather try and put it up and let people evaluate if it's correct: The ellipse would look something like the below image: Since the ellipse is rotated along Y axis it will form circles(of vary. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Major axis is vertical. To rotate the graph of the parabola about the origin, we rotate each point individually. (An ellipse where a = b is in fact a circle. For ellipses not centered at the origin, simply add the coordinates of the center point (e, f) to the calculated (x, y). h is x-koordinate of the center of the ellipse. We shall now study the Cartesian representation of the hyperbola and the ellipse. fftial Calculus Grinshpan Rotated Ellipse The implicit equation x2 xy +y2 = 3 describes an ellipse. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. What is the volume of the solid? Step 2: Determine the boundaries of the integral Since the rotation is | {
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is the volume of the solid? Step 2: Determine the boundaries of the integral Since the rotation is around the y-axis, the boundaries will be between y = 0 and y = 1 Step 4: Evaluate integrals to find volume Step 1:. Ellipse configuration panel. The higher the value from 0 through 89. This equation defines an ellipse centered at the origin. If $$\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\displaystyle D<0$$, a parabola for $$\displaystyle D = 0$$, and a hyperbola for $$\displaystyle D>0$$. However, I could not find anywhere an equation for a spheroid that does not have its axis or revolution along the x,y, or z axis. Rewrite the equation 2x^2+√3 xy+y^2−2=0 in a rotated x^′ y^′-system without an x^′ y^′-term. We can apply one more transformation to an ellipse, and that is to rotate its axes by an angle, θ, about the center of the ellipse, or to tilt it. Constructing (Plotting) a Rotated Ellipse. Use the information provided to write the equation of the ellipse in standard form. For the maps we consider, the axes of the distortion ellipse are in the north/south and the east/west directions. Rotation of axis After rotating the coordinate axes through an angle theta, the general second-degree equation in the new x'y'-plane will have the form __________. How to make an ellipse No one knows for sure when the ellipse was discovered, but in 350 BCE the Ancient Greeks knew about the ellipse as a member of the group of two-dimensional geometric figures called conic sections. This equation has vertices at (5, –1 ± 4), or (5, 3) and (5, –5). the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. Far deirun figlio e cartesian equation of rotated ellipse jackpot. R = distance between axis and rotation mass (ft, m) The moment of all other moments of inertia of an object are calculated from the the sum of the moments. For ellipses not centered at the origin, simply add the coordinates of the center point (e, f) to the calculated (x, y). The | {
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at the origin, simply add the coordinates of the center point (e, f) to the calculated (x, y). The parameters of an ellipse are also often given as the semi-major axis, a, and the eccentricity, e, 2 2 1 a b e =-. I was able to find the equation of an ellipse where its major axis is shifted and rotated off of the x,y, or z axis. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). has offshore analysis. (25) Here, σ′ 1 is the 1-sigma confidence value along the minor axis of the ellipse, and σ′ 2 is that along the major axis (σ′ 2 ≥ σ′ 1). P(at2, 2at) tangent We shall use the formula for the equation of a straight line with a given gradient, passing through a given point. You will probably has to project the data, create the ellipse and project back to WGS to make it work; I think the snippet provided by Darren does not include rotation (bearing). Rewrite the equation 2x^2+√3 xy+y^2−2=0 in a rotated x^′ y^′-system without an x^′ y^′-term. major axis is along y-axis. We can come up with a general equation for an ellipse tilted by θ by applying the 2-D rotational matrix to the vector (x, y) of coordinates of the ellipse. It has co-vertices at (5 ± 3, –1), or (8, –1) and (2, –1). 4 we saw that the graph is a hyperbola when AC < 0. Vertical Major Axis Example. They are located at (h±c,k) or (h,k±c). The ellipse that is most frequently studied in this course has Cartesian equation; where. Note: When tracing feature is ON, shading feature is OFF. A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. The objective is to rotate the x and y axes until they are parallel to the axes of the conic. Find the points at which this ellipse crosses the x -axis and show that the tangent lines at these points are parallel. I wish to plot an ellipse by scanline finding the values for y for each value of x. The orthonormality of the axis directions and Equation (1) imply | {
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values for y for each value of x. The orthonormality of the axis directions and Equation (1) imply x i = U i (P C). You don't have to use power in contrast to the Columbus' egg. The Steiner ellipse can be extended to higher dimensions with one more point than the dimension. Rotate the hyperbola : In the above graph, the preimage is in blue and the image (rotated) is. Equation 3 thus becomes: rz n S(y, cos & - x, sin &)2 - N Eq. Writing Equations of Ellipses Centered at the Origin in Standard Form. Log InorSign Up. Tracing for Cartesian Equations: ON OFF Video To trace a graph, click on the radio button to the right of the input equation. B is the. The following 12 points are on this ellipse: (p 3;0); (0; p 3); ( p 3; p 3); ( p 3;. Eliminate the parameter : Consider. Two fixed points inside the ellipse, F1 and F2 are called the foci. y axis [see Figs. An ellipse is a circle scaled (squashed) in one direction, so an ellipse centered at the origin with semimajor axis a and semiminor axis b < a has equation. Values between 89. The ellipse points are P = C+ x 0U 0 + x 1U 1 (1) where x 0 e 0 2 + x 1 e 1 2 = 1 (2) If e 0 = e 1, then the ellipse is a circle with center C and radius e 0. Example : Given ellipse : 4 2 (x − 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. Elliptic cylinders are also known as cylindroids , but that name is ambiguous, as it can also refer to the Plücker conoid. It is important to know the differences in the equations to help quickly identify the type of conic that is represented by a given equation. It's "just" twice the rotation: 2 is a regular number so doubles our rotation rate to a full -180 degrees in a unit of time. The Steiner ellipse has the minimal area surrounding a triangle. Earth’s orbit has an eccentricity of less than 0. It has co-vertices at (5 ± 3, –1), or (8, –1) and (2, –1). A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. This form of the | {
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through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. This form of the ellipse has a graph as shown below. The point $$\left( {h,k} \right)$$ is called the center of the ellipse. a is the ellipse axis which is parallell to the x-axis when rotation is zero. A circle if A = C 2. Example : Given ellipse : 4 2 (x − 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. In other words, from the centre to the points where the bounding box touches the ellipse's line. x¿y¿-system x¿-axis. The equation of such an ellipse we can write in the usual form 2 2 + 2 =1 (1) The slope of the tangent line to this ellipse has evidently the form (Dvořáková, et al. ' Draw an ellipse centered at (cx, cy) with dimensions' wid and hgt rotated angle degrees. The Steiner ellipse can be extended to higher dimensions with one more point than the dimension. It is a procedure for drawing an approximation to an ellipse using 4 arc sections, one at each end of the major axes (length a) and one at each end of the minor axes (length b). If the center of the ellipse is at the origin, the equation simplifies to (x 2 /a 2) + (y 2 /b 2)=1. We can apply one more transformation to an ellipse, and that is to rotate its axes by an angle, θ, about the center of the ellipse, or to tilt it. Example of the graph and equation of an ellipse on the. Equation of ellipse from its focus, directrix, and eccentricity Last Updated: 20-12-2018 Given focus(x, y), directrix(ax + by + c) and eccentricity e of an ellipse, the task is to find the equation of ellipse using its focus, directrix, and eccentricity. Then write the equation in standard form. Major axis is horizontal. , 2015) ´= − 2√1− 𝑥2 2 (2) For the tangent point of the line with the slope 𝑔∝ and our ellipse then holds 𝑇= √1− 𝑇 2 2 (3) 124. By the way the correct rotation. It has the following form: (x - c₁)² / a² + (y - c₂)² / b² = 1. Thus, the standard equation of an ellipse is x 2 a 2 + y 2 b 2 = 1. Then write the equation in standard | {
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the standard equation of an ellipse is x 2 a 2 + y 2 b 2 = 1. Then write the equation in standard form. It has co-vertices at (5 ± 3, –1), or (8, –1) and (2, –1). Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a ≥ b > 0, major axis on x-axis, minor axis on y-axis). If the data is uncorrelated and therefore has zero covariance, the ellipse is not rotated and axis aligned. This gives a surface composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. Writing Equations of Ellipses Centered at the Origin in Standard Form. Determine the foci and vertices for the ellipse with general equation 4x^2+9y^2-48x+72y+144=0. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x − h and y by y − k in the equation of the curve. By a suitable choice of coordinate axes, the equation for any conic can be reduced to one of three simple r forms: x 2 / a 2 + y 2 / b 2 = 1, x 2 / a 2 − y 2 / b 2 = 1, or y 2 = 2px, corresponding to an ellipse, a hyperbola, and a parabola, respectively. If you are given an equation of ellipse in the form of a function whose value is a square root, you may need to simplify it to make it look like the equation of an ellipse. STRAIN ELLIPSE. The eccentricity of an ellipse is e =. 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter‐clockwise rotation (θ) of the ellipse: à L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : ê T ; 6 F : ê U ; 6 p h. GeoGebra Math Apps Get our free online math tools for graphing, geometry, 3D, and more!. Rotation of Axes 3 Coordinate Rotation Formulas If a rectangular xy-coordinate system is rotated through an angle to form an ^xy^- coordinate system, then a point P(x;y) will have coordinates P(^x;y^) in the new system, where (x;y)and(^x;y^) are related byx =^xcos − y^sin and y | {
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coordinates P(^x;y^) in the new system, where (x;y)and(^x;y^) are related byx =^xcos − y^sin and y =^xsin +^ycos : and x^ = xcos +ysin and ^y = −xsin +ycos : EXAMPLE 1 Show that the graph of the equation xy = 1. Solution The equation of an ellipse usually appears when the plot of the from MECHANICAL MAE351 at Korea Advanced Institute of Science and Technology. Other forms of the equation. Keyword-suggest-tool. It follows that 0 £e< 1 and p> 0, so that anellipse in polar coordinates with one focus at the origin and the other onthe positive x-axis is given by. and through an angle of 30°. The equation of the ellipse in the rotated coordinates is. is along the ellipse’s major axis, the correlation matrix is σ′ = σ′2 1 0 0 σ′2 2. The ellipse may be rotated to a di erent orientation by a 2 2 rotation matrix R= 2 4 cos sin sin cos 3 5 The major axis direction (1;0) is rotated to (cos ;sin ) and the minor axis direction (0;1) is rotated to ( sin ;cos ). 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter‐clockwise rotation (θ) of the ellipse: à L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : ê T ; 6 F : ê U ; 6 p h. The center of the circle used to be at the origin. Approximately sketch the ellipse - the major axis of the ellipse is x-axis. The equation stated is going to have xy terms, and so there needs to be a suitable rotation of axes in order to get the equation in the standard form suitable for the recommended definite integration. The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. H(x, y) = A x² + B xy + C y² + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. Introduce some delay in function(in ms). How | {
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pixels that fulfill the equation as much as possible. Introduce some delay in function(in ms). How might I go about deriving such an equation. For a plain ellipse the formula is trivial to find: y = Sqrt[b^2 - (b^2 x^2)/a^2] But when the axes of the ellipse are rotated I've never been able to figure out how to compute y (and possibly the extents of x). Let us consider a point P(x, y) lying on the ellipse such that P satisfies the definition i. Therefore, equations (3) satisfy the equation for a non-rotated ellipse, and you can simply plot them for all values of b from 0 to 360 degrees. Learn how each constant and coefficient affects the resulting graph. (iii) is the equation of the rotated ellipse relative to the centre. Differential Equations (10) Discrete Mathematics (4) Discrete Random Variable (5) Disk Washer Cylindrical Shell Integration (2) Division Tricks (1) Domain and Range of a Function (1) Double Integrals (3) Eigenvalues and Eigenvectors (1) Ellipse (1) Empirical and Molecular Formula (2) Enthalpy Change (2) Expected Value Variance Standard. It can be a parametric formulation (e. (1) Ellipse (2) Rotated Ellipse (3) Ellipse Representing Covariance. Rotated Ellipse The implicit equation x2 xy +y2 = 3 describes an ellipse. (1) which is in the form Ax2+ Bxy+ Cy2= 1, with Aand Cpositive. Writing Equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ into standard form by rotating the axes. Let the coordinates of F 1 and F 2 be (-c, 0) and (c, 0) respectively as shown. (a) Find the points at which this ellipse crosses the x-axis. Major axis is vertical. By using this website, you agree to our Cookie Policy. H(x, y) = A x² + B xy + C y² + D x + E y + F = 0 The basic principle of the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that | {
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the incremental line tracing algorithms (I wouldn't call them scanline) is to follow the pixels that fulfill the equation as much as possible. Ellipse graph from standard equation. Find dy dx. If and are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. Except for degenerate cases, the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 x¿y¿-term x¿ 2 4 + y¿ 1 = 1. asked • 04/02/15 find the equation of the image of the ellipse x^2/4 + y^2/9 when rotated through pi/4 about origin. The Danish author and scientist Piet Hein (1905-1996) dealt with the super ellipse in great detail (book 4). For any ellipse, 0 < e < 1. 1) and (e, f) = (e. Rewrite the equation 2x^2+√3 xy+y^2−2=0 in a rotated x^′ y^′-system without an x^′ y^′-term. In other words, the equation of the plane through the center of the circle sloping away from the drawing plane with slope m is given by (3. Solve them for C, D, E. Example : Given ellipse : 4 2 (x − 3) 2 + 5 2 y 2 = 1 b 2 X 2 + a 2 Y 2 = 1 a 2 > b 2 i. For the Earth–sun system, F1 is the position of the sun, F2 is an imaginary point in space, while the Earth follows the path of the ellipse. The equation is (x - h) squared/a squared plus (y - k) squared/a squared equals 1. This makes the analysis somewhat easier. Rotated Parabolas and Ellipse. Rotate the ellipse by applying the equations: RX = X * cos_angle + Y * sin_angle RY = -X * sin_angle + Y * cos_angle. The Steiner ellipse has the minimal area surrounding a triangle. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. 3 Major axis, minor axis and rotated angle of a ellipse Find the major axis, minor axis and rotated angle, where the major axis is twice of the longest radius and the minor axis is also twice of the shortest radius. Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle. | {
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this represents elliptically polarized light in which the major axis of the ellipse makes an angle. This ellipse is called the distortion ellipse. − <: Ellipse − >: Hyperbola; Firstly, if B is not zero then the graph represents a rotated conic. Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a ≥ b > 0, major axis on x-axis, minor axis on y-axis). In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. How might I go about deriving such an equation. A conic in a rotated coordinate system takes on the form of , where the prime notation represents the rotated axes and associated coefficients. point is expressed as (yi = yi cos O - x, sin 0). none of these. I have the verticles for the major axis: d1(0,0. for the ellipse tting is to transform Eq. and through an angle of 30°. Activity 4: Determining the general equation of an ellipse/ Determining the foci and vertices of an ellipse. Donate or volunteer today! Site Navigation. An ellipse is a unique figure in astronomy as it is the path of any orbiting body around another. Then: (Canonical equation of an ellipse) A point P=(x,y) is a point of the ellipse if and only if Note that for a = b this is the equation of a circle. I am trying to find an algorithm to derive the 4 angles, from the centre of a rotated ellipse to its extremities. The radii of the ellipse in both directions are then the variances. Solved Examples Q 1: Find out the coordinates of the foci, vertices, lengths of major and minor axes, and the eccentricity of the ellipse 9x 2 + 4y 2 = 36. tive number has a square root. Log InorSign Up. Solve them for C, D, E. Parabola if A or C = 0 therefore AC = 0 B. 4 we saw that the graph is a hyperbola when AC < 0. I first solved the equation of the ellipse for y, | {
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we saw that the graph is a hyperbola when AC < 0. I first solved the equation of the ellipse for y, getting y= '. The length a always refers to the major axis. Erase the previous Ellipse by drawing the Ellipse at same point using black color. com Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse. Because A = 7, and C = 13, you have (for 0 θ < π/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. For a given chord or triangle base, the. (3), we get (x 0+D=(2A 0))2 (F0=A0) + (x0+E=(2C0))2 (F0=C0) = 1; (6). A ray of light passing through a focus will pass through the other focus after a single bounce (Hilbert and Cohn-Vossen 1999, p. Then the equation of the ellipse in this new coordinate system becomes. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). Notes College Algebra teaches you how to find the equation of an ellipse given a graph. (An ellipse where a = b is in fact a circle. First I want to look at the case when , and. The ellipse that is most frequently studied in this course has Cartesian equation; where. Substituting in the values for x and y above, we get an equation for the new coordinates as a function of the old coordinates and the angle of rotation: x' = x × cos (β) - y × sin (β) y' = y × cos (β) + x × sin (β). 4 we saw that the graph is a hyperbola when AC < 0. The major axis is parallel to the X axis. The ellipse can be rotated. A constructional method for drawing an ellipse in drafting and engineering is usually referred to as the "4 center ellipse" or the "4 arc ellipse". Figure 3: Polarization Ellipse. If an ellipse is rotated about one of its principal axes, a spheroid is the result. Determine the minimum and maximum X and Y limits for the ellipse. At first blush, these are really strange exponents. ) (11 points) The equation x2−xy+y2 = 3 represents a “rotated ellipse”—that is, an ellipse whose axes are not parallel to the | {
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x2−xy+y2 = 3 represents a “rotated ellipse”—that is, an ellipse whose axes are not parallel to the coordinate axes. y axis [see Figs. is along the ellipse’s major axis, the correlation matrix is σ′ = σ′2 1 0 0 σ′2 2. A circle in 3D is parameterized by six numbers: two for the orientation of its unit normal vector, one for the radius, and three for the circle center. The equation of the ellipse we discussed in class is 9 x2 - 4 xy + 6 y2 = 5. the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. Log InorSign Up. I know that i can draw it using ellipse equation, then rotate it, compute points and connect with lines. Erase the previous Ellipse by drawing the Ellipse at same point using black color. HELP?! Rotate the axes to eliminate the xy-term in the equation. If the major axis lies along the y-axis, a and b are swapped in the equation of an ellipse (below). Once we have those we can sketch in the ellipse. Center of ellipse will be the mid point of first and second point always. Accordingly, we can find the equation for any ellipse by applying rotations and translations to the standard equation of an ellipse. General Equation. To rotate the graph of the parabola about the origin, we rotate each point individually. When a>b, we have a prolate spheroid, that is, an ellipse rotated around its major axis; when a K > -1 = ellipse, -1 = parabolic, and K < -1 is hyperbolic; R is the radius of curvature. Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (a ≥ b > 0, major axis on x-axis, minor axis on y-axis). If $$\displaystyle D = b^2- 4ac$$, then it's an ellipse for $$\displaystyle D<0$$, a parabola for $$\displaystyle D = 0$$, and a hyperbola for $$\displaystyle D>0$$. Apply a square completion method to Eq. Here is a short (and probably inaccurate, because I don't really understand the math) explanation for how it works. Rotate the ellipse by applying the equations: RX = X * cos_angle + Y * sin_angle RY = -X * | {
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it works. Rotate the ellipse by applying the equations: RX = X * cos_angle + Y * sin_angle RY = -X * sin_angle + Y * cos_angle. Matrix transformations are affine and map a point such as that to the expected point on the rotated ellipse, but these transformations don't work like that. For the Earth–sun system, F1 is the position of the sun, F2 is an imaginary point in space, while the Earth follows the path of the ellipse. Then the equation of the ellipse in this new coordinate system becomes. We have also seen that translating by a curve by a fixed vector ( h , k ) has the effect of replacing x by x − h and y by y − k in the equation of the curve. Draw the Ellipse at calculated point using white color. To verify, here is a manipulate, which plots the original -3. Question 625240: If the ellipse defined by the equation 16x^2+4y^2+96x-8y+84=0 is translated 6 units down and 7 units to the left, write the standard equation of the resulting ellipse Answer by Edwin McCravy(18045) (Show Source):. phi is the rotation angle. Two fixed points inside the ellipse, F1 and F2 are called the foci. 829648*x*y - 196494 == 0 as ContourPlot then plots the standard ellipse equation when rotated, which is. Expanding the binomial squares and collecting like terms gives. If it is rotated about the major axis, the spheroid is prolate, while rotation about the minor axis makes it oblate. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse. The rotated axes are denoted as the x′ axis and the y′ axis. Rotation Creates the ellipse by appearing to rotate a circle about the first axis. Therefore, equations (3) satisfy the equation for a non-rotated ellipse, and you can simply plot them for all values of b from 0 to 360 degrees. A Rotated Ellipse In this handout I have used Mathematica to do the plots. the sum of distances of P from F 1 and F 2 in the plane is a constant 2a. The equations of tangent and normal to the ellipse\frac{{{x^2}}}{{{a^2}}} + | {
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plane is a constant 2a. The equations of tangent and normal to the ellipse\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$at the point$$\left( {{x_1},{y_1}} \right)$$are$$\frac. Repeat from Step 1. Let x' and y' be the new set of axes along the principal axes of the ellipse. lationship between two images is pure rotation, i. And for a hyperbola it is: x 2 a 2 − y 2 b 2 = 1. Because A = 7, and C = 13, you have (for 0 θ < π/2) Therefore, the equation in the x'y'-system is derived by making the following substitutions. The major axis of this ellipse is horizontal and is the red segment from (-2, 0) to (2, 0). The amount of correlation can be interpreted by how thin the ellipse is. In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle. Here is a reference to plotting an ellipse, without rotation of the major axis from the horizontal: Ellipse in a chart. Here a > b > 0. Sketch the graph of Solution. He provided me with some equations that I combined with a neat ellipse display program written by Wayne Landsman for the NASA Goddard IDL program library to come up with a "center of mass" ellipse fitting program, named Fit_Ellipse. Put ellipse equation in. Thus an ellipse may be drawn using two thumbtacks and a string. The velocity equation for a hyperbolic trajectory has either + , or it is the same with the convention that in that case a is negative. E, qua, euros' Sale, per forza di guerra. The Rotated Ellipsoid June 2, 2017 Page 1 Rotated Ellipsoid An ellipse has 2D geometry and an ellipsoid has 3D geometry. Moment of inertia is defined with respect to a specific rotation axis. A more general figure has three orthogonal axes of different lengths a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. Cartesian Equations of | {
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a, b and c, and can be represented by the equation x 2 /a 2 + y 2 /b 2 + z 2. Cartesian Equations of the ellipse and hyperbola. x2 a2 + y2 b2 = 1. Ellipse drawing tool. If the center of the ellipse is at point (h, k) and the major and minor axes have lengths of 2a and 2b respectively, the standard equation is. The chord perpendicular to the major axis at the center is the minor axis. If σ 6= τ the set Q is a hyperbola when F 6= 0. The longer axis, a, is called the semi-major axis and the shorter, b, is called the semi-minor axis. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Applying the methods of Equation of a Transformed Ellipsenow leads to the following equation for a standard ellipse which has been rotated through an angle α. Standard Equations of Ellipse. (Since I wasn't asked for the length of the minor axis or the location of the co-vertices, I don't need the value of b itself. Im making a small sample, kinda like line rider except with less functionablilty and an ellipse. A constructional method for drawing an ellipse in drafting and engineering is usually referred to as the "4 center ellipse" or the "4 arc ellipse". 3) Calculate the lengths of the ellipse axes, which are the square root of the eigenvalues of the covariance matrix: A E C R = H L A E C A J R = H Q A O : ? ; 4) Calculate the counter‐clockwise rotation (θ) of the ellipse: à L 1 2 Tan ? 5 d l 1 = O L A ? P N = P E K p I l 2 T U : ê T ; 6 F : ê U ; 6 p h. +- hat j b. -The equation x2 − xy + y2 = 3 represents a "rotated" ellipse, which means the axes of the ellipse are not parallel to the coordinate axes (feel free to graph the ellipse on wolframalpha to get a picture). This way we only draw one object (instead of a thousand) and x and y are now the arrays of all of these points (or coordinates) for the ellipse. If it is rotated about the major axis, the spheroid is prolate, while rotation about the | {
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ellipse. If it is rotated about the major axis, the spheroid is prolate, while rotation about the minor axis makes it oblate. Express the equation in the standard form of a conic section. Several exam. This gives a surface composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. In this section, we will discuss the equation of a conic section which is rotated by. Given an ellipse on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-h)²/a²+(y-k)²/b²=1. The objective is to rotate the x and y axes until they are parallel to the axes of the conic. This video derives the formulas for rotation of axes and shows how to use them to eliminate the xy term from a general second degree polynomial. Here is a simple calculator to solve ellipse equation and calculate the elliptical co-ordinates such as center, foci, vertices, eccentricity and area and axis lengths such as Major, Semi Major and Minor, Semi Minor axis lengths from the given ellipse expression. Question 625240: If the ellipse defined by the equation 16x^2+4y^2+96x-8y+84=0 is translated 6 units down and 7 units to the left, write the standard equation of the resulting ellipse Answer by Edwin McCravy(18045) (Show Source):. The following equation on the polar coordinates (r, θ) describes a general ellipse with semidiameters a and b, centered at a point (r 0, θ 0), with the a axis rotated by φ relative to the polar axis:. 5 (a) with the foci on the x-axis. The distance between the center and either focus is c, where c 2 = a 2 - b 2. ; If and are equal and nonzero and have the same sign, then the graph may be a circle. An ellipse is a circle scaled (squashed) in one direction, so an ellipse centered at the origin with semimajor axis a and semiminor axis b < a has equation. Since (− 6 3) 2 − 4 ⋅ 7 ⋅ 13 = − 346 < 0 and A ≠ C since 7 ≠ 13, the equation satisfies the conditions to be an ellipse. 1) and we are back to equations (2). Here is a reference to | {
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the conditions to be an ellipse. 1) and we are back to equations (2). Here is a reference to plotting an ellipse, without rotation of the major axis from the horizontal: Ellipse in a chart. First, notice that the equation of the parabola y = x^2 can be parametrized by x = t, y = t^2, as t goes from -infinity to infinity; or, as a column vector, [x] = [t] [y] = [t^2]. | {
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# What is the chance of a sample in white noise to be a local extremum?
I draw a long sequence of values from a uniform or normal distribution. Observing the resulting sequence, what is the chance of any point in the sequence (aside from the first and last where it is not defined) to be a local extremum? I.e. either a local maximum (the point is larger than both its neighbors) or a local minimum (the point is smaller than both its neighbors).
At first glance, you would think that once you've drawn the 1st point, the expectancy of the 2nd to be smaller\bigger than the 1st is 50% and then the 3rd is bigger\smaller in 50% hence 0.5*0.5 + 0.5*0.5 = 0.5 chance. However there is additional information in the 2nd point being larger than the 1st and hence a bigger (than 50%) chance that the 3rd point is also smaller than the 2nd.
A matlab code which calculates these chances result a chance of 2/3 (0.66%) :
sum(diff(sign(diff(normrnd(0,1,1,10000))))~=0)/10000 % Normal distribution
sum(diff(sign(diff(rand(1,10000))))~=0)/10000 % Uniform distribution
Iv'e tried to apply a combinatoric reasoning to explain this result : As being extremum is defined by 3 points only, let's narrow the problem to observe only 3 points. There are 6 possible orderings to 3 points, in 4 of which the second point is a local extremum, hence the 2/3 chance.
Is this a correct reasoning? Are there other better ways to approach the problem (Integrating the probability density function)? | {
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• By probability you here mean the expectancy of the number of local extrema relative to the number of points in each realization, correct? Sep 7 '15 at 18:14
• Indeed. The expected proportion of points which are local extrema from all the points in an instance of a randomized sequence. Sep 7 '15 at 19:23
• You have the right answer and the right reasoning. Another way to look at things: if you have a cube with the position in the cube dictated by the uniform distribution, then its individual coords are uniformly distributed. The planes that satisfy the equations $x=y$, $y=z$ and $x=z$ cut the cube into six equal parts. Two of them satisfy your condition. This uses the fact that distribution is irrelevant, as pointed out by Ian and A. Donda. Sep 7 '15 at 20:33
I have only a partial theoretical answer, plus another numeric simulation.
As pointed out in a comment to Ian's answer, the solution does not depend on the distribution of the data (as long as it is continuous), so I'm simply using uniformly distributed data on [0, 1].
First, I can confirm that in simulations the expected frequency of local extrema appears to not depend on the length $n$ of sequences, and appears to be exactly 2/3. Here is Matlab code to demonstrate this:
N = 1e6; % number of realizations
n = 100; % length of sequence
ext = diff(sign(diff(rand(n, N)))) ~= 0; % extrema
mean(mean(ext)) % relative frequency across points, estimated expectancy across realizations | {
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First, why exactly 2/3? As the original poster pointed out, the answer of course depends on the probability of occurrences of different orderings, for simplicity for $n = 3$. There are six such orderings: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because the underlying three random variables are independent and identically distributed, any permutation of the three does not change their joint distribution. The probability distribution across orderings therefore has to be invariant under these permutations. This means this distribution must be uniform, and we therefore have a probability of 1/6 for each ordering.
Since four of the orderings lead to a local extremum, namely those with a 1 or a 3 in the middle position, we have a probability of 4/6 = 2/3 for a local extremum.
As pointed out by Ian, the same probability does not necessarily apply to the situation with $n > 3$. If there is a local maximum at one point, there cannot be a local maximum at each of the neighboring points, i.e. we have a negative local correlation w.r.t. the occurrence of a local maximum. It is however possible that there is a local minimum at the next position. I do not see a way to sort out these correlations theoretically, but I looked at them in the simulated data:
R = corr(ext');
The correlation matrix looks like this:
It appears that the autocorrelation is stationary. Averaging across diagonals we can extract the autocorrelation function
lags = -(n - 3) : n - 3;
r = nan(size(lags));
for k = 1 : numel(lags)
r(k) = mean(diag(R, lags(k)));
end
with this result: | {
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with this result:
According to this we do have a small negative autocorrelation (-0.125) at lag 1, but an even smaller positive autocorrelation (0.025) at lag 2. Beyond that, correlations are smaller than $\pm$0.001. So my only guess is that the negative autocorrelation at lag 1 is somehow perfectly counterbalanced by the positive autocorrelation at lag 2 so that the average probability of occurrence of a local extremum stays at 2/3.
Generalization of the order-statistics approach: Just as we enumerated the possible orderings for $n = 3$ above, we can do so also for larger $n$ computationally, and then determine the probability of a local extremum under the assumption (following the same argument as above) that all orderings are equally possible. Here is an implementation in Matlab which uses the Symbolic Math Toolbox to perform exact arithmetic on rational numbers:
for n = 3 : 10
uo = perms(1 : n);
ext = diff(sign(diff(uo'))) ~= 0;
ratio = sprintf('%d / %d', sum(sum(ext)), numel(ext));
fprintf('%d : %s = %s\n', n, ratio, char(sym(ratio)))
end
The result is
3 : 4 / 6 = 2/3
4 : 32 / 48 = 2/3
5 : 240 / 360 = 2/3
6 : 1920 / 2880 = 2/3
7 : 16800 / 25200 = 2/3
8 : 161280 / 241920 = 2/3
9 : 1693440 / 2540160 = 2/3
10 : 19353600 / 29030400 = 2/3
The first number in the ratios is the number of local extrema, across all $n - 2$ positions and all $n!$ possible orderings, the second is the number of possible local extrema, $(n - 2) ~ n!$.
So now we have the computational but exact proof that the probability is exactly 2/3 for $n$ up to 10. Beyond that, the number of possible orderings becomes so large that it takes very long to evaluate.
I have the feeling this could lead to a proof by induction for arbitrary $n$, but do not know yet how to do so.
Some thoughts here: | {
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Some thoughts here:
Consider a sequence of $N$ iid random variables $X_i$. We have $N-2$ Bernoulli random variables $\{ B_i \}_{i=2}^{N-1}$ which are $1$ when $X_i$ is an extremum and $0$ otherwise. It is not immediately obvious that the $B_i$ are even independent; certainly the version where $B_i$ is $1$ when $X_i$ is a maximum are not independent! So there may be some nontrivial coupling to be considered here.
We can remove this coupling (if it exists) by restricting attention to the case $N=3$ for a moment. Denote the PDF of the variables separately by $f$. Then the probability that $B_2=1$ is
$$\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{\max \{ x,z \}}^\infty f(x) f(y) f(z) dy dx dz + \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^{\min \{ x,z \}} f(x) f(y) f(z) dy dx dz.$$
This can be simplified by splitting over which of $x,z$ is the maximum/minimum. In the uniform case it is straightforward to compute analytically.
In general this probability is not independent of the distribution used. For example, in the Bernoulli case, this quantity is $p^2(1-p)+p^2(1-p)=p-p^2$, so it is larger for $p$ closer to $1/2$. So while I think you are right that it is $2/3$ in the uniform case (I seem to recall a geometric proof that each of the $n!$ "ordered tetrahedra" of the $n$-cube have equal probability under the uniform distribution), I would expect it to be different in general. | {
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• The property of being a local extremum only depends on the ordering of data. It is therefore invariant under strictly increasing transformations of the data, which means every continuous distribution can be transformed into the uniform distribution without changing extrema. I therefore believe that the probability in question should not depend on the distribution. Sep 7 '15 at 18:16
• @A.Donda Indeed you're right; concretely, if $X$ is continuously distributed with CDF $F$, then $F$, considered with codomain $(0,1)$, is invertible. Moreover $F^{-1}$ is an increasing function $F^{-1}(X)$ is $U(0,1)$ distributed. This will also work if $X$ is not continuous but rather merely has no discrete part. That's an interesting discrepancy, which in view of the central limit theorem is surprising, at least to me.
– Ian
Sep 7 '15 at 18:26
• I'd be interested to know what you think of my partial answer. Sep 7 '15 at 19:34
• Minor error in my previous comment: $F(X)$ is $U(0,1)$. $F^{-1}(X)$ needn't make sense.
– Ian
Sep 8 '15 at 15:18 | {
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# What does the math symbol $\propto$ mean?
I came across this symbol in my engineering class and I have never seen it before. Anyone know this?
• You know, you can always raise your hand and ask :) Likely others in the class were also not familiar with it. – Jair Taylor Aug 31 '19 at 18:32
• Well.. More like reading it in my book and came across it :) – C. K. Aug 31 '19 at 18:42
It typically means proportional to. Such that
If $$y=cx$$ for some constant $$c$$ we say
$$y\propto x$$ so that when x grows, y grows proportionally by the ratio $$c$$
Alternatively inverse proportionality is when
$$y=c\frac{1}{x}$$ so that when x gets smaller, y gets bigger proportionally by $$c$$
$$y\propto \frac{1}{x}$$ to my knowledge there isn’t a symbol specifically for inverse proportionality and $$y\propto \frac{1}{x}$$ is used instead
• Is there a symbol for inverse proportionality as well? – C. K. Aug 31 '19 at 18:34
• No I don’t think so typically $$y\propto \frac{1}{x}$$ is used – Colin Hicks Aug 31 '19 at 18:43
• This solved my question. Thanks – C. K. Aug 31 '19 at 18:44
As others have pointed out, this symbol means "is proportional to." That is, $$a \propto b$$ means that there is some constant $$C$$ such that $$a = Cb.$$
That being said, in the interests of "teaching a man to fish", it is worth pointing out that it is often possible to "reverse engineer" the meaning of specific mathematical symbols using DeTeXify. In this case, drawing the mysterious symbol gives
Either \propto or \varpropto seems to give the correct symbol. TeX often uses "var" to indicate a variation of a symbol (for example \epsilon $$\epsilon$$ vs \varepsilon $$\varepsilon$$; \theta $$\theta$$ vs \vartheta $$\vartheta$$), so "propto" seems like a reasonable guess as to what this symbol should be called. Googling this term gives a large number of results, many of which are relevant. | {
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• This is awesome! I'll use this in the future if I come across more mystic symbols! – C. K. Aug 31 '19 at 18:45
• Glad I could be of use. Honestly, DeTeXify is one of my favorite things on the interwebs. There is also a free-to-use desktop version (you can pay something like 8 bucks to remove shareware reminders, if you desire---I think it is worth it to support a snazzy bit of hardware). – Xander Henderson Aug 31 '19 at 18:48
$$\propto$$ means "proportional to."
This symbol is often used, primarily in physics, to denote direct proportionality. so the RHS and LHS of your expression only differ by some scalar multiple. | {
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# Homework Help: Finding eigenvectors of [[1,-1,-1],[-1,1,-1],[-1,-1,1]]
1. Jul 15, 2012
### thetrystero
he eigenvalues of the 3x3 matrix [[1,-1,-1],[-1,1,-1],[-1,-1,1]] are 2,2, and -1.
how can i compute the eigenvectors?
for the case lambda=2, for example, i end up with an augmented matrix [[-1,-1,-1,0],[-1,-1,-1,0],[-1,-1,-1,0]] so i'm stuck at this point.
much appreciated.
2. Jul 15, 2012
### Ray Vickson
So, you need to solve the linear system
$$x_1 + x_2 + x_3 = 0\\ x_1 + x_2 + x_3 = 0\\ x_1 + x_2 + x_3 = 0$$
There are lots of solutions. In fact, you should be able to find two linearly independent solution vectors, corresponding to the double eigenvalue 2.
RGV
3. Jul 15, 2012
### HallsofIvy
I prefer to work from the basic definitions (perhaps I just never learned these more sophisticated methods!):
Saying that 2 is an eigenvalue of this matrix means there exist a non-zero vector such that
$$\begin{bmatrix}1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}x - y- z\\ -x+ y- z \\ -x- y+ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y\\ 2z\end{bmatrix}$$
which gives the three equations x- y- z= 2x, -x+ y- z= 2y, -x- y+ z= 2z which are, of course, equivalent to -x- y- z= 0, -x- y- z= 0, -x- y- z= 0. Those three equations are the same. We can, for example, say that z= -x- y so that any vector of the form <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1> is an eigenvector. Notice that the eigenvalue, 2, not only has algebraic multiplicity 2 (it is a double root of the characteristic equation) but has geometric multiplicity 2 (the space of all corresponding eigenvalues is 2 dimensional). | {
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Similarly, the fact that -1 is an eigenvalue means there are x, y, z, satisfying x- y- z= -x, -x+ y- z= -y, -x- y+ z= -z which are, of course, equivalent to 2x- y- z= 0, -x+ 2y- z= 0, -x- y+ 2z= 0. If we subtract the second equation from the first, we eliminate z to get 3x- 3y= 0 so y= x. Putting that into the third equation, 2x+ 2z= 0 so z= -x.
Any eigenvector corresponding to eigenvalue -1 is of the form <x, x, -x>= x<1, 1, -1>.
4. Jul 15, 2012
### thetrystero
by that reasoning, can i not have <1,-1,0> and <-1,1,0> as my two solutions for eigenvalue 2? but wolframalpha says i need to have the case where y=0.
also, i think the solution for eigenvalue -1 is <1,1,1>
5. Jul 16, 2012
### Ray Vickson
Your two listed vectors (for eigenvalue 2) are just multiples of each other. You need two *linearly independent* eigenvectors, such as <1,-1,0> and <1,0,-1>, or <0,-1,1> and <1,-1/2, -1/2>, etc. There are infinitely many possible pairs of vectors <x1,y1,z1> and <x2,y2,z2> that are linearly independent and satisfy the equation x+y+z=0. Any such pair will do.
RGV
Last edited: Jul 16, 2012
6. Jul 16, 2012
### thetrystero
yes, i had thought of that, but found it uncomfortable that of all the many possibilities, both my professor and wolframalpha chose the cases y=0 and z=0 as solutions, so i was wondering what made these two special compares to the others.
7. Jul 17, 2012
### Ray Vickson
There is nothing special about these choices, except for the fact that they both have one component = 0 so are, in a sense, the simplest possible. However, you could equally take x=0 and y=0 or x=0 and z=0.
RGV
8. Jul 17, 2012
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RGV
8. Jul 17, 2012
### HallsofIvy
If you have a vector that depends upon parameters, say, <x, y, -x- y> as I have above, then choosing x= 0, y= 1 gives you <0, 1, -1> and choosing x= 1, y= 0 gives <1, 0, -1>. That is, in effect, the same as writing <x, y, -x- y>= <x, 0, -x>+ <0, y, -y>= x<1, 0, -1>+ y<0, 1, -1>, showing that any such vector is a linear combination of <1, 0, -1> and <0, 1, -1>. | {
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# How can I know the analytic continuation exists in certain cases?
As pointed in Does the analytic continuation always exists? we know it doesn't always exist.
But: take the $\Gamma$ function: the first definition everyone meet is the integral one: $$z\mapsto\int_{0}^{+\infty}t^{z-1}e^{-t}\,dt$$ which defines an holomorphic function on the half plane $\{\Re z>0\}$. Moreover we immediately get the functional equation: $$\Gamma(z+1)=z\Gamma(z)\;,\;\;\;\forall\; \Re z>0.$$ This equation is used to extend the function on the whole complex plane (minus the negative integers)... but: WHY CAN WE DO THIS?!
We know that there is an holomorphic function $\Gamma$ which can be expressed as the integral on that half plane. Why are we allowed to write $$\Gamma\left(\frac12\right)=-\frac12\Gamma\left(-\frac12\right)$$ for example? LHS is defined, RHS, NOT!!! But where's the problem? Simply let's define $\Gamma\left(-\frac12\right)$ in such a way... but why can we do this? How can I know that this function I named $\Gamma$ which is holomorphic on the above half plane admits an extension?
• You are just using the same name for two differenti objects. – N74 Apr 13 '16 at 20:11
Once you have the functional equation for $\Gamma$, we can define a new function $\tilde \Gamma$ defined on the half-plane $\operatorname{Re} z > -1$ (except $z=0$) by $$\tilde \Gamma(z) = \frac1z \Gamma(z+1).$$ It's clear that $\tilde \Gamma$ is holomorphic on $\{ \operatorname{Re} z > -1 \} \setminus \{ 0 \}$, and coincides with $\Gamma$ on $\operatorname{Re} z > 0$ (because of the functional equation). In other words, $\tilde \Gamma$ is an analytic continuation of $\Gamma$ to $\{ \operatorname{Re} z > -1 \} \setminus \{ 0 \}$. So we may as well call $\tilde \Gamma$ by $\Gamma$.
Repeating the above construction, we can define a "new $\Gamma$-function" on successively larger sets until we get something defined and holomorphic on the whole complex plane except the non-positive integers. | {
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• I don't agree. you are totally obfuscating the process of analytic continuation this way : what about the analytic continuation on a path of intersecting disks ? how do you know that the analytic continuation doesn't depend on the path (the fact that $\Gamma(s)$ is meromorphic ?) – reuns Apr 13 '16 at 21:47
The following are citations from the classic Applied and Computational Complex Analysis, Vol. I by P. Henrici. The chapter 3: Analytic Continuation provides a thorough treatise of the theme. Here we look at two aspects, which should help to clarify the situation.
At first we take a look when two functions $f(z)$ and $g(z)$ are analytic continuations from each other.
Theorem 3.2d: (Fundamental lemma on analytic continuation)
Let $Q$ be a set with point of accumulation $q$ and let $R$ and $S$ be two regions such that their intersection contains $Q$ and $q$ is connected. If $f$ is analytic on $R$, $g$ is analytic on $S$, and $f(z)=g(z)$ for $z\in Q$, then $f(z)=g(z)$ throughout $R\cap S$ and $f$ and $g$ are analytic continuations of each other.
We observe, we need at least a set $Q$ with an accumulation point where analytic functions $f$ and $g$ have to coincide. This set is part of the intersection of two regions $R$ and $S$ where $f$ and $g$ are defined. Finally we conclude that throughout $R\cap S$ the functions coincide.
The second aspect sheds some light at functional relationships in connection with analytic continuation. We can read in
Section 3.2.5: Analytic Continuation by Exploiting Functional Relationships
Occasionally an analytic continuation of a function $f$ can be obtained by making use of a special functional relationship satisfied by $f$. Naturally this method is restricted to those functions for which such relationships are known.
He continues with example 15 which seems that P. Henrici had precisely a user with OPs question in mind.
Example 15:
Let the function $g$ possess the following properties: | {
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Example 15:
Let the function $g$ possess the following properties:
• (a) $g$ is analytic in the right half-plane: $R:\Re (z)>0$
• (b) For all $z\in R, zg(z)=g(z+1)$
We assert that $g$ can be continued analytically into the whole complex plane with the exception of the points $z=0,-1,-2,\ldots$.
We first continue $g$ into $S:\Re (z)>-1,z\neq 0$. For $z\in S$,let $f$ be defined by \begin{align*} f(z):=\frac{1}{z}g(z+1) \end{align*} For $z\in S, \Re(z+1)>0$. Hence by virtue of (a) $f$ is analytic on $S$. In view of (b) $f$ agrees with $g$ on the set of $R$. Since $S$ is a region, $f$ represents the analytic continuation of $g$ from $R$ to $S$. We note that $f$ satisfies the functional relation $f(z+1)=zf(z)$ on the whole set $S$.
Denoting the extended function again by $g$, we may use the same method to continue $g$ analytically into the set $\Re(z)>-2,z\neq 0,-1$, and thus step by step into the region $z\neq 0,-1,-2,\ldots$.
Of course this example addresses the Gamma Function $\Gamma(z)$ which is treated in detail in chapter 8, vol. 2. He then continues with further methods of analytic continuation, such as the principle of continuous continuation and the symmetry principle. | {
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# Orthonormal basis
Consider $\mathbb{R}^3$ together with inner product $\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \rangle = 2x_1 y_1+x_2 y_2+3 x_3 y_3$. Use the Gram-Schmidt procedure to find an orthonormal basis for $W=\text{span} \left\{(-1, 1, 0), (-1, 1, 2) \right\}$.
I don't get how the inner product $\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \rangle = 2 x_1 y_1+x_2 y_2+3 x_3 y_3$ would affect the approach to solve this question.. When I did the gram-schmidt, I got $v_1=(-1, 1, 0)$ and $v_2=(0, 0, 2)$ but then realized that you have to do something with the inner product before finding the orthonormal basis. Can someone please help me?
Update: So far I got $\{(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0), (0, 0, \frac{2}{\sqrt{12}})\}$ as my orthonormal basis but I'm not sure if I am doing it right with the given inner product.
-
please try to learn TeX syntax. I did it for you in this case. – Siminore Jun 15 '12 at 13:16
@Alice You got the same solution I got :) To convince yourself it's right, just compute the pairwise inner products between your solution vectors, and you will find they are orthnormal! – rschwieb Jun 15 '12 at 13:29
The choice of inner product defines the notion of orthogonality.
The usual notion of being "perpendicular" depends on the notion of "angle" which turns out to depend on the notion of "dot product".
If you change the way we measure the "dot product" to give a more general inner product then we change what we mean by "angle", and so have a new notion of being "perpendicular", which in general we call orthogonality.
So when you apply the Gram-Schmidt procedure to these vectors you will NOT necessarily get vectors that are perpendicular in the usual sense (their dot product might not be $0$).
Let's apply the procedure.
It says that to get an orthogonal basis we start with one of the vectors, say $u_1 = (-1,1,0)$ as the first element of our new basis. | {
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Then we do the following calculation to get the second vector in our new basis:
$u_2 = v_2 - \frac{\langle v_2, u_1\rangle}{\langle u_1, u_1\rangle} u_1$
where $v_2 = (-1,1,2)$.
Now $\langle v_2, u_1\rangle = 3$ and $\langle u_1, u_1\rangle = 3$ so that we are given:
$u_2 = v_2 - u_1 = (0,0,2)$.
So your basis is correct. Let's check that these vectors are indeed orthogonal. Remember, this is with respect to our new inner product. We find that:
$\langle u_1, u_2\rangle = 3(-1)(0) + (1)(0) + 2(0)(2) = 0$
(here we also happened to get a basis that is perpendicular in the traditional sense, this was lucky).
Now is the basis orthonormal? (in other words, are these unit vectors?). No they arent, so to get an orthonormal basis we must divide each by its length. Now this is not the length in the usual sense of the word, because yet again this is something that depends on the inner product you use. The usual Pythagorean way of finding the length of a vector is:
$||x||=\sqrt{x_1^2 + ... + x_n^2} = \sqrt{x . x}$
It is just the square root of the dot product with itself. So with more general inner products we can define a "length" via:
$||x|| = \sqrt{\langle x,x\rangle}$.
With this length we see that:
$||u_1|| = \sqrt{2(-1)(-1) + (1)(1) + 3(0)(0)} = \sqrt{3}$
$||u_2|| = \sqrt{2(0)(0) + (0)(0) + 3(2)(2)} = 2\sqrt{3}$
(notice how these are different to what you would usually get using the Pythagorean way).
Thus an orthonormal basis is given by:
$\{\frac{u_1}{||u_1||}, \frac{u_2}{||u_2||}\} = \{(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0), (0,0,\frac{1}{\sqrt{3}})\}$
-
Thanks so much for the thorough explanation! :) – Alice Jun 15 '12 at 13:36
Perhaps you carried out the Gram-Schmidt algorithm using the ordinary inner product? I think that is the only way you could have gotten through without using the given inner product :) | {
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Anyhow, you need to use the given inner product at each step of the orthonormalization procedure. Changing the inner product will change the output of the algorithm, because different inner products yield different lengths of vectors and report different "angles" between vectors.
For example, when you begin with the first step (normalizing $(-1,1,0)$, you should compute that $\langle(-1,1,0),(-1,1,0)\rangle=3$, and so the first vector would be $\frac{1}{\sqrt{3}}(-1,1,0)$.
-
Thanks a lot! I understand it now. Oh, I was also wondering if it asks to find the orthogonal projection of (1, 1, 1) onto the same W, I guess we still apply the given inner product? I got (1/3, 1/3, 1) for the projection and I want to make sure I'm doing it right.. – Alice Jun 15 '12 at 13:37
@alice I got a negative sign on the middle entry when I tried a moment ago... can you check your work to see if that's missing in your computation? – rschwieb Jun 15 '12 at 13:44
Oh, oops, I did forget the negative sign in my calculation. Thanks again :) – Alice Jun 15 '12 at 13:46 | {
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# Linear
viewed 16538 times and licensed 1105 times
Calculates the linear regression parameters and evaluates the regression line at arbitrary abscissas
Controller: CodeCogs
C++
## Class Linear
Linear regression is a method to best fit a linear equation (straight line) of the form to a collection of points , where is the slope and the intercept on the axis.
The algorithm basically requires minimisation of the sum of the squared distance from the data points to the proposed line. This is achieved by calculating the derivative with respect to a and b and setting these to zero.
Let us define the following
Then the slope is
the intercept on the Y axis
Below you will find the regression graph for a set of arbitrary points, which were also used in the forthcoming example. The regression line, displayed in red, has been calculated using this class.
### Example 1
The following example displays the slope, Y intercept and regression coefficient for a certain set of 7 points.
#include <codecogs/maths/approximation/regression/linear.h>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double x[7] = { 1.5, 2.4, 3.2, 4.8, 5.0, 7.0, 8.43 };
double y[7] = { 3.5, 5.3, 7.7, 6.2, 11.0, 9.5, 10.27 };
Maths::Regression::Linear A(7, x, y);
cout << " Slope = " << A.getSlope() << endl;
cout << "Intercept = " << A.getIntercept() << endl << endl;
cout << "Regression coefficient = " << A.getCoefficient() << endl;
cout << endl << "Regression line values" << endl << endl;
for (double i = 0.0; i <= 3; i += 0.6)
{
cout << "x = " << setw(3) << i << " y = " << A.getValue(i);
cout << endl;
}
return 0;
}
Output:
Slope = 0.904273
Intercept = 3.46212
Regression coefficient = 0.808257
Regression line values | {
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Regression coefficient = 0.808257
Regression line values
x = 0 y = 3.46212
x = 0.6 y = 4.00469
x = 1.2 y = 4.54725
x = 1.8 y = 5.08981
x = 2.4 y = 5.63238
x = 3 y = 6.17494
### Authors
Lucian Bentea (August 2005)
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login.
## Members of Linear
#### Linear
Linear( int n double* x double* y )[constructor]
Initializes the class by calculating the slope, intercept and regression coefficient based on the given constructor arguments.
### Note
The slope should not be infinite.
n The number of initial points in the arrays x and y x The x-coordinates of points y The y-coordinates of points
#### GetValue
doublegetValue( double x )
x the abscissa used to evaluate the linear regression function
#### GetCoefficient
doublegetCoefficient( )
The regression coefficient indicated how well linear regression fits to the original data. It is an expression of error in the fitting and is defined as:
This varies from 0 (no linear trend) to 1 (perfect linear fit). If and , then r is considered to be equal to 1.
## Linear Once
doubleLinear_once( int n double* x double* y double a )
This function implements the Linear class for one off calculations, thereby avoid the need to instantiate the Linear class yourself.
### Example 2
The following graph fits a straight line to the following values:
x = 1 y = 0.22
x = 2 y = 0.04
x = 3 y = -0.13
x = 4 y = -0.17
x = 5 y = -0.04
x = 6 y = 0.09
x = 7 y = 0.11
There is an error with your graph parameters for Linear_once with options n=7 x="1 2 3 4 5 6 7" y="0.22 0.04 -0.13 -0.17 -0.04 0.09 0.11" a=1:7 .input
Error Message:Function Linear_once failed. Ensure that: Invalid C++
### Parameters
n The number of initial points in the arrays x and y x The x-coordinates of points y The y-coordinates of points a The x-coordinate for the output location | {
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### Returns
the interpolated y-coordinate that corresponds to a.
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login. | {
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Almost every vector space we have encountered has been infinite in size (an exception is Example VSS). But some are bigger and richer than others. Dimension, once suitably defined, will be a measure of the size of a vector space, and a useful tool for studying its properties. You probably already have a rough notion of what a mathematical definition of dimension might be --- try to forget these imprecise ideas and go with the new ones given here.
## Dimension
Definition D (Dimension) Suppose that $V$ is a vector space and $\set{\vectorlist{v}{t}}$ is a basis of $V$. Then the dimension of $V$ is defined by $\dimension{V}=t$. If $V$ has no finite bases, we say $V$ has infinite dimension.
This is a very simple definition, which belies its power. Grab a basis, any basis, and count up the number of vectors it contains. That's the dimension. However, this simplicity causes a problem. Given a vector space, you and I could each construct different bases --- remember that a vector space might have many bases. And what if your basis and my basis had different sizes? Applying Definition D we would arrive at different numbers! With our current knowledge about vector spaces, we would have to say that dimension is not "well-defined." Fortunately, there is a theorem that will correct this problem.
In a strictly logical progression, the next two theorems would precede the definition of dimension. Many subsequent theorems will trace their lineage back to the following fundamental result.
Theorem SSLD (Spanning Sets and Linear Dependence) Suppose that $S=\set{\vectorlist{v}{t}}$ is a finite set of vectors which spans the vector space $V$. Then any set of $t+1$ or more vectors from $V$ is linearly dependent. | {
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The proof just given has some monstrous expressions in it, mostly owing to the double subscripts present. Now is a great opportunity to show the value of a more compact notation. We will rewrite the key steps of the previous proof using summation notation, resulting in a more economical presentation, and even greater insight into the key aspects of the proof. So here is an alternate proof --- study it carefully.
\begin{proof} {\bf (Alternate Proof of Theorem SSLD)} We want to prove that any set of $t+1$ or more vectors from $V$ is linearly dependent. So we will begin with a totally arbitrary set of vectors from $V$, $R=\setparts{\vect{u}_j}{1\leq j\leq m}$, where $m>t$. We will now construct a nontrivial relation of linear dependence on $R$.
Each vector $\vect{u_j}$, $1\leq j\leq m$ can be written as a linear combination of $\vect{v}_i$, $1\leq i\leq t$ since $S$ is a spanning set of $V$. This means there are scalars $a_{ij}$, $1\leq i\leq t$, $1\leq j\leq m$, so that
\begin{align*} \vect{u}_j&=\sum_{i=1}^{t}a_{ij}\vect{v_i}&&1\leq j\leq m \end{align*}
Now we form, unmotivated, the homogeneous system of $t$ equations in the $m$ variables, $x_j$, $1\leq j\leq m$, where the coefficients are the just-discovered scalars $a_{ij}$,
\begin{align*} \sum_{j=1}^{m}a_{ij}x_j=0&&1\leq i\leq t \end{align*}
This is a homogeneous system with more variables than equations (our hypothesis is expressed as $m>t$), so by Theorem HMVEI there are infinitely many solutions. Choose one of these solutions that is not trivial and denote it by $x_j=c_j$, $1\leq j\leq m$. As a solution to the homogeneous system, we then have $\sum_{j=1}^{m}a_{ij}c_{j}=0$ for $1\leq i\leq t$. As a collection of nontrivial scalars, $c_j$, $1\leq j\leq m$, will provide the nontrivial relation of linear dependence we desire, | {
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\begin{align*} \sum_{j=1}^{m}c_{j}\vect{u}_j &=\sum_{j=1}^{m}c_{j}\left(\sum_{i=1}^{t}a_{ij}\vect{v_i}\right) &&\text{ Definition TSVS}\\ &=\sum_{j=1}^{m}\sum_{i=1}^{t}c_{j}a_{ij}\vect{v}_i &&\text{ Property DVA}\\ &=\sum_{i=1}^{t}\sum_{j=1}^{m}c_{j}a_{ij}\vect{v}_i &&\text{ Property CMCN}\\ &=\sum_{i=1}^{t}\sum_{j=1}^{m}a_{ij}c_{j}\vect{v}_i &&\text{Commutativity in $\complex{}$}\\ &=\sum_{i=1}^{t}\left(\sum_{j=1}^{m}a_{ij}c_{j}\right)\vect{v}_i &&\text{ Property DSA}\\ &=\sum_{i=1}^{t}0\vect{v}_i &&\text{$c_j$ as solution}\\ &=\sum_{i=1}^{t}\zerovector &&\text{ Theorem ZSSM}\\ &=\zerovector &&\text{ Property Z} \end{align*}
That does it. $R$ has been undeniably shown to be a linearly dependent set. \end{proof} Notice how the swap of the two summations is so much easier in the third step above, as opposed to all the rearranging and regrouping that takes place in the previous proof. In about half the space. And there are no ellipses ($\ldots$).
Theorem SSLD can be viewed as a generalization of Theorem MVSLD. We know that $\complex{m}$ has a basis with $m$ vectors in it (Theorem SUVB), so it is a set of $m$ vectors that spans $\complex{m}$. By Theorem SSLD, any set of more than $m$ vectors from $\complex{m}$ will be linearly dependent. But this is exactly the conclusion we have in Theorem MVSLD. Maybe this is not a total shock, as the proofs of both theorems rely heavily on Theorem HMVEI. The beauty of Theorem SSLD is that it applies in any vector space. We illustrate the generality of this theorem, and hint at its power, in the next example.
Example LDP4: Linearly dependent set in $P_4$.
Theorem SSLD is indeed powerful, but our main purpose in proving it right now was to make sure that our definition of dimension (Definition D) is well-defined. Here's the theorem.
Theorem BIS (Bases have Identical Sizes) Suppose that $V$ is a vector space with a finite basis $B$ and a second basis $C$. Then $B$ and $C$ have the same size. | {
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Theorem BIS tells us that if we find one finite basis in a vector space, then they all have the same size. This (finally) makes Definition D unambiguous.
## Dimension of Vector Spaces
We can now collect the dimension of some common, and not so common, vector spaces. \begin{theorem}{DCM}{Dimension of $\complex{m}$}{complex vector space!dimension} The dimension of $\complex{m}$ (Example VSCV) is $m$. \end{theorem} \begin{proof} Theorem SUVB provides a basis with $m$ vectors. \end{proof}
Theorem DP (Dimension of $P_n$) The dimension of $P_{n}$ (Example VSP) is $n+1$.
\begin{theorem}{DM}{Dimension of $M_{mn}$}{matrix vector space!dimension} The dimension of $M_{mn}$ (Example VSM) is $mn$. \end{theorem} \begin{proof} Example BM provides a basis with $mn$ vectors. \end{proof}
Example DSM22: Dimension of a subspace of $M_{22}$.
Example DSP4: Dimension of a subspace of $P_4$.
Example DC: Dimension of the crazy vector space.
It is possible for a vector space to have no finite bases, in which case we say it has infinite dimension. Many of the best examples of this are vector spaces of functions, which lead to constructions like Hilbert spaces. We will focus exclusively on finite-dimensional vector spaces. OK, one infinite-dimensional example, and then we will focus exclusively on finite-dimensional vector spaces.
Example VSPUD: Vector space of polynomials with unbounded degree.
## Rank and Nullity of a Matrix
For any matrix, we have seen that we can associate several subspaces --- the null space (Theorem NSMS), the column space (Theorem CSMS), row space (Theorem RSMS) and the left null space (Theorem LNSMS). As vector spaces, each of these has a dimension, and for the null space and column space, they are important enough to warrant names.
Definition NOM (Nullity Of a Matrix) Suppose that $A$ is an $m\times n$ matrix. Then the nullity of $A$ is the dimension of the null space of $A$, $\nullity{A}=\dimension{\nsp{A}}$. | {
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Definition ROM (Rank Of a Matrix) Suppose that $A$ is an $m\times n$ matrix. Then the rank of $A$ is the dimension of the column space of $A$, $\rank{A}=\dimension{\csp{A}}$.
Example RNM: Rank and nullity of a matrix.
There were no accidents or coincidences in the previous example --- with the row-reduced version of a matrix in hand, the rank and nullity are easy to compute.
Theorem CRN (Computing Rank and Nullity) Suppose that $A$ is an $m\times n$ matrix and $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ nonzero rows. Then $\rank{A}=r$ and $\nullity{A}=n-r$.
Every archetype (appendix A) that involves a matrix lists its rank and nullity. You may have noticed as you studied the archetypes that the larger the column space is the smaller the null space is. A simple corollary states this trade-off succinctly. (See technique LC.)
Theorem RPNC (Rank Plus Nullity is Columns) Suppose that $A$ is an $m\times n$ matrix. Then $\rank{A}+\nullity{A}=n$.
When we first introduced $r$ as our standard notation for the number of nonzero rows in a matrix in reduced row-echelon form you might have thought $r$ stood for "rows." Not really --- it stands for "rank"!
## Rank and Nullity of a Nonsingular Matrix
Let's take a look at the rank and nullity of a square matrix.
Example RNSM: Rank and nullity of a square matrix.
The value of either the nullity or the rank are enough to characterize a nonsingular matrix.
Theorem RNNM (Rank and Nullity of a Nonsingular Matrix) Suppose that $A$ is a square matrix of size $n$. The following are equivalent.
1. A is nonsingular.
2. The rank of $A$ is $n$, $\rank{A}=n$.
3. The nullity of $A$ is zero, $\nullity{A}=0$.
With a new equivalence for a nonsingular matrix, we can update our list of equivalences (Theorem NME5) which now becomes a list requiring double digits to number.
Theorem NME6 (Nonsingular Matrix Equivalences, Round 6) Suppose that $A$ is a square matrix of size $n$. The following are equivalent. | {
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1. $A$ is nonsingular.
2. $A$ row-reduces to the identity matrix.
3. The null space of $A$ contains only the zero vector, $\nsp{A}=\set{\zerovector}$.
4. The linear system $\linearsystem{A}{\vect{b}}$ has a unique solution for every possible choice of $\vect{b}$.
5. The columns of $A$ are a linearly independent set.
6. $A$ is invertible.
7. The column space of $A$ is $\complex{n}$, $\csp{A}=\complex{n}$.
8. The columns of $A$ are a basis for $\complex{n}$.
9. The rank of $A$ is $n$, $\rank{A}=n$.
10. The nullity of $A$ is zero, $\nullity{A}=0$. | {
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# What do we actually prove using induction theorem?
Here is the picture of the page of the book, I am reading:
$$P_k: \qquad 1+3+5+\dots+(2k-1)=k^2$$ Now we want to show that this assumption implies that $P_{k+1}$ is also a true statement: $$P_{k+1}: \qquad 1+3+5+\dots+(2k-1)+(2k+1)=(k+1)^2.$$ Since we have assumed that $P_k$ is true, we can perform operations on this equation. Note that the left side of $P_{k+1}$ is the left side of $P_k$ plus $(2k+1)$. So we start by adding $(2k+1)$ to both sides of $P_k$: \begin{align*} 1+3+\dots+(2k-1) &= k^2 &P_k\\ 1+3+\dots+(2k-1)+(2k+1) &= k^2+(2k+1) &\text{Add $(2k+1)$ to both sides.} \end{align*} Factoring the right side of this equation, we have $$1+3+\dots+(2k-1)+(2k+1) =(k+1)^2 \qquad P_{k+1}$$ But this last equation is $P_{k+1}$. Thus, we have started with $P_k$, the statement we assumed true, and performed valid operations to produce $P_{k+1}$, the statement we want to be true. In other words, we have shown that if $P_k$ is true, then $P_{k+1}$ is also true. Since both conditions in Theorem 1 are satisfied, $P_n$ is true for all positive integers $n$.
In the top line there is written that we have to show that 'the assumption $P_k$ is true implies $P_{k+1}$ is true'.
And what I think is that: as long as we know that the state of a proposition being true for any positive integer $k$ after Base number implies that proposition is true for integer $k+1$, we have to show that $P_k$ is true. However I don't have an idea yet how to show the truth of $P_k$.
So, my first question is that who is right? My book or me? And if I am right then how can I show the truth of $P_k$?
2) This may be taken as the second question but it is also bit annoying.
In the second paragraph it is written, "Since we have assumed that $P_k$ is true, we can perform operations on it". | {
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Why is it necessary for an equation to be true for performing operations on it? Well, this is not much annoying as compared to the first question because whole the induction theory depends upon that.
• Who is right about what? You haven't made a claim that contradicts the book, Feb 8 '16 at 13:32
• I have tried to retype the text from your picture. It would be rather difficult task, considering the low quality of the image. Luckily Google and Google Books returns some places with the same text. Could you perhaps add to your post a more precise reference than "the book I am reading"? Feb 8 '16 at 16:27
• You prove the $P_1$ is true. We don't know that $P_n$ is true for all $n$. Only that it is true for $n = 1$. Then we prove that IF it is true for a specific $P_k$ then it is true for $P_{k+1}$. Once we prove these two things: 1) $P_1$ is true. and 2) $P_k \implies P_{k+1}$. Then we can conclude and we have proven 3) $P_n$ is true for all n. Your comment. If we prove it is true for all $P_k$ then it is true for $P_{k+1}$ is irrelevant as we are not assuming and certainly haven't proven it is true for all $P_k$. We have assumed it for ONE $P_k$. Feb 8 '16 at 17:25
• @MarkJoshi Induction works on many structures. Induction of the form "start at zero and add one" also works on finite fields, that is arithmetic $\pmod p$. It is not correct at all to say that natural numbers are precisely the set for which induction works. Feb 9 '16 at 6:02
• Once there is enough evidence to make you believe something is true, it can suddenly become much easier to understand why it is true. Given the multitude of experts here (and the authoer himself) telling you the book is right I recommend, as strange as it sounds, to reread the chapter on induction trying to believe it (and everyone on this site) is right and understand why that is, rather than trying to argue why it is wrong. Feb 9 '16 at 13:25
Induction is, intuitively, an outline of an infinite proof. | {
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Induction is, intuitively, an outline of an infinite proof.
You first prove $P(0)$, the base case.
Then you prove $P(1)$ follows from $P(0)$.
Then you prove that $P(2)$ follows from $P(1)$.
Et cetera.
In general, if you know that $P(k+1)$ follows from $P(k)$, and you know that $P(0)$ is true, then you know how to prove $P(n)$ for any natural number. For example, $P(4)$ follow because $P(0)\implies P(1)\implies P(2)\implies P(3)\implies P(4)$.
So you don't need to know how to show $P(k)$ to show that $P(k)$ implies $P(k+1)$. Once you've shown this implication, and $P(0)$, we know how to write a non-inductive finite proof of $P(10000)$ or $P(10^{99})$.
Your book is right. You have to show that $$P_k \implies P_{k+1}$$
Because if you prove that and since you said it is true for base number, the proposition is automatically proved true for every $n \in S$.
Let us see how!
To prove "$P_k \implies P_{k+1}$" means you are proving that a particular proposition has a quality of being true for $k+1$ because of being true for $k$.
Improvement (Edit): Just as you do in case of domino effect. Suppose it is reported that the cornered domino has been pushed over. And then you find out that dominoes are arranged such that whenever $k^{th}$ domino falls, $(k+1)^{th}$ falls. You would at once say that all dominoes must fall.
Remember that these two conditions of induction theorem simultaneously imply that a certain proposition is true.
i) It is true for base number
ii) It is true for an integer $k$ implies it is true for integer $k+1$ after it.
For the proposition in your picture, if you would take base number $2$ then after doing all that (done in your book), you would prove that proposition is true for integer $2$ and integers after it. But since we usually ask to prove that conjecture for every $n \in \mathbb N$, we take base number $1$ not $2$ because the smallest number of set $\mathbb N$ is $1$. | {
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• It may help to think that you are supposed to prove the arrow in-between the two propositions. Feb 9 '16 at 4:00
(Wanted to post as a comment, but it is too long :/)
When we prove something by induction we prove that our claim is correct for a base case (for example, n=1). Afterwards we assume (not proving, only assuming) that our claim stands for some arbitrary value k and than, based on the assumption we prove it holds for k+1.
Note that based on the proof of base case and the proof that if $P_k$ is true we show that $P_{k+1}$ is true. Example: you prove that some claim is correct for $n=1$, then you assume it is correct for some n=k. Now you prove for n=k+1.
We proved for n=1 and using the assumption that the claim stands for n=k we proved for n=k+1. In this case, we know that the claim stands for n=1, thus we can now say it stands for n=2. Afterwards we can say that it stands for n=3 and so on, hence it stands for any n.
That is how we use induction - you don't prove that $P_k$ is true, you assume it it true - that is a big difference.
If my explanation is not coherent enough, you might want to google "Induction" and read a little more.
Good luck. | {
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Good luck.
• "...that our claim stands for some arbitrary $k$ and then, based on the assumption we prove that claim holds for $k+1$" . Okay.. if we prove the claim for $k+1$ . At what place in the picture, I can see such prove? Feb 8 '16 at 14:12
• All that shown is how they are assuming that $P_k$ is true and how they use it to prove that $P_{k+1}$ is true. Feb 8 '16 at 14:23
• After reading this "$P_k$ holds implies $P_{k+1}$ holds", I guess that we should show truth of $P_k$ and not of $P_{k+1}$ because it is an implication and once we prove $P_k$, then according to our claim ($P_k \implies P_{k+1}$), $P_{k+1}$ would get proved automatically. Feb 8 '16 at 14:30
• @Man_Of_Wisdom, the induction assumption is that $P_k$ is true, we don't need to prove it. We show that because it is true (take it as a fact), than $P_{k+1}$ is also true. We know that "$P_k$ holds", we want to show that $P_{k+1}$ holds. Feb 8 '16 at 14:43
• You show the truth of $P_1$. Than you show that if it's true for $P_k$ which may or may not be the same as $P_1$ then it must be true for $P_{k+1}$. From that we know that $P_1 \implies P_2$. And therefore $P_2 \implies P_3$. Now we can hire an undergraduate to sit in an office to continue "$...\implies P_{2,345}$ and $P_{2,345} \implies P_{2,346}$ and $P_{2346} \implies ...$". Or we can conclude: 1) ($P_1$ is true). and 2) ($P_k \implies P_{k+1}): \implies$ 3) $P_n$ is true for all $n$. Feb 8 '16 at 17:31
## What do we actually prove using induction theorem?
Probably an example is needed here. Let's suppose you want to prove the little Gauss. Suppose we want to prove this Theorem for all $$n\in\mathbb{N}$$:
$$1+2+3+ \cdots+ n = \sum_{k=1}^n k = \frac{n(n+1)}{2}$$ | {
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$$1+2+3+ \cdots+ n = \sum_{k=1}^n k = \frac{n(n+1)}{2}$$
The statement $$A(n)$$ is thus given by: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ With induction this is really simple. Step one lets suppose $$n=1$$: $$1 = \sum_{k=1}^1 k = \frac{1(1+1)}{2} = \frac{n(n+1)}{2}$$ Done the statement is true for $$n=1$$. The good part about induction is we can select the number for which we want to prove it so we could also have shown with $$n=0$$. So dont try to prove it for $$n$$ beeing any number but prove it for one number you selected, almost everytime $$n=0$$ or $$n=1$$.
Now we have to prove that if the statement is true for $$n$$ it is also true for $$n+1$$. Meaning if $$A(n)$$ is true, $$A(n+1)$$ is also true:
$$\underbrace{1+ 2+ \cdots + n}_{\sum_{k=1}^n k} + n+1 = \sum_{k=1}^n k + n+1 \overset{A(n) \text{ is true}}{=} \frac{n(n+1)}{2} + n+1 = \frac{n^2+n+2n+2}{2} = \frac{n^2+3n+2}{2} = \frac{(n+1)(n+2)}{2}$$
So now we know the statement is true for $$n+1$$ or $$A(n+1)$$ is true.
### Prove without induction
Proving this statement without induction is not that simple. You can do it like this. Suppose $$n\in \mathbb{N}$$, then we can look at the sum and rearange that:
$$n$$ is even
$$1+2+\cdots + n = (1 + n) + (2+n-1) + \cdots + \left(\frac{n}{2} + \frac{n}{2} +1\right)=\frac{n}{2}(n+1) = \frac{n(n+1)}{2}$$
and for $$n$$ is odd
$$1+2+\cdots + n = (1+n) + (2+n-1) + \cdots + \left(\frac{n-1}{2} + \frac{n+1}{2} + 1\right) + \frac{n+1}{2} = (n+1)\frac{n-1}{2} + \frac{n+1}{2} = \frac{n(n+1)}{2}$$
### Conclusion
So let's conclude, we can use induction here but we can go without it too. The first though is much simpler because it uses the induction and if you get your head around it, it's easier to verify that it is true. The other one is using induction internally for the rearanging and you have to think alot if this is true or if I made a mistake.
## What does the operation part mean? | {
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## What does the operation part mean?
The funny thing is that induction can be generalized in a lot of things. If we look at the set $$\mathbb{N}$$ than we can define it as follows $$1$$ is in $$\mathbb{N}$$ and if $$n\in\mathbb{N}$$ than $$n+1$$ is also in $$\mathbb{N}$$, meaning lets call that $$(\star)$$:
1. $$1 \in \mathbb{N}$$
2. $$\forall n\in\mathbb{N} \Rightarrow n+1\in\mathbb{N}$$
If you look at that you will find this is looking a lot like induction. So basically you can say if you have the statement $$A(n)$$ and you want to show that it is true for all $$n\in\mathbb{N}$$ you do the following, you determine the set $$\mathfrak{A}$$ for which the statement is true. This set is given by: $$\mathfrak{A} = \{n\in\mathbb{N} | A(n) \text{ is true}\}$$ And now if we know $$1\in\mathfrak{A}$$ and for all $$n\in\mathfrak{A}$$ it follows $$n+1\in\mathfrak{A}$$ then it follows from $$(\star)$$: $$\mathfrak{A} = \mathbb{N}$$ so the statement is true for all $$n\in\mathbb{N}$$
## More generalized induction shemes
The example where this is generalized is the "principle of the good set" (this is the german name translated directly), it's used in measure theory and works with sets that are way bigger then $$\mathbb{N}$$ in fact with sets that have the cardinality of $$\mathbb{R}$$. And it works as follows. A $$\sigma$$-Algebra $$\mathcal{A}$$ of a set $$\Omega$$ fullfills the following axioms:
1. $$\Omega \in \mathcal{A}$$
2. $$A\in \mathcal{A} \Rightarrow \Omega\setminus A\in\mathcal{A}$$
3. $$A_1,A_2,\cdots \in \mathcal{A} \Rightarrow \bigcup_{k=1}^\infty A_k \in \mathcal{A}$$ | {
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Now we have a function $$\sigma$$ that generates $$\sigma$$-Algebras for a given subset $$\mathcal{G}\subset \mathscr{P}(\Omega)$$ just using the axioms 1-3 just more often than $$\mathbb{R}$$ has elements. Like $$\mathbb{N}$$ is generated by $$\{1\}$$, if we take $$1$$ and then we add $$1$$ onto it put the result $$2$$ in our set. Then we take $$2$$ and add $$1$$ onto it and put it in our set until we cannot add any more element because we have all of them already. Back to the example, $$\mathcal{G}$$ is called a generator if $$\sigma(\mathcal{G}) = \mathcal{A}$$, this also means $$\Omega \in \mathcal{G}$$. Now if we want to show a statement $$B(A)$$ is true for all $$A\in\mathcal{A}$$ we just show:
1. $$B(G)$$ is true for all $$G\in \mathcal{G}$$
2. $$B(A)$$ is true then it is also true for $$B(\Omega\setminus A)$$
3. $$B(A_1),\ B(A_2),\ \cdots$$ is true than its also true for $$B\left(\bigcup_{k=1}^\infty A_k\right)$$
So here 1. shows that we know the statement is true for at least a generating set of $$\mathcal{A}$$, like in induction we know the statement is true for $$\{1\}$$ which generates $$\mathbb{N}$$ using $$n+1$$. And then we show it for the operations 2. and 3. like we would in induction where we show it for $$n+1$$.
## Conclusion
I hope I didn't confuse you too much. But I am certain that after some time you will accept inductions and learn how to use them. And if you once return here or have to explain it to somebody else, after using it a lot of times. You will fully understand it. People often find induction counter-intuitiv, I have some ideas why but I think that would really go to far. So my hint, accept it first, use it, try to explain it und you will have understood this and similar shemes.
Intuitively you can view this as a chain reaction, because once you show that $\color{green}{P_0}$ is true, then by applying $\color{blue}{P_k \Rightarrow P_{k+1}}$ repeatedly you can reach any number $n \geq 0$ and see that $P_n$ is indeed true. | {
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$$\color{green}{P_0} \color{blue}{\Rightarrow} P_1 \color{blue}{\Rightarrow} \dots \color{blue}{\Rightarrow} P_n \color{blue}{\Rightarrow} \dots .$$
At no point are you directly proving that $P_1$, $P_2$, $\dots$ are true, instead they follow from base case and induction step.
• What is $P_k \implies P_{k+1}$? I think my question actually asks about this. Feb 8 '16 at 14:17
• It is the implication, you are proving $P_{k+1}$ under assumption of $P_{k}$, i.e. you are assuming that $P_{k}$ is true, and you show that then $P_{k+1}$ is true. Once you prove that, you can apply this for case $k=0$ (because you know that $P_0$ is true). By that you immediately get that $P_1$ is true. And you can repeat that...
– Sil
Feb 8 '16 at 14:20
• You must combine it with the base case, in which you proved that $P_0$ is true. Both base case AND induction step then imply that $P_1$ is true. But now you have $P_1$ is true and you can use it same way with induction step, to arrive at $P_2$ being true, etc... Reaching any $n$ if you repeat the process.
– Sil
Feb 8 '16 at 14:25
• And that is exactly what you do, in case of $P_0$. You have $P_0$ and $P_0 \implies P_1$, therefore $P_1$. The trick is that you ONLY need to show this for $P_0$, everything else follows from the $P_k \implies P_{k+1}$.
– Sil
Feb 8 '16 at 14:48
• You prove $P_k$ is true when $k = 1$. You do not need to assume that $k \ne 1$ or that $P_k$ is true for any value of $k$ other than 1. But you don't want to assume that $k$ has any characteristic that pertains only to 1. You want to prove that it is true for some $k$ of which you don't know the value of, then it must also be true of $k + 1$. You know it is true for some $k$ because it is true for $k = 1$. It might also be true for $k = 7,921$ but you don't know that. You can't say I know it is true for $P_1$ so I will prove it for $P_2$ because you want something more general. Feb 8 '16 at 17:39 | {
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The part about assuming $P_k$ is an application of basic rules of propositional logic.
Suppose you have a proposition of the form $A \implies B$. One way you might be able to prove this proposition is by using the rule for Introduction of an Implication:
1. Assume that $A$ is true.
2. Using $A$ as a statement already known to be true, prove $B$.
3. "Discharge" the assumption that $A$ is true.
4. Conclude that $A \implies B$ is true.
Of course this works only if you are able to come up with a proof of $B$ in the context of Step 2. But while you are in that context, you already know $A$ is true, and you do not have to prove it.
Once you have proven $A \implies B$, if you later find out that $A$ actually is true (not just assumed to be true in the limited context of Step 2, above), you can apply a rule called Modus Ponens:
1. Given that $A \implies B$ is known to be true.
2. Given that $A$ is known to be true.
3. Conclude that $B$ is true.
None of these rules of logic has used anything from the method of induction, but the method of induction uses those rules of logic.
We prove $P_0$.
We prove $P_k \implies P_{k+1}$ for any possible value of $k$, using the rule for Introduction of an Implication.
Now we know $P_0$, and since $P_k \implies P_{k+1}$ for any $k$, including the case $k=0$, we have $P_0 \implies P_1$. Apply Modus Ponens:
1. Given that $P_0 \implies P_1$ is true.
2. Given that $P_0$ is true.
3. Conclude that $P_1$ is true.
Now we know $P_1$, and $P_k \implies P_{k+1}$ is still true in the case $k=1$, so we can apply Modus Ponens again:
1. Given that $P_1 \implies P_2$ is true.
2. Given that $P_1$ is true.
3. Conclude that $P_2$ is true.
We can repeat this process as many times as needed to show $P_3$, then $P_4$, then $P_5$. | {
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We can repeat this process as many times as needed to show $P_3$, then $P_4$, then $P_5$.
The principle of induction merely summarizes this long, repetitive, boring process so that we can relatively quickly show $P_n$ for any given non-negative integer $n$. And since we can always do this no matter what non-negative value we choose for $n$, we conclude $P_n$ is true for all $n$.
By the way, I would write at least one of the sentences in the book a little differently. Instead of "perform operations on this equation" I would write "use algebraic manipulation of $P_k$ to prove that other equations are true". That is,
Since we have assumed that $P_k$ is true, we can use algebraic manipulation of $P_k$ to prove that other equations are true.
I hope this clears up one of the points of confusion. | {
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• According to point no. 2 we have to prove that $P_{k+1}$ is true based on the assumption that $P_k$ is true. The author shows us that $$P_{k}+T_{k+1}=P_{k+1}$$. Is it the way to implement point no.2? And, if it iss then how? I mean in what way proving $P_{k}+T_{k+1}=P_{k+1}$ is equivalent to proving $P_{k+1}$ under the assumption?? Feb 9 '16 at 10:41
• What does $P_k + T_{p+1}$ mean? $P_k$ is an equation. Does it make sense to write $(5^2 = 25) + 3$? I would never write math that way. But do you agree that if $a$ and $b$ are numbers and $a=b$, then $a$ and $b$ are the same number, so $a + 3$ is the same as $b +3$, that is, $a+3=b+3$? That's what we mean by "add ___ to both sides of the equation." And yes, when you start with a true equation, then (the left side plus something) is equal to (the right side plus the same something), which is how the author derived $P_{k+1}$. Feb 9 '16 at 12:52
• DavidK first of all, adding something to an equation always means it is added to its both sides. In other words, it can never mean to add something to only side of equation because two sides make up an equation. So $P_k+T_{k+1}$ is not wrong anyway. It is not necessary for something to be right. When it is only "not wrong", we can adopt it. Feb 9 '16 at 18:41
• Secondly, you said that since we assumed $P_k$ is true, we add $T_{k+1}$ to its both sides and then if it becomes $P_{k+1}$, it is proved that $P_{k+1}$ is true. And to support it, you said that if something is added to a true equation such as $3^{2}=9$ then the new equation for example $3^{2}-1=9-1$ is necessarily true (because it is the property of equation). In case of $3^{2}=9$ we are certain that the equation is true but in case of $P_k$ we are at the sake of assumption. These two things are different. I am sure I am missing something but please explain more accurately. Feb 9 '16 at 18:54 | {
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• On the second point, that's the whole point of Introduction of an Implication. To prove $P\implies Q$, assume $P$ is true and show that $Q$ is true under this assumption. Of course now all you have actually shown is that $Q$ is true if $P$ is true; but that is what $P\implies Q$ means. Feb 9 '16 at 19:09 | {
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There is another way of thinking of this.
Throw out the standard induction theorem, and replace it with this one:
Any nonempty set of natural numbers has a smallest element.
This seems obviously true, right? This is called the well-ordering principle, and it is equivalent to induction. Here's how you do induction with the well-ordering principle (WOP for short):
1. Prove that $P_0$ holds (or $P_1$, if you don't consider zero to be a "natural number").
2. Take this set: $$S = \{x | x \in \mathbb{N} \wedge \neg P_x \}$$ That's the set of natural numbers for which our proposition does not hold.
3. Assume $S$ is non-empty. Then, by WOP, it has a least element, which we can call $k+1$. In particular, we know that $\neg P_{k+1}$ (we know that the proposition does not hold for $k+1$, since it's an element of $S$).
4. By step (1), we know $k+1 > 0$ (or $k+1 > 1$, if you started with $P_1$). So $k$ is a natural number. If we had not proven step (1), $k$ could be equal to negative one (or zero), which would invalidate the rest of the proof.
5. Prove that $\neg P_{k}$; that is, prove that the proposition is not true for $k$, based on our knowledge that it is not true for $k+1$.
6. Since $k < k + 1$ and $k \in S$, we have a contradiction ($k + 1$ isn't the smallest element of the set). That means our assumption in step (3) must be false. The set of counterexamples is empty, meaning the proposition must hold for every number.
Note that these statements are equivalent: $$P_k \implies P_{k+1} \\ \neg P_{k+1} \implies \neg P_k$$ If you can prove one, you can prove the other quite easily. Standard induction asks you to prove the first, while WOP-based induction requires you to prove the second. | {
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In the end, it's just a matter of notation. But you might find one form of induction more intuitive than the other. If you are going to use this method, you should note that we usually talk about $k$ and $k - 1$ instead of $k+1$ and $k$; I used the latter because it corresponds more obviously to standard induction.
"as long as we know that the state of a proposition being true for any positive integer k after Base number implies that proposition is true for integer k+1, we have to show that Pk is true. "
1) No, you don't. If I wanted to show that IF Godzilla were 10 times as large in all dimensions as a T. Rex THEN Godzilla would be weigh 1,000 times as much, I do NOT need to prove that Godzilla is 10 times as large in all dimensions as a T. Rex. I CAN'T prove Godzilla is 10 times as large in all dimensions as a T. Rex because Godzilla isnt 10 times as large in all dimensions as a T. Rex.
and
2) You already have shown $P_k$ is true for some k. You have shown it is true for $k = 1$. Reread that statement. It does not say "we will assume this is true for all $P_k$ and show that means it is true for all $P_{k+1}$". It says "we will show that if it is true for (one) $P_k$ than that means it is true for $P_{k+1}$".
Thus we have proven an infinite loop: If it is true for k, then it is true for k + 1: Relabel k+1 as k and it is true for k + 1 (which is the old k + 2): Relabel and repeat indefinitely and we have proven it is true for all $n \ge k$. Now we just need to demonstrate that it is true for a single base case k. Oh, wait. It's true for k = 1. Therefore it is true for $n \ge 1$.
As for you second question....
"Why is it necessary for an equation to be true for performing operations on it?"
It isn't. You can perform operations on untrue equations. You just can't assume any of the results are true.
Let me write down: "2 + 2 = 5". Let me perform operations on them:
$7^{2+2} - 37*5 = 7^5 - 37*{2 + 2}$
$2401 - 185 = 16,807 - 74 -74$
$2,216=16,659$ | {
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$7^{2+2} - 37*5 = 7^5 - 37*{2 + 2}$
$2401 - 185 = 16,807 - 74 -74$
$2,216=16,659$
I can do that. But I can't get any meaningful results.
====
In a very abstract and philosophical way a proof by induction is a tiny bit like a proof by contradiction.
In both you assume a statement without proof and you reach a conclusion that would be true if your assumption were.
In a proof by induction you might prove that following: "If any even number is divisible by two then they all are divisible by two."
And in a proof by contradiction you might prove the following: "If any odd number is divisible by two then they all are divisible by two."
What's the difference in these statements? They are both shown in the exact same way ("if $2|n$ then $2|n + 2$ and $2|n - 2$"). They are both equally valid. And furthermore, they are both TRUE.
But in the proof by induction you verify a single confirmation: "2 is divisible by two. Therefore all even numbers are."
While in a proof by contradiction you contradict with a contradiction: "3 is not divisible by two. Therefore no odd numbers are."
Think about Peano system (see footnotes) for defining natural numbers. You take an initial number, which you call 1, and then you say:
• Element 1 exists in the set.
• Any successor of an element n in the set, say $suc(n)$, exists in the set.
With these axioms, you construct the whole set of natural numbers. You don't need to prove such elements exist in a deductive was as you think, but you must construct the proof and you have the building blocks for that (while in the deductive method, the path is the inverse).
So, the analogous of Peano system applies for the induction: | {
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So, the analogous of Peano system applies for the induction:
• Prove that $P(1)$ exists, where 1 is just the base case you want to start the proof with. This is the equivalent of the Peano system for the value 1.
• The constructive step: Prove that $P(n+1)$ is true when $P(n)$ was true. In this case, you move forward in the demonstration (induction actually means move towards the end) to construct the successive cases in the proof.
So you initially prove $P(1)$. If you prove $P(n) => P(n+1)$ then you can start traversing infinitely:
• $P(2) because P(1)$
• $P(3) because P(2)$
• ...
You never prove the $n$ case, because $P(n)$ is in hypotetic case which derives ultimately from the case $P(1)$, and formerly being $P(n)$, it was $P(n+1)$ in former... iterations.
So, the way to think why you should assume $P(n)$ is true, is because:
• In an initial case, $P(1)$ has been proven true by you. If you didn't, then you should start for it.
• Since the construction is upward, instead of downward, you should not think that you have to prove the n-case, but that you have alreadr proven it.
Footnotes:
• You are not forced to start from 1 and upwards. You can take any number you want, and so you will restrict the proof on it and further.
• Other induction processes can traverse entire integer set instead of just natural numbers. Such proof system involves prooving $n+1$ and $n-1$. | {
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# Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$
How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$
where $$\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$$ , $$|x|\leq1$$
I came across this integral while I was working on $$\displaystyle \displaystyle\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\$$ and here is how I established a relation between these two integrals:
$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\underbrace{\int_1^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx}_{x\mapsto 1/x}$$
$$=\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\int_0^1 \frac{\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$ $$\left\{\color{red}{\text{add the integral to both sides}}\right\}$$
$$2\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+\int_0^1 \frac{\operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)}{1+x^2}\ dx$$
$$\{\color{red}{\text{use}\ \operatorname{Li}_3(x)-\operatorname{Li}_3(1/x)=2\zeta(2)\ln x-\frac16\ln^3x+i\frac{\pi}2\ln^2x}\}$$
$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx+2\zeta(2)\underbrace{\int_0^1\frac{\ln x}{1+x^2}\ dx}_{-G}-\frac16\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}+i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2x}{1+x^2}\ dx}_{2\beta(3)}$$
$$=\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-2\zeta(2)G+\beta(4)+i\pi \beta(3)$$
Then
$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\frac12\int_0^\infty\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx-\zeta(2)G+\frac12\beta(4)+i\frac{\pi}2 \beta(3)\tag{1}$$
where $$\displaystyle\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\$$ is the the Dirichlet beta function.
So any idea how to evaluate any of these two integrals?
Thanks. | {
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So any idea how to evaluate any of these two integrals?
Thanks.
• What are you asking? – clathratus Jun 13 '19 at 2:17
• Note the formula you got from the triglogarithmic identity involved you being cavalier about branch cuts; you are missing a factor of $\frac{16 i \pi^4}{512}$ – Brevan Ellefsen Jun 13 '19 at 4:04
• @clathratus I am not sure if you are being sarcastic or not but I'll take as not. If you want to know what I'm asking just ignore the body and read the title please. – Ali Shather Jun 13 '19 at 4:28
• @BrevanEllefsen wolfram gives the closed form. – Ali Shather Jun 13 '19 at 4:30
• @BrevanEllefsen I dont think I'm missing any. To make sure, just take x=1/2 and compare the two sides. – Ali Shather Jun 13 '19 at 4:45
Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.
$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $$n=2$$ we get
$$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2 u}{1-xu}\ du$$
we can write
$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du$$ $$=\frac12\int_0^1\ln^2u\left(-\frac12\left(\frac{\pi u}{1+u^2}+\frac{2\ln(-u)}{1+u^2}\right)\right)\ du,\quad \color{red}{\ln(-u)=\ln u+i\pi}$$
$$=-\frac{\pi}{4}\underbrace{\int_0^1\frac{u\ln^2u}{1+u^2}\ du}_{\frac3{16}\zeta(3)}-\frac12\underbrace{\int_0^1\frac{\ln^3u}{1+u^2}\ du}_{-6\beta(4)}-i\frac{\pi}2\underbrace{\int_0^1\frac{\ln^2u}{1+u^2}\ du}_{2\beta(3)}$$
Then
$$\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx=-\frac{3\pi}{64}\zeta(3)+3\beta(4)-i\pi\beta(3)\tag{2}$$
Bonus:
By combining $$(1)$$ in the question body and $$(2)$$, the imaginary part $$i\pi\beta(3)$$ nicely cancels out and we get
$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=2\beta(4)-\zeta(2)G-\frac{3\pi}{128}\zeta(3)$$ | {
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$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=2\beta(4)-\zeta(2)G-\frac{3\pi}{128}\zeta(3)$$
where $$\beta(4)$$ $$=\frac{1}{768}\psi^{(3)}(1/4)-\frac{\pi^4}{96}$$
• The formula \begin{align}\Re\int_0^\infty\frac{\operatorname{Li}_{3}(x)}{1+x^2}\ dx&=\frac12\int_0^1\ln^2u\left(\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx\right)\ du\end{align} is weird. In the right side it's $\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$ not its real part. There is no indication for the values of x involved in the formula about polylogarithm in the beginning of your post. Anyway your closed-form matches numeric evaluation. – FDP Nov 22 '19 at 8:45
• its edited now. thanks for pointing this out. – Ali Shather Nov 22 '19 at 8:58
• The formula for polylogarithm in the beginning of your post is valid for $|x|<1$ only. If you take $x=2$ the denominator is $1-2u$ and $1-2u=0$ when $u=1/2\in [0;1]$ – FDP Nov 22 '19 at 15:42
• Yes .. actually i got the right formula yesterday and I'll fix the problem today. – Ali Shather Nov 22 '19 at 16:18
• When you consider the integral $$\int_0^\infty\frac{x}{(1-ux)(1+x^2)}\ dx$$ as a CauchyPrincipalValue you don't get the imagniary part. It arises from the $\epsilon$ contour around that singularity. – Diger Nov 22 '19 at 22:05
For a different solution, use the first result from A simple idea to calculate a class of polylogarithmic integrals by using the Cauchy product of squared Polylogarithm function by Cornel Ioan Valean.
Essentially, the main new results in the presentation are: | {
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