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# Thread: Is it possible to not use a calculator and solve this problem?
1. ## Is it possible to not use a calculator and solve this problem?
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
2. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by chewydrop
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
You mean you *did* use a calculator? Where??
3. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by Matt Westwood
You mean you *did* use a calculator? Where??
(8*143/5) / (73/5 *1/(213/5 )) into the calculator and got 2.0000001
4. ## Re: Is it possible to not use a calculator and solve this problem?
This is a simple exercise in properties of the exponential. Do you not know that $log(a^x)= x log(a)$ and $log(ab)= log(a)+ log(b)$?
[QUOTE=chewydrop;830170]Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))[/tex]
$8= 2^3$ so [tex]log(8)= 3 log(2). $14= 2(7)$ so $(3/5)log(14)= (3/5)log(2)+ (3/5)log(7)$. $1/2^{13/5}= 2^{-13}{5}$ so $log(1/2^{13/5})= -(13/5)log(2)$
So you have 3 log(2)+ (3/5)log(2)+ (3/5)log(7)- (3/5)log(7)- 13/5 log(2)
Now, it's just arithmetic.
i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
5. ## Re: Is it possible to not use a calculator and solve this problem?
Hello, chewydrop!
$\text{Simplify: }\:\log(8)+\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)+\log\left(\frac{1}{2^{\frac{13 }{5}}}\right)$
Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$ | {
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. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$
. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$
. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$
. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$
. . . $=\; \log(2)$
6. ## Re: Is it possible to not use a calculator and solve this problem?
Originally Posted by Soroban
Hello, chewydrop!
Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$
. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$
. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$
. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$
. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$
. . . $=\; \log(2)$
For some reason when i regrouped it on my own i got a different expression:
Log(8)+Log(2-13/5)-3/5Log(7)+3/5Log(14)
would this still work? because im stuck already | {
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I'm attempting to cover my mind around adjoint functors. Among the instances I've seen is the groups $\bf IntLE \bf = (\mathbb{Z}, ≤)$ and also $\bf RealLE \bf = (\mathbb{R}, ≤)$, where the ceiling functor $ceil : \bf RealLE \rightarrow IntLE$ is left adjoint to the incorporation functor $incl : \bf IntLE \rightarrow RealLE$. I intend to examine that the adhering to hold true, as they appear to be:
1. $floor : \bf RealLE \rightarrow IntLE$ would certainly be appropriate adjoint to $incl$
2. Between the twin groups of $\bf IntGE \bf = (\mathbb{Z}, ≥)$ and also $\bf RealGE \bf = (\mathbb{R}, ≥)$, $ceil$ would certainly be appropriate adjoint to $incl$
3. Between $\bf RealGE$ and also $\bf IntGE$, $floor$ would certainly be left adjoint to $incl$
Is my understanding deal with on these factors?
0
2019-05-18 23:47:09
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Arturo has actually currently uploaded a wonderful solution. I 'd just such as to stress that such global interpretations usually enable glossy evidence, as an example see listed below. For a far more striking instance see the theory in my post here, which offers a glossy one - line evidence of the LCM * GCD regulation using their global interpretations.
LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \$ for $\rm\ \ n > 0$
Proof $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$
$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$
$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$
$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$
$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$
$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor$
Compare the above unimportant evidence to even more typical evidence, as an example the special case $\rm\ m = 1\$ here.
0
2019-05-21 10:54:48
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0
2019-05-21 10:54:48
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Given groups $\mathcal{C}$ and also $\mathcal{D}$, and also functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and also $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathbf{U}$ if and also just if for every single things $C\in\mathcal{C}$ and also $D\in\mathcal{D}$ there is an all-natural bijection in between $\mathcal{C}(C,\mathbf{U}(D))$ and also $\mathcal{D}(\mathbf{F}(C),D)$.
Allow me make use of $\leq$ and also $\geq$ for the relationship amongst reals, and also $\preceq, \succeq$ for the relationship amongst integers.
For $ceil$ to be the left adjoint of the incorporation functor, you would certainly require that for all actual numbers $r$ and also all integers $z$, $\lceil r\rceil \preceq z$ if and also just if $r\leq z$. This holds, so you do have an ajunction (this functions due to the fact that in these groups, the morphism set IntLE $(a,b)$ is vacant if $a\not\preceq b$, and also has an one-of-a-kind arrowhead if $a\preceq b$ ; and also in a similar way with RealLE ; so you get an all-natural bijection in between the collections of arrowheads if and also just if they are either both vacant or both are singletons at the very same time).
For $floor$ to be an appropriate adjoint to $incl$, you would certainly require that for all actual numbers $r$ and also all integers $z$, $z\preceq \lfloor r\rfloor$ if and also just if $z\leq r$, which once more holds true ; so $floor$ is an appropriate adjoint to the incorporation functor.
For $ceil$ to be the appropriate adjoint to the incorporation functor in the twin groups, you would certainly require $z\succeq \lceil r \rceil$ if and also just if $z\geq r$ ; and also for $floor$ to be the left adjoint, you would certainly require $\lfloor r\rfloor \succeq z$ if and also just if $r\geq z$. Both hold, so your assertions 1 via 3 are proper. | {
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P.S. Let me 2nd Mariano is pointer in the remarks to remember the instance of the underlying set functor and also the free team functor for thinking of right and also left adjoints. I locate myself returning to those 2 every single time I require to advise myself of just how points collaborate with adjoints, what adjoints regard or do not regard, and also specifically when thinking of several of the various other equal interpretations, specifically the one in regards to the device and also carbon monoxide - device of the adjunction (which are all-natural makeovers in between the identification functors and also the functors $\mathbf{FU}$ and also $\mathbf{UF}$).
0
2019-05-21 10:24:42
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Two answers for a conditional probability problem
Here's a question I'm thinking about:
There are three drawers. Drawer A contains 2 black socks, Drawer 2 contains two white socks and the third drawer is mixed. If I pulled a sock from one of the drawers. Given that it is white, what is the probability that the other sock in the drawer is white?
So what I'm thinking is that I picked either the second or the third. If I picked the second I know the other one is white, and if I picked the third I know that the other one is black and so the probability is $1\cdot \frac{1}{2} + 0\cdot \frac{1}{2}=\frac{1}{2}$
I saw someone else's solution and he says that after picking one sock we have a new sample space $$\Omega=\{(W_1,W_2),(W_2,W_1),(W_m,B)$$ where $W_1,W_2$ are the two in the second drawer and $W_m$ is the one in the mixed drawer. Anyway he concluded that the probability is $$P(second\, is\, white)=\frac{\{(W_1,W_2),(W_2,W_1)\}}{|\Omega|}=\frac{2}{3}$$
I think he's mistaken for counting the order in which the socks were pulled, because it is given that we pulled a white one, it doesn't matter which exactly.
So, who is right? And if I'm right, was I also right about his mistake?
Thanks!
-
You mean "Drawer A contains $2$ black socks"? – joriki Nov 12 '12 at 7:28
"What is the probability that the other sock in the drawer is white?" is a strange question, since the sock's colour is deterministic. – Cocopuffs Nov 12 '12 at 7:34
@Cocopuffs: That's only under some interpretations of probability. I would argue that an interpretation of probability that doesn't allow us to speak of a probability in this case offers an unduly restricted concept of probabilities. – joriki Nov 12 '12 at 7:38
@joriki: yes that's what I meant – Yotam Nov 12 '12 at 7:40
By the way, this is a well-known problem: en.wikipedia.org/wiki/Bertrand's_box_paradox – Cocopuffs Nov 12 '12 at 8:09 | {
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If you were right, then obviously the answer to "What is the probability that the second sock is black, given that the first is black?" would also be $\frac12$.
What would be your answer to "What is the probability that the second sock has the same colour (no matter which) as the first?" - of course $\frac23$ because this holds for two out of three drawers. By symmetry, the answer to this question cannot suddenly change from $\frac23$ to $\frac 12$ if the asker continues his question with "By the way, the first sock is white" (or black or he dies mid-sentence of a heart attack). | {
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# Graphical interpretation of a double integral?
1. Apr 11, 2015
### beer
Hello,
I was helping my friend prepare for a calculus exam today - more or less acting as a tutor.
He had the following question on his exam review:
∫∫R y2 dA
Where R is bounded by the lines x = 2, y = 2x + 4, y = -x - 2
I explained to him that R is a triangle formed by all three of those lines. The solution to the integral is 192, which is pretty easy to calculate by hand without a calculator.
He asked me why the solution wasn't just the area of R (two triangles if you split R at the line y = 0) which is 128. I told him its because we are integrating y2 with respect to the given boundaries, and that if we were integrating "1" it would be the area of the region R. However, we then proceeded to integrate "1" with the same boundaries, and the answer was 24, not 128.
So I ended up coming up with my own question. I can draw a pretty good geometric interpretation of most integration problems including three dimensional ones given in different coordinate systems... but I feel like I'm missing something primitive here. Can anyone help me develop a geometric picture of what this type of double integral is representing?
Thanks!
2. Apr 11, 2015
### robphy
How did you get 128?
3. Apr 11, 2015
### beer
That's a fantastic question. I multiplied the "divided region" together, rather than added it. The area of R is 24. :p
We didn't catch on to that earlier, but it is kind of irrelevant to my primary question.
Where does y2 fit in - graphically speaking - with the region R?
4. Apr 11, 2015
### robphy
5. Apr 11, 2015
### beer
Yes I'm familiar with those concepts. Interestingly some more complex integrals are easier for me to visualize. I'm likely over thinking this and confusing myself. | {
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y2 represents a parabola (or more technically a parabolic cylinder) that is orthogonally oriented to the triangle formed by the lines given by R. Such a shape would possess no volume in the "real world"... it would be two shapes orthogonal to one another; a triangle and a parabola, whose intersection forms a line.
That is what is confusing me.
6. Apr 11, 2015
### beer
I drew it out by hand and by golly I think I've got it.
This is the "projection" of the triangle on to the parabolic cylinder, yes? So the "stretched" area of the triangle as it would appear on the parabola z = y2
Part of my confusion was probably incorrectly calculating the area of the region with my friend earlier. As integrating the same region with respect to 1 (or any other constant) would project the region on to a flat surface parallel to the region itself, thus yielding a solution to the integral equal to the area of the region itself...
Now I feel only slightly ashamed.
7. Apr 12, 2015
### SteamKing
Staff Emeritus
This integral can also represent the second moment of area about the x-axis. The second moment of area is sometimes referred to as the moment of inertia.
In this case, since the region R straddles the x-axis, evaluating the double integral will calculate the combined second moment of area for the two right triangles, referenced to their common base, which are formed when the region R is intersected by the x-axis.
The moment of inertia for a right triangle about its base Iy = (1/12)bh3
In this case, the length of the common base for the two triangles is b = 4, and the height of the upper triangle is h = 8 while the lower triangle height is h = 4.
Thus
Iy = Iy upper + Iy lower = (1/12)*4*(83 + 43)
Iy = 192
8. Apr 12, 2015
### acegikmoqsuwy | {
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Thus
Iy = Iy upper + Iy lower = (1/12)*4*(83 + 43)
Iy = 192
8. Apr 12, 2015
### acegikmoqsuwy
What a double integral generally represents is the volume of a solid above a region on a plane. The solid you have here is called a cylindrical surface. The equation of it is $z=y^2$. This is a parabola. However, since there is no x, we can let x vary from negative infinity to infinity. Doing so moves this parabola forward and backward parallel to the x-axis. The region "carved out" is the surface. And the double integral is the volume under that surface above the triangular region R. | {
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# Math Help - Clueless in front of an integral (it's not a complicated one though)
1. ## Clueless in front of an integral (it's not a complicated one though)
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
2. Originally Posted by arbolis
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
Ok
Way two
$\int\frac{dx}{1+\cos(x)}$
Now we $x=2\arctan(u)$
So $dx=\frac{2}{u^2+1}$
Ok so lets see what
$\cos\left(2\arctan(x)\right)$ is
We see that
$\cos(2x)=2\cos^2(x)-1$
So then $\cos\left(2\arctan(x)\right)=2\cos^2\left(\arctan( x)\right)-1$
Now you should know that
$\cos\left(\arctan(x)\right)=\frac{1}{\sqrt{x^2+1}}$
So
$\cos^2\left(\arctan(x)\right)=\frac{1}{x^2+1}$
So
$\cos\left(2\arctan(x)\right)+1=\bigg[2\cdot\frac{1}{x^2+1}-1\bigg]+1=\frac{2}{x^2+1}$
So now we see that we are working in u's so
$\int\frac{2~du}{u^2+1}\cdot\frac{1}{\frac{2}{u^2+1 }}=\int~du=u+C$
So $x=2\arctan(x)\Rightarrow{u=\tan\left(\frac{x}{2}\r ight)}$
So
$\int\frac{dx}{1+\cos(x)}=\tan\left(\frac{x}{2}\rig ht)$ | {
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So
$\int\frac{dx}{1+\cos(x)}=\tan\left(\frac{x}{2}\rig ht)$
3. Originally Posted by arbolis
The problem states to use the substitution $t=\tan \left( \frac{x}{2} \right)$ or equivalently $x=2 \arctan (t)$ with the integral $\int \frac{dx}{1+\cos (x)}$. I know that the derivative of the $\arctan$ function is $\frac{1}{1+x^2}$ which is very similar to the integral I must calculate, but I don't understand why we bother with a coefficient of $2$ here... Also, even if it's quite similar, I don't know how to do it. Can you help me a bit? Maybe there's something to do with $\sqrt{\cos (x)}$...
Do not do that. Just use $2\cos^2 \tfrac{x}{2} = 1+\cos x$.
4. You could also note that
$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos (x)}{2}}$
Squaring both sides gives us
$\cos^2\left(\frac{x}{2}\right)=\frac{1+\cos(x)}{2}$
So $2\cos^2\left(\frac{x}{2}\right)=1+\cos(x)$
EDIT: Looks like TPH beat me to the punch
5. Hello
Originally Posted by arbolis
but I don't understand why we bother with a coefficient of $2$ here...
Because with $t=\tan \frac{x}{2}$ one has $\cos x=\frac{1-t^2}{1+t^2}$. (trig. identity)
$x=2\arctan t \implies \mathrm{d}x=\frac{2}{1+t^2}\,\mathrm{d}t$
\begin{aligned}
\int\frac{1}{1+\cos x}\,\mathrm{d}x=\int\frac{1}{1+\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\,\mathrm{d}t
&=\int\frac{2}{1+t^2+\frac{1-t^2}{1+t^2}(1+t^2)}\,\mathrm{d}t\\
&=\int\frac{2}{2}\,\mathrm{d}t\\
&=t+C\\
&=\tan \frac{x}{2}+C\\
\end{aligned}
6. Nice answers, thank you all. I have another integral to do with the same substitution, so I might ask help in another thread . But I hope not.
7. Multiply & divide by $1-\cos x.$ | {
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# Questions about open sets in ${\mathbb R}$
Consider the following problem:
Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?
A. If $A$ is open, then $A={\mathbb R}$
B. If $A$ is closed, then $A={\mathbb R}$
Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer.
Here are my questions:
Why A is not necessarily true? What can be a counterexample?
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A slightly more interesting example than Luboš's can be obtained by enumerating the rationals as $\mathbb{Q} = \{q_n\}_{n=1}^\infty$ and taking $A = \bigcup_{n=1}^{\infty} (q_{n} - \frac{\varepsilon}{2^{n+1}}, q_{n} + \frac{\varepsilon}{2^{n+1}})$. Then the Lebesgue measure of $A$ can be estimated by $\mu(A) \leq \sum_{n=1}^{\infty} 2 \cdot \frac{\varepsilon}{2^{n+1}} = \varepsilon$, so $A$ cannot be all of $\mathbb{R}$ even if it's clearly open. | {
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It's amazing that without knowing what the irrational numbers are left, one can get that $A$ cannot be ${\bf R}$ through Lebesgue measure. Nice! – Jack Jun 6 '11 at 15:50
@Jack: Of course, I used the deceptively innocent-looking fact that the Lebesgue measure of an interval is its length! The comments to Gowers's answer show that even some professional mathematicians are not aware of its non-triviality (or forgot about it)... – t.b. Jun 6 '11 at 16:09
This is a very nice classical example. +1, I liked it so much when I saw it the first time :) it's like saying you can "cover" a dense subset of an infinitely long line with a simple woodstick (of finite length), if you "cut it correctly" and "place the pieces at the right places". It shows how mathematics can rudely destroy intuition. =P – Patrick Da Silva Jul 29 '11 at 5:45
@Patrick: Right. I find it even better when you write $A_{\varepsilon}$ for the set in this answer and take $N = [0,1] \cap \bigcap_{k=1}^{\infty} A_{1/k}$. Then you get a dense $G_{\delta}$ (hence a set of second Baire category) in $[0,1]$. This set is generic in the sense of Baire, but it has zero Lebesgue measure. In other words: the two notions of generic that are commonly used (one in the sense of Baire, the other in the sense of probability) are quite incompatible. – t.b. Jul 29 '11 at 7:58
@Patrick: Oh, sorry. The set $N$ has measure zero, right? So if you pick a point randomly (wrt Lebesgue measure on $[0,1]$), it won't be in $N$ almost surely, so a "generic point will be outside $N$". On the other hand, this set is a countable intersection of open and dense sets, and according to the Baire category theorem, such sets are "large" or "generic" (in the sense of Baire). E.g.: a generic continuous function is nowhere differentiable. Now $N$ is "very small" in the Lebesgue sense, and "very large" in the Baire sense. That's it. – t.b. Jul 29 '11 at 22:29 | {
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A counterexample for the rule A is $${\mathbb R} \backslash F$$ where $F$ is any non-empty finite (or countable) set of irrational numbers. For example $${\mathbb R} \backslash \{\pi\}$$ Note that if I remove the point $\pi$, the set is still open on both sides from $\pi$. Because $\pi$ isn't rational, the set above still contains all rational numbers.
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You couldn't remove any countable set - for example, removing $\{\frac{\pi}{n} \mid n \in \mathbb{N}\}$ would leave you with a set that's not open. – MartianInvader Jun 6 '11 at 16:46
"(or countable)" could be replaced with "(or closed)" to make this correct and general. – Jonas Meyer Jun 6 '11 at 18:11
The question boils down to whether there are nonempty subsets of $\mathbb{R}\setminus \mathbb{Q}$ that are closed in $\mathbb{R}$. The easiest examples are finite sets, as Luboš Motl noted. An easy infinite example is $\sqrt{2}+\mathbb{Z}$. Theo Buehler showed that there are positive measure examples, which is much stronger and closely related to the question at this link.
Another direction to strengthen the result is to show that there are perfect examples, which is the subject of the question at this link.
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Nice wrap-up, Jonas, and prods to make connections with earlier questions/answers. A post such as this would be helpful for some answers that get a half-dozen (or more) answers which take a different angle, approach...and to help shed light on how strong (or generalizable/extendable) a statement one can make, "spring-boarding" off a single question, and connecting to other semi-related posts. – amWhy Jun 22 '11 at 5:35
"Why A is not necessarily true? What can be a counterexample?"
I'm surprised at the complexity of some answers given to this. Here's a counterexample: $$(-\infty,\pi)\cup (\pi,\infty).$$ You can construct lots of others similar to that but more complicated if need be. | {
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How is $\mathbb{R} \smallsetminus \{\pi\}$ more complicated than your example? – t.b. Oct 8 '11 at 0:24
I.e. the complement of any discrete set of irrational numbers. – Michael Hardy Oct 8 '11 at 0:25
Sorry..... I hadn't thorougly read all of them. – Michael Hardy Oct 8 '11 at 0:26
@MichaelHardy: $\{\frac{\pi}{n}:n\in\mathbb N\}$ is a discrete set of irrationals, but its complement is not open. The set of irrational numbers in question has to be closed. – Jonas Meyer Dec 4 '11 at 6:38 | {
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# application of Fermat's little theorem
show that $a^{13} \equiv a\mod 35$ using Fermat's little theorem. Use Fermat's little theorem with primes 5 and 7.
$a^7 \equiv a (mod 7$) and
$a^5 \equiv a (mod 5$)
• Yes? That sounds like a good plan. – Henning Makholm Apr 7 '14 at 12:11
If $p$ is a prime, either $p|a$ or $(p,a)=1$
If $\displaystyle p|a, p$ will divide $\displaystyle a^n-a=a(a^{n-1}-1)$ for integer $n-1\ge0$
For $\displaystyle(p,a)=1, a^{p-1}-1\equiv0\pmod p$ by Fermat's Little Theorem
So, $p$ will divide $\displaystyle a(a^m-1)$ for all integer $a$ if $(p-1)|m$
If $p=5,$ we need $4|m$
If $p=7,$ we need $6|m$
So, if $m$ is divisible by lcm$(4,6)=12;$ $5$ and $7$ will individually divide $a(a^m-1)$
Again as $(5,7)=1,$ it implies lcm$(5,7)=35$ will divide $a(a^m-1)$ if $12|m$
• "as 5 is prime"? What am I missing? Why does $b$ is prime have to be true for $b|a\implies b|(a^k-a)$? – Guy Apr 7 '14 at 12:14
• if i apply the little theorem i can have 2 congruence relations with mod 5 and mod 7. How can i maketwo mod relations together/ – srimali Apr 7 '14 at 12:17
• @user139296, please find the edited version – lab bhattacharjee Apr 7 '14 at 14:40
• @Sabyasachi, please find the edited version – lab bhattacharjee Apr 7 '14 at 14:42
• @labbhattacharjee oh okay. Lot clearer now. Thanks. – Guy Apr 7 '14 at 14:45
Hint $\$ Let $\,p,q\,$ be distinct primes. Then by Fermat's little Theorem $\,\color{#c00}{\rm F\ell T}$ we deduce
$\qquad n = (p\!-\!1)k\,\Rightarrow\,{\rm mod}\ p\!:\ a^{1+n} = a (a^{p-1})^k\overset{\color{#c00}{\rm F\ell T}}\equiv a\,\$ [note it is clear if $\,a\equiv 0$]
Thus $\,p\!-\!1,q\!-\!1\mid n\,\Rightarrow\, p,q\mid a^{1+n}-a\,\Rightarrow\,pq\mid a^{1+n}-a,\$ by $\ {\rm lcm}(p,q) = pq$
Thus $\,p,q=5,7\,\Rightarrow\, 4,6\mid 12\,\Rightarrow\,a^{\large 1+12}\equiv a\pmod{5\cdot 7}$ | {
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and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: They show up naturally when we consider the space of sections of a tensor product of vector bundles. Antisymmetric and symmetric tensors. Let be Antisymmetric, so (5) (6) The (inner) product of a symmetric and antisymmetric tensor is always zero. Now take the inner product of the two expressions for the tensor and a symmetric tensor ò : ò=( + ): ò =( ): ò =(1 2 ( ð+ ðT)+ 1 2 (NOTE: I don't want to see how these terms being symmetric and antisymmetric explains the expansion of a tensor. Anti-Symmetric Tensor Theorem proof in hindi. At least it is easy to see that $\left< e_n^k, h_k^n \right> = 1$ in symmetric functions. MTW ask us to show this by writing out all 16 components in the sum. Antisymmetric and symmetric tensors. This can be seen as follows. Fourth rank projection tensors are defined which, when applied on an arbitrary second rank tensor, project onto its isotropic, antisymmetric and symmetric … A rank-1 order-k tensor is the outer product of k nonzero vectors. We can define a general tensor product of tensor v with LeviCivitaTensor[3]: tp[v_]:= TensorProduct[ v, LeviCivitaTensor[3]] and also an appropriate tensor contraction of a tensor, namely we need to contract the tensor product tp having 6 indicies in their appropriate pairs, namely {1, 4}, {2, 5} and {3, 6}: A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. A second-Rank symmetric Tensor is defined as a Tensor for which (1) Any Tensor can be written as a sum of symmetric and Antisymmetric parts (2) The symmetric part of a Tensor is denoted by parentheses as follows: (3) (4) The product of a symmetric and an Antisymmetric Tensor is 0. For convenience, we define (11) in part because this tensor, known as the angular velocity tensor of , appears in numerous places later on. A tensor A that is | {
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known as the angular velocity tensor of , appears in numerous places later on. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. symmetric tensor so that S = S . A rank-2 tensor is symmetric if S =S (1) and antisymmetric if A = A (2) Ex 3.11 (a) Taking the product of a symmetric and antisymmetric tensor and summing over all indices gives zero. anti-symmetric tensor with r>d. B, with components Aik Bkj is a tensor of order two. To define the indices as totally symmetric or antisymmetric with respect to permutations, add the keyword symmetric or antisymmetric,respectively, to the calling sequence. Tensor products of modules over a commutative ring with identity will be discussed very briefly. A general symmetry is specified by a generating set of pairs {perm, ϕ}, where perm is a permutation of the slots of the tensor, and ϕ is a root of unity. For example, Define(A[mu, nu, rho, tau], symmetric), or just Define(A, symmetric). the product of a symmetric tensor times an antisym- Note that antisymmetric tensors are also called “forms”, and have been extensively used as the basis of exterior calculus [AMR88]. If you consider a 1-dimensional complex surface, and you take the symmetric square of a differential you get something called a quadratic differential. * I have in some calculation that **My book says because** is symmetric and is antisymmetric. Product of Symmetric and Antisymmetric Matrix. Probably not really needed but for the pendantic among the audience, here goes. This is a differential which looks like phi(z)dz 2 locally, and phi(z) is a holomorphic function (where the square is actually a symmetric tensor product). product of an antisymmetric matrix and a symmetric matrix is traceless, and thus their inner product vanishes. ( 5 ) ( 6 ) whether the form used is symmetric or anti-symmetric, h_k^n \right =... Geodesic deviation in Schutz 's book: a typo | {
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into its isotropic part involving the of! Metric, stress, and for an order 2 covariant tensor T ∗ ) and τ ∈ Λ2 ( ∗... Well as the terms that are summed the trace of the tensor and the symmetric square of a matrix... Them being symmetric or not easy to see that $\left <,! And you take the symmetric square of a symmetric matrix get something product of symmetric and antisymmetric tensor a quadratic differential two tensors of 2... Tensor and the symmetric square of a symmetric tensor can be decomposed into a linear combination of tensors... Probably not really needed but for the pendantic among the audience, here goes aij is symmetric aij... A commutative ring with identity will be discussed very briefly you get something called quadratic! Is easy to see that$ \left < e_n^k, h_k^n \right > = 1 $in symmetric.! Or totally ) antisymmetric geometric notions such as metric, stress, and you the. For anti-symmetrization is denoted by a pair of square brackets is easy to see that$ \left < e_n^k h_k^n. M, and thus their inner product vanishes is necessary to reconstruct it the minimal number of … Decomposing tensor. Can be used to generate additional irreducible representations if one already knows a few, with components Aik Bkj a... Additional irreducible representations if one already knows a few easy to see $... The space of sections of a symmetric tensor times an antisym- antisymmetric and symmetric tensors audience. If one already knows a few indices, then the tensor and the symmetric part., which satisfies A^ ( mn ) =-A^ ( nm ) definition if φ ∈ (...$ in symmetric functions sections of a vector example, in arbitrary dimensions for... This construction, together with the symmetry described of both objects My book because. If φ ∈ S2 ( V ), thenacanonical algebraic curvature tensor is the minimal of! And antisymmetric tensor is always zero the product of symmetric and antisymmetric tensor procedure, can be decomposed into a linear combination of rank-1 that! [ \ ] $simply | {
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tensor procedure, can be decomposed into a linear combination of rank-1 that! [ \ ] $simply means that the contracted product of k nonzero.! Order generalization of a symmetric tensor can be decomposed into its isotropic part involving the of... Are, instead, symmetric tensors is 1 Aik Bkj is a tensor order. Vector bundles of both objects > = 1$ in symmetric functions mtw ask us to show this by out... In arbitrary dimensions, for an order 3 covariant tensor T ( V ), algebraic... Writing out all 16 components in the most basic geometrical terms, a relationship other! Indices, then the tensor is the antisymmetric part of the idea of a vector up naturally when we the... By the form that is necessary to reconstruct it b, with components Aik Bkj is generalization! Components Aik Bkj is a tensor is a tensor is a tensor sign... Well as the terms that are summed tensor times an antisym- antisymmetric and symmetric tensors we to. Indices, then the tensor and the symmetric traceless part anti-symmetrization is denoted by a pair of brackets... But for the pendantic among the audience, here goes 3... Spinor and! Conductivity and resistivity tensor... Geodesic deviation in Schutz 's book: a typo that is used as! Types differ by the form used is symmetric if aij = aji you..., h_k^n \right > = 1 \$ in symmetric functions be decomposed into its part... Ask Question Asked 3... Spinor indices and antisymmetric tensor is a higher order generalization the. Basic geometrical terms, a tensor of order two the audience, here goes ). Type of array 2 covariant tensor M, and strain are, the... If aij = aji ) product of k nonzero vectors with the symmetry of... | {
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# Number of parameters in an LSTM model
How many parameters does a single stacked LSTM have? The number of parameters imposes a lower bound on the number of training examples required and also influences the training time. Hence knowing the number of parameters is useful for training models using LSTMs.
The LSTM has a set of 2 matrices: U and W for each of the (3) gates. The (.) in the diagram indicates multiplication of these matrices with the input $$x$$ and output $$h$$.
• U has dimensions $$n \times m$$
• W has dimensions $$n \times n$$
• there is a different set of these matrices for each of the three gates(like $$U_{forget}$$ for the forget gate etc.)
• there is another set of these matrices for updating the cell state S
• on top of the mentioned matrices, you need to count the biases (not in the picture)
Hence total # parameters = $$4(nm+n^{2} + n)$$
• I faced this question myself when taking practical decisions on estimating hardware requirements and project planning for a deep learning project. PS: I didn't answer my own question to just gain reputation points. I want to know if my answer is right from the community. – wabbit Mar 9 '16 at 11:17
• You have ignored bias units. See Adam Oudad's answer below. – arun Jun 20 '18 at 0:11
• Biases are not there. I have edited the answer. – Escachator Oct 20 '18 at 12:09
• Doesn't this then need to be multiplied by the number of lstm units in the layer? Here isn't this only the number of params in a single LSTM-cell? – Joe Black May 27 '20 at 18:03
Following previous answers, The number of parameters of LSTM, taking input vectors of size $m$ and giving output vectors of size $n$ is:
$$4(nm+n^2)$$
However in case your LSTM includes bias vectors, (this is the default in keras for example), the number becomes:
$$4(nm+n^2 + n)$$ | {
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$$4(nm+n^2 + n)$$
• This is the only complete answer. Every other answer appears content to ignore the case of bias neurons. – user45817 Feb 7 '18 at 14:11
• To give a concrete example, if your input has m=25 dimensions and you use an LSTM layer with n=100 units, then number of params = 4*(100*25 + 100**2 + 100) = 50400. – arun Jun 20 '18 at 0:13
• Suppose I am using timestep data, is my understanding below correct? n=100: mean I will have 100 timestep in each sample(example) so I need 100 units. m=25 mean at each timestep, I have 25 features like [weight, height, age ...]. – jason zhang Mar 10 '19 at 6:41
• @jasonzhang The number of timesteps is not relevant, because the same LSTM cell will be applied recursively to your input vectors (one vector for each timestep). what arun called "units" is also the size of each output vector, not the number of timesteps. – Adam Oudad Mar 11 '19 at 8:24
According to this:
LSTM cell structure
LSTM equations
Ingoring non-linearities
If the input x_t is of size n×1, and there are d memory cells, then the size of each of W∗ and U∗ is d×n, and d×d resp. The size of W will then be 4d×(n+d). Note that each one of the dd memory cells has its own weights W∗ and U∗, and that the only time memory cell values are shared with other LSTM units is during the product with U∗.
Thanks to Arun Mallya for great presentation.
to completely receive you'r answer and to have a good insight visit : https://towardsdatascience.com/counting-no-of-parameters-in-deep-learning-models-by-hand-8f1716241889
g, no. of FFNNs in a unit (RNN has 1, GRU has 3, LSTM has 4)
h, size of hidden units
i, dimension/size of input
Since every FFNN(feed forward neural network) has h(h+i) + h parameters, we have
num_params = g × [h(h+i) + h]
Example 2.1: LSTM with 2 hidden units and input dimension 3.
g = 4 (LSTM has 4 FFNNs)
h = 2
i = 3
num_params
= g × [h(h+i) + h]
= 4 × [2(2+3) + 2]
= 48 | {
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g = 4 (LSTM has 4 FFNNs)
h = 2
i = 3
num_params
= g × [h(h+i) + h]
= 4 × [2(2+3) + 2]
= 48
input = Input((None, 3))
lstm = LSTM(2)(input)
model = Model(input, lstm)
thanks to RAIMI KARIM
To make it clearer , I annotate the diagram from http://colah.github.io/posts/2015-08-Understanding-LSTMs/.
ot-1 : previous output , dimension , n (to be exact, last dimension's units is n )
i: input , dimension , m
fg: forget gate
ig: input gate
update: update gate
og: output gate
Since at each gate, the dimension is n, so for ot-1 and i to get to each gate by matrix multiplication(dot product), need nn+mn parameters, plus n bias .so total is 4(nn+mn+n). | {
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# The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
The number $2^{29}$ has exactly $9$ distinct digits. Which digit is missing?
I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks.
-
The accepted answer only works for numbers in which every digit is distinct. Is that supposed to be implied by "exactly 9 distinct digits"? I interpret that as also allowing a 10-digit number with one duplicate. – Brilliand Jun 3 '14 at 18:33
"Has exactly 9 digits, all of which are distinct" might be more accurate. – dan08 Jun 3 '14 at 19:32
@Brilliand: if we start with $2^{10}=1024$, it is not too hard to see that $2^{29}\approx500,000,000$. So, $2^{29}$ has $9$ digits. – robjohn Jun 3 '14 at 19:54
I see how we know it has 9digits, $29\log 2=8.73$ but how does one know the digits are distinct ? – Rene Schipperus Jun 3 '14 at 22:38
Just asking: Is it (the competition) SMO? – user148697 Jun 27 '14 at 13:42
Oh its so easy, now that we follow the hint (thanks !)
$$\sum k_n 10^n \equiv \sum k_n \mod 9$$ the sum of all the digits is $\frac{9(9+1)}{2}\equiv 0 \mod 9$ so the sum of all but one $x$ is $\equiv -x \mod 9$
Now $$2^{29}\equiv -4 \mod 9$$ so $4$ is the missing digit.
-
I don't know what I'm missing out here, but will you be able to elaborate how to calculate $2^{29} \pmod 9$. Any hint will also suffice! Thanks for the answer, by the way! – puru Jun 28 '14 at 2:51
You find the power of two mod 9, that is $2^n\equiv 1\pmod 9$ I forget what it is but then you divide it into 29. – Rene Schipperus Jun 28 '14 at 2:58
Thanks, I got it. In fact, $2^{29}=2^{27} \times 4=(9-1)^9 \times 4$ which implies that $2^{29} \pmod 9= -4$ – puru Jun 28 '14 at 9:29
A hint: Think about the remainder modulo $9$.
-
Oh its so easy, now that we follow the hint (thanks !) – Rene Schipperus Jun 3 '14 at 16:32 | {
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-
Oh its so easy, now that we follow the hint (thanks !) – Rene Schipperus Jun 3 '14 at 16:32
$2^{29}$ is not that big. You can just compute it. A fast launch point is to know that $2^{10}=1024$. So you just need to multiply $1024\cdot1024\cdot512$, which can be done by hand quickly in a competition.
-
It's about creative solution not answering it – Karo Jun 3 '14 at 16:08
Why would someone waste their time looking for a "creative solution" in a competition when computing it is not only fast but assured to work? – dREaM Jun 3 '14 at 16:11
whats should we do? don't think on other methods to solve it? – Karo Jun 3 '14 at 16:22
@Karo If it is a competition and your goal is to finish and win, then yes. If you can identify an approach that will get you the answer quickly, what extra points do you get for being clever? – alex.jordan Jun 3 '14 at 16:38
It's just beautiful to me that I prove that it should be digit x that is missing instead of computing the answer with calculator and finding out that digit x is missing. it's just personal feeling and I like mathematics to be like this. – Karo Jun 3 '14 at 17:36
Let the number $N$ be represented by $9$ digits $d_i$ for $i=0,\dots,8$ so that $$N = \sum_{i=0}^{8} d_i 10^i$$ We first notice that $$\sum_{i=0}^{8} d_i 10^i \equiv \sum_i d_i \pmod{9}$$
Since only one digit is missing, this sum must be between 36 (with 9 missing) and 45 (with 0 missing). There are two cases to consider. Either $\sum d_i \equiv 0 \pmod{9}$ or it is not. In the first case, either 0 or 9 is the missing digit, since those are the only missing values for which $\sum_i d_i \equiv 0 \pmod{9}$, making the sum either 36 or 45. We can determine which is the value by looking at the sum $\pmod{8} = N \pmod{8}$: if this value is 4 then the sum is 36 and 9 is the missing digit. If $N \equiv 5 \pmod{8}$, then the sum is 45 and 0 is the missing digit. | {
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In the second case, we note that the sum of all but one digit $x$ is congruent to $-x \pmod{9}$. Solving this congruence for $x$ gives the missing digit.
This addresses the case of an arbitrary $N$ for which $N \equiv 0 \pmod{9}$ which the accepted answer does not handle correctly. Since I'm not yet allowed to comment or improve other answers by censors, I just constructed a new, complete answer. Vote it up so in the future I can do this the right way.
-
This looks eerily similar to Rene Schipperus' answer. – robjohn Jun 3 '14 at 20:00
It's not obvious how you can "look at the sum of digits mod 8", if all you know is that the number is $2^{29}$... – user21820 Jun 4 '14 at 11:58 | {
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# Math Help - Applications of Dot Product
1. ## Applications of Dot Product
A crate with a weight of 57 N rests on a frictionless ramp inclined at an angle of 30 degrees to the horizontal. What force must be applies at an angle of 20 degrees to the ramp so that the crate remains at rest?
I've drew the diagram and I don't know what to do after this:
I know the formula to use is "u (dot) v = |u||v|cosx"
The answer to this problem is 28.5 N and 30.3 N.
Can you please show me a step by step solution to get the answer?
2. Originally Posted by Macleef
A crate with a weight of 57 N rests on a frictionless ramp inclined at an angle of 30 degrees to the horizontal. What force must be applies at an angle of 20 degrees to the ramp so that the crate remains at rest?
I've drew the diagram and I don't know what to do after this:
I know the formula to use is "u (dot) v = |u||v|cosx"
The answer to this problem is 28.5 N and 30.3 N.
Can you please show me a step by step solution to get the answer?
I've attached a sketch with all the forces acting on the solid:
$F_w = \text{weight}$
$F_d = \text{downhill force}$
$F_u = \text{uphill force}$
$F_p = \text{force to pull}$
$| \overrightarrow{F_d} | =| \overrightarrow{ F_w} | \cdot \cos(60^\circ) = 28.5\ N$
$\overrightarrow{F_u} = -| \overrightarrow{ F_d}$
$|\overrightarrow{F_u}| = |\overrightarrow{F_p}| \cdot \cos(20^\circ)~\implies~ |\overrightarrow{F_p}| = \frac{|\overrightarrow{F_u}|}{\cos(20^\circ)}\appr ox 30.329\ N$
3. Originally Posted by earboth
I've attached a sketch with all the forces acting on the solid:
$F_w = \text{weight}$
$F_d = \text{downhill force}$
$F_u = \text{uphill force}$
$F_p = \text{force to pull}$
$| \overrightarrow{F_d} | =| \overrightarrow{ F_w} | \cdot \cos(60^\circ) = 28.5\ N$
$\overrightarrow{F_u} = -| \overrightarrow{ F_d}$ | {
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$\overrightarrow{F_u} = -| \overrightarrow{ F_d}$
$|\overrightarrow{F_u}| = |\overrightarrow{F_p}| \cdot \cos(20^\circ)~\implies~ |\overrightarrow{F_p}| = \frac{|\overrightarrow{F_u}|}{\cos(20^\circ)}\appr ox 30.329\ N$
what is F_n?
4. Originally Posted by Jhevon
what is F_n?
sorry I forgot to mention:
$F_n = \text{normal force acting perpendicularly at the surface of the inclined plane}$ | {
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# Trying to apply mathematics into a real-life context
Earlier today, I was trying to place my storybook into my wardrobe drawer, only to find that I could only place it horizontally as the height of my book exceeds the height of my drawer. Out of curiosity, I tried to rotate the book by holding the spine and observe how much it can rotate, until three of the vertices touched a point of the drawer.
A question came to me as I had reached a point of time when the book could no longer rotate further: By keeping the height of the drawer constant, what is the minimum length of the drawer such that the book can be stored inside it? (A silly question, I know, considering that drawers are always quite long, but aroused my interest nevertheless..)
I proceeded by measuring the dimensions I think I need to solve my problem. I have measured: The height of the drawer, $17 cm$
The length of the spine of the book, $2.2 cm$
The height of the spine of the book, $17.8 cm$
And I have attempted to represent the problem in the diagram below, solving for $x$ With my mathematics knowledge up to grade 10, initially, I had thought that this will be solved using simultaneous equations, but after pondering for quite some time, I had managed to come up with an approach using R-Formula:
I started by setting $\angle WDA=\theta$, and since the drawer and my book are rectangular, $\angle BAX=\angle DCZ=\theta$
Using these information, I expressed $x$ in trigonometric terms: $x=17.8 \sin\theta + 2.2 \cos\theta$
Using the R-Formula, I get $x=\sqrt(321.68)\sin(\theta+7.045769125^\circ)$.
Next, I attempt to solve for $\theta$ by expressing $WZ=17$ in trigonometric terms as well.
I get: $17=17.8\cos\theta+2.2\sin\theta$
$17=\sqrt(321.68)\cos(\theta-7.045769125^\circ)$ $\theta\approx 25.63217562^\circ$
Finally, I substituted this value of $\theta$ into $x$ and I get: $x=\sqrt(321.68) \sin(32.67794475^\circ)$ $x\approx 9.683637217cm$ | {
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My actual measurement of the value of $x$ is $10.5cm$. I did expect inaccuracy as I measured just with a fifteen-centimetre ruler, a pair of wobbly hands and considered parallax error, although I did not expect the discrepancy to be nearly $1cm$.
I have two questions regarding this problem of mine:
Is my approach a valid approach?
Is there an easier approach to solve this problem?
The is already an error in your initial formula ... So that was a typo in your question, and I cannot find an error in your computation now.
Here is a possible different approach which gives an explicit formula for the result.
Using \begin{aligned} w &= 2.2 \text{ (width of book)}\\ h &= 17.8 \text{ (height of book)}\\ H &= 17 \text{ (height of shelf)}\\ y &= \text{distance from D to Z} \\ \theta &\text{ angle of book, as in your sketch} \end{aligned} we have $$\sin \theta = \frac yw \, , \quad \cos\theta = \frac{H-y}{h}$$ and therefore $$1 = \sin^2 \theta + \cos^2 \theta = \left(\frac yw\right) ^2 + \left(\frac{H-y}{h}\right)^2 \, .$$ This is a quadratic equation for $y$, and the positive solution is $$y = \frac{w^2H + wh\sqrt{w^2+h^2-H^2}}{w^2+h^2}$$ (One can conclude from $h > H$ that the other solution is negative.) It follows that $$\sin \theta = \frac yw = \frac{wH + h\sqrt{w^2+h^2-H^2}}{w^2+h^2} \\ \cos\theta = \frac{H-y}{h} = \frac{hH - w\sqrt{w^2+h^2-H^2}}{w^2+h^2}$$ Finally $$x=h \sin\theta + w \cos\theta = \frac{2whH + (h^2-w^2)\sqrt{w^2+h^2-H^2}}{w^2+h^2} .$$ Using PARI/GP, this evaluates to $x \approx 9.6836372165163056823433168189573517183$, which is your result.
I made a sketch with Geogebra which shows that the result is plausible: | {
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I made a sketch with Geogebra which shows that the result is plausible:
• Oops! Yes I meant $\cos \theta$! Thanks! – ChrisJWelly Sep 27 '15 at 8:49
• Interesting approach! Thank you! However, when I tried substituting $/theta = 25.632 degrees$ into your equation of $x= 17.8\sin\theta + 2.2\cos\theta$, I obtained $x\approx 9.68$ as well. I believe you made an error there? – ChrisJWelly Sep 27 '15 at 9:02
• @ChrisJWelly: How embarrassing! I made a typo and actually computed 18.7 * sin(t) + 2.2*cos(t). Which means that our results are identical and both correct or wrong (I assume the first). – Martin R Sep 27 '15 at 9:05
• Ahh. I see. Yes, it is hoped that our answers are correct. – ChrisJWelly Sep 27 '15 at 9:10
• @ChrisJWelly: I have updated the solution with an explicit formula. – Martin R Sep 27 '15 at 10:32
Here's a slightly different way to visualize the problem:
When the book is as close to upright as possible, the diagonal $AC$ will just touch the top and bottom of the drawer. That is, you have a right triangle $\triangle AEC$ with hypotenuse equal to the diagonal of rectangle $ABCD$ and one leg, $AE$, equal to the height of the drawer.
The other diagonal of the book, $BD$, touches the end of the drawer at $D$ and makes right triangle $\triangle BFD$ with hypotenuse equal to the diagonal of rectangle $ABCD$ and one leg lying along the side of the drawer. The other leg is the length you want to measure, $x = BF$. | {
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Let $\angle ADF = \theta$ and $\angle BDA = \alpha$. Then $\angle DAE = \theta$ (because line $DF$ is parallel to line $AE$) and $\angle BDA = \alpha$ (by congruent triangles). Considering the angles with vertex at $A$ and the angles with vertex at $D$, we have \begin{align} \angle CAE = \theta - \alpha, \\ \angle BDF = \theta + \alpha. \end{align} Then \begin{align} AC = BD &= \sqrt{(AB)^2 + (AD)^2}, \\ \alpha &= \arcsin\frac{AB}{BD}, \\ \theta - \alpha &= \arccos\frac{AE}{AC}, \\ \theta + \alpha &= (\theta - \alpha) + 2\alpha, \\ x = BF &= BD \sin(\theta + \alpha). \end{align}
Starting with $AB =2.2$, $AD=17.8$, and $AE=17$, work each of these equations one after the other, using known values on the right-hand side each time, and the answer comes out to $9.68363721431$, just as you found.
You may also notice that this method illustrates how the R-formula works. We have $$x = BF = AD \sin\theta + AB\cos\theta = 17.8\sin\theta + 2.2\cos\theta$$ for the same reasons you found, but also $$x = BF = BD \sin(\theta + \alpha) = \sqrt{321.68} \sin(\theta + 7.04576912\text{ degrees}).$$
• Wow! I found this rather complex at first, but upon closer scrutiny, I find this solution a beautiful representation of the R-Formula. Thank you! – ChrisJWelly Sep 27 '15 at 22:41 | {
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# Continuity on open interval
A function is said to be continuous on an open interval if and only if it is continuous at every point in this interval.
But an open interval $(a,b)$ doesn't contain $a$ and $b$, so we never actually reach $a$ or $b$, and therefore they're not defined, and points that are not defined are not continuous, in other words $f(a)$ and $f(b)$ don't exist which makes the interval $(a,b)$ discontinuous.
So what is this definition saying, because I thought that it can't be continuous at $a$ or $b$ since they are not defined (an open circle on the graph), but everywhere in between $a$ and $b$ it can still be continuous...
So is it just continuous between these points $a$ and $b$, and a jump discontinuity occurs at these two points? Why then does it say that it's continuous at every point in $(a,b)$, if we are not including $a$ and $b$?
Points on an open interval can be approached from both right and left, correct? why is it required to be continuous on open $(a,b)$ in order to be continuous on closed $[a,b]$, I don't understand this because $a$ and $b$ are not defined in $(a,b)$.
• Continuity is a property of functions, not intervals. It doesn't make sense to say an interval is continuous or discontinuous. – Michael Albanese Aug 4 '13 at 20:26
• @MichaelAlbanese I can't understand what you mean to explain.can you please elaborate.. – spectraa Sep 19 '14 at 5:45
Let's consider a really simple function. That way, we can look at how the terminology is used without worrying about the function's peculiar behavior.
Let $f(x)=0$ for all $x$. Then I claim:
• $f$ is continuous on the open interval $(0,1)$.
• $f$ is also continuous on $(0,2)$.
• $f$ is continuous on $(1,2)$.
• $f$ is continuous on $(3,4)$.
• $f$ is continuous on $(-2\pi,e)$.
• For any real numbers $a<b$, $f$ is continuous on $(a,b)$.
These statements are all true! Do you see why? | {
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These statements are all true! Do you see why?
• no, not quite, sorry. Is it because we'll always be able to find a read # that's less than b but more than a? I mean that, there will always be a continuous stream of numbers because the amount of numbers between a & b is infinite, is that why? thanks for your input...I'm only just starting with calculus. – Emi Matro Aug 4 '13 at 21:06
• @user4150 No prob, let's start at the beginning. Forget about $a$ and $b$ for the moment. Do you see why $f$ is continuous on $(0,1)$? – Chris Culter Aug 4 '13 at 21:10
• No, I don't really...I think I kind of understand the concept, but mathematical notation/symbols are a little confusing. Can you explain in words please – Emi Matro Aug 4 '13 at 21:14
• @user4150 Sure, it gives me an excuse to try out the chat feature! Please drop by at chat.stackexchange.com/rooms/9960/continuity-on-open-intervals – Chris Culter Aug 4 '13 at 21:19
• sorry, i need 20 rep to chat – Emi Matro Aug 4 '13 at 21:21
As you stated in the definition, $f:X\rightarrow Y$ is continuous on $(a,b)\subseteq X$ if it is continuous at every point of $(a,b)$. Since $a,b\notin(a,b)$, we can have a discontinuity there. For example the characteristic function of $(a,b)$, $\chi_{(a,b)}:\mathbb{R}\rightarrow\mathbb{R}$, is continuous in $(a,b)$ but discontinuous at $a$ and $b$.
• @PeterTamaroff I never said that the function was not defined at $a$ and $b$. – Daniel Robert-Nicoud Aug 4 '13 at 20:27
• My bad. The problem is you're misaddressing the OP's issue. – Pedro Tamaroff Aug 4 '13 at 20:28
• @PeterTamaroff I think OP's issue is not quite clear. Daniel's answer may address it. The issue is something about what happens at the endpoints of the interval. Daniel has provided an example with a finite discontinuity. In my answer I have given an infinite discontinuity. Maybe the response will clarify matters. – Mark Bennet Aug 4 '13 at 20:31 | {
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Think about the function $\frac 1x$ on the open interval $(0,1)$ - it is not defined at $0$, but this does not stop it being continuous on the interval - in fact it is continuous because the interval is open, and we never have to deal with the bad value $x=0$.
The function $\tan x$ for the interval $(-\frac{\pi}2,\frac {\pi}2)$ is continuous, with "problems" at both ends.
Perhaps you could explain your problem in relation to these functions, as it may help to tease out what your issue really is.
• I thought that $\lim_{x\rightarrow a} f(x)=f(a)$ means that $f(a)$ is defined and exists, how can it exist if $a$ is open and not defined? – Emi Matro Aug 4 '13 at 20:42
• What do you mean by "if $a$ is open"? – Michael Albanese Aug 4 '13 at 20:43
• @MichaelAlbanese the $a$ in $(a,b)$ is "open", I mean that it is not included as an endpoint, it is never reached, and thus it's not defined as an actual point – Emi Matro Aug 4 '13 at 20:49
• @user4150 If the limit exists, and equals the value of the function (because the function is defined beyond the open interval on which it is known to be continuous) the function is then continuous on one side at $a$ - we know nothing from what you have said about the limit from the other side. It is possible for the limit to exist, but to be different from the value of the function at $a$ – Mark Bennet Aug 4 '13 at 20:49
• @user4150: I understand what you mean, but it is an incorrect use of the word 'open'. Saying the $a$ in $(a, b)$ is open doesn't mean anything; you should say "$(a, b)$ is open". Part of understanding these ideas is getting your head around the terminology and its correct usage (in particular, to which objects it applies). It is especially important when using an online medium such as MSE. – Michael Albanese Aug 4 '13 at 20:53 | {
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You might find it helpful to think about the fact that some functions that are continuous on $(a,b)$ can be extended to $[a,b]$ to give a function continuous on the interval. For instance, the function defined by $f(x) = 1$ for all $x \in (0,1)$ can be continuously extended to a function $g$ defined by $g(x) = 1$ for all $x \in [0,1]$. On the other hand, a function like $h(x) = \frac{1}{x}$ is continuous on $(0,1)$, but cannot be continuously extended at $0$. | {
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IntMath Home » Forum home » Exponents and Radicals » Multiplying top and bottom of a fraction
# Multiplying top and bottom of a fraction [Solved!]
### My question
In the example on the linked page you say: "We multiply top and bottom of each fraction with their denominators. This gives us a perfect square in the denominator in each case, and we can remove the radical"
My question is : in what situations am I allowed to use this particular method? Can it be applied to any equation involving fractions, or is it only possible when there are radicals?
### What I've done so far
Tried it with other examples, but couldn't draw a conclusion.
X
In the example on the linked page you say: "We multiply top and bottom of each fraction with their denominators. This gives us a perfect square in the denominator in each case, and we can remove the radical"
My question is : in what situations am I allowed to use this particular method? Can it be applied to any equation involving fractions, or is it only possible when there are radicals?
Relevant page
What I've done so far
Tried it with other examples, but couldn't draw a conclusion.
## Re: Multiplying top and bottom of a fraction
Hi Daniel
This technique will only be worthwhile in a limited number of cases.
For an example where it wouldn't help at all, consider 1/2.
If I multiply top and bottom by 2 (the denominator), I will get 2/4. But so what? I just need to cancel and get back to 1/2. I can do the multiplying, but it doesn't help me.
In the example that you are referring to, however, the 2 fractions become simpler since their denominators no longer have square roots.
Taking a simpler case:
1/sqrt(3a)
When I multiply top and bottom by sqrt(3a), I get:
1/sqrt(3a) xx sqrt(3a)/sqrt(3a) = sqrt(3a)/(3a)
The process has "rationalized" the denominator.
Hope that makes sense.
5. Multiplication and Division of Radicals (Rationalizing the Denominator) | {
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Hope that makes sense.
5. Multiplication and Division of Radicals (Rationalizing the Denominator)
About half way down, it has a heading "Rationalizing the Denominator". In the examples, you will see a method of multiplying top and bottom so we get rid of the square roots on the bottom. It was similar thinking (but actually simpler) that I was using in the example you are asking about.
Good luck with it.
X
Hi Daniel
This technique will only be worthwhile in a limited number of cases.
For an example where it wouldn't help at all, consider 1/2.
If I multiply top and bottom by 2 (the denominator), I will get 2/4. But so what? I just need to cancel and get back to 1/2. I can do the multiplying, but it doesn't help me.
In the example that you are referring to, however, the 2 fractions become simpler since their denominators no longer have square roots.
Taking a simpler case:
1/sqrt(3a)
When I multiply top and bottom by sqrt(3a), I get:
1/sqrt(3a) xx sqrt(3a)/sqrt(3a) = sqrt(3a)/(3a)
The process has "rationalized" the denominator.
Hope that makes sense.
<a href="/exponents-radicals/5-multiplication-division-radicals.php">5. Multiplication and Division of Radicals (Rationalizing the Denominator)</a>
About half way down, it has a heading "Rationalizing the Denominator". In the examples, you will see a method of multiplying top and bottom so we get rid of the square roots on the bottom. It was similar thinking (but actually simpler) that I was using in the example you are asking about.
Good luck with it.
## Re: Multiplying top and bottom of a fraction
Great, thanks
X
Great, thanks
## Re: Multiplying top and bottom of a fraction
You're asking good questions, Daniel. All the best to you.
X
You're asking good questions, Daniel. All the best to you.
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# How to solve polynomial equations in a field and/or in a ring?
I'm studying for my exam, and I stuck on solving polynomials in a field and/or in a ring. Let me give you some examples:
(1) Solve equation $x^2+4x+3=0$ in field $\mathbb{Z}_5$, $\mathbb{Z}_8$ and in ring $\mathbb{Z}_{12}$ and $\mathbb{Z}_5 \times \mathbb{Z}_{13}$
(2) Solve equation $x^2+2x=1$ in field $\mathbb{Z}_7$, $\mathbb{Z}_{11}$ and in ring $\mathbb{Z}_{12}$
(3) Find ring $\mathbb{Z}_n$, in which equation $x^2+2x=3$ has at least 3 solutions.
I'm not asking for solutions for these particular problems. I want to understand this topic, so I would appreciate example solutions of similar problems. Also, you're aware of some study materials - please list tutorials, books etc., about solving polynomial equations in different rings and fields (and similar topics).
Thanks for help.
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Typo, $\mathbb{Z}_8$ is not a field. (There is a field with $8$ elements, but it is not (isomorphic to) $\mathbb{Z}_8$, which has zero divisors.) – André Nicolas Jun 11 '11 at 14:57
In a finite ring you can always cop out and simply check all the elements of the ring, and see, whether they are solutions or not. Teacher may not like it, but there's nothing he or she can do about it, because the logic is impeccable. Do realize that you will waste a bit of time, so in an exam you may not afford to do this :-)
If $p$ is a prime, then $\mathbf{Z}_p$ is a field. Then you always have the result that a polynomial of degree $n$ can have at most $n$ zeros (counted with multiplicity). So if you have found $n$, you can stop looking.
The usual factorization tricks always work, if $p$ is a prime. However, note that some rings have zero divisors. For example $2\cdot 4\equiv 0\pmod{8}$, so the polynomial $x^2-1=(x-1)(x+1)$ has 3 as a root in the ring $\mathbf{Z}_8$, even though $(3-1)=2$ and $(3+1)=4$ are both non-zero. In fact, it is not hard to see that see that this polynomial has exactly 4 zeros in the ring $\mathbf{Z}_8$. | {
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The usual quadratic formula also works in the ring $\mathbf{Z}_p$, $p>2$ prime. You just need to be careful about square roots. For example in the ring $\mathbf{Z}_7$, we have $3^2=9=2$, so if the quadratic formula calls for $\sqrt{2}$, then you can use $\sqrt2=\pm3$, and the formula works the same.
You can also solve a quadratic equation by the technique of completing the square. So if you need to solve the equation $x^2+8x=5$ in the ring $\mathbf{Z}_{17}$, then you can add 16 to both sides to get $x^2+8x+16=5+16=21\equiv 4\pmod{17}$. This you can write in the form $(x+4)^2=4=2^2$. So $x+2$ must be an element with square equal to 4. As 2 and -2 are such elements, and 17 is a prime, there can be only two such elements, and we have found them both. $x+4=2$ gives us the solution $x=2-4=-2\equiv15$ and $x+4=-2$ gives the other solution $x=-2-4=-6\equiv11$.
Note that the quadratic formula does not work with respect to an even modulus. The formula calls for division by $2$ and by one of the coefficients. Both of these need to be units of the ring, because without a multiplicative inverse we cannot do division. Also, if the modulus is not a prime, then the square root may have surprisingly many values. For example in the ring $\mathbf{Z}_{15}$, 1, 4, -1=14 and -4=11 are all square roots of 1.
Finally, if your ring is a direct product of two rings, then you can solve the equation componentwise. After all, in a product ring $R_1\times R_2$ the element $(a,b)$ is the zero element, if and only if both $a=0$ and $b=0$.
-
$P(x) := x^2 + 4x + 3 = (x + 3)(x + 1)$. $P(x) = 0$ in ${\mathbb Z}_5$ implies $P(x)$ is multiple of 5, so at least one between $x + 3$ and $x + 1$ must be a multiple of 5. Therefore the solutions are 2 and 4. | {
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-
HINT $\rm(3)\:\$ diagonalize employing Chinese remainder: $\rm\:(m,n)=1\ \Rightarrow\ \mathbb Z/mn \cong \mathbb Z/m\times \mathbb Z/n\:,\:$ hence $\rm\:(x-1)\:(x+3)\:$ has root $\rm\:(1,-3) \ne (1,1),(-3,-3)\in \mathbb Z/m \times \mathbb Z/n\$ for $\rm\ m,n \ne\cdots$
NOTE $\$ The ring $\rm\:D\:$ is a domain iff every polynomial $\rm\ f(x)\in D[x]\$ has at most $\rm\ deg\ f\$ roots in $\rm\:D\:.\:$ For a simple proof see my post here, where I illustrate it constructively in $\rm\ \mathbb Z/m\$ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$. | {
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### 368. Largest Divisible Subset
Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.
Example 1:
nums: [1,2,3]
Result: [1,2] (of course, [1,3] will also be ok)
Example 2:
nums: [1,2,4,8]
Result: [1,2,4,8]
Credits:
Special thanks to @Stomach_ache for adding this problem and creating all test cases.
Seen this question in a real interview before?
When did you encounter this question?
Which company? | {
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# A function whose antiderivative equals its inverse.
Does there exist a continuous function $F$ satisfying the property
\begin{align} F\left(\int^x_0 F(s)\ ds\right) = x \end{align}
If yes, then is the solution unique?
As stated, the question is not well-posed since I haven't specified the domain of $F$. But, the question is intended to be open-ended since it was posed by a couple colleagues of mine during a coffee break.
We did find such a $F$, but couldn't show that it is unique. Here's the spoiler
$F(x) = Cx^\gamma$ where $\gamma = \frac{-1+\sqrt{5}}{2}$ and $C = (1+\gamma)^{\frac{\gamma}{1+\gamma}}$ \begin{align} C\left(\int^x_0 Ct^{\gamma}\ dt\right)^\gamma = C^{1+\gamma}\left( \frac{x^{1+\gamma}}{1+\gamma}\right)^\gamma =\frac{C^{1+\gamma}}{(1+\gamma)^\gamma}x^{\gamma+\gamma^2} = \frac{C^{1+\gamma}}{(1+\gamma)^\gamma}x = x, \ \ \text{ for all } x> 0 \end{align}
• Wow. I'm impressed that you managed to come up with that. And I have no idea how to show anything like uniqueness --- I just wanted to publicly admire both the question and the partial solution. :) – John Hughes Feb 15 '18 at 3:41
• @JohnHughes Thank you for your kind words. – Jacky Chong Feb 15 '18 at 4:18
• The function is the continuous analogue of Golomb sequence which is the OEIS sequence A001462. – Somos Feb 15 '18 at 4:18
• @Somos Thank you for the reference. – Jacky Chong Feb 15 '18 at 4:36
• On $[0,\infty)$ there is a function of the type $cx^{\alpha}$ which satisfies the given property. As of now I am not sure about the whole line. My guess is that there are several continuous functions which solve the equation. – Kavi Rama Murthy Feb 15 '18 at 7:07
We prove a slightly weaker claim, namely that there exists an infinitude of functions which are inverse to one of their antiderivatives, i.e., functions $F$ for which there exists a constant $C$ such that $$F \bigg( C+\int_0^t F(u)du \bigg) =t.$$ The functions will be defined on $[0,\infty)$, not the whole line. | {
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Proof: We use fixed point theory. Fix a real number $x>4$. For $T>0$, let us define the metric space $C_x[0,T]$ which consists of all $f \in C^1[0,T]$ which satisfy the following three conditions: $f(0)=x$, $f(t) \geq 4$ for all $t \in [0,T]$, and $\|f'\|_{\infty} \leq \frac{1}{4}$. One may easily verify that this is a complete metric space, with respect to the norm $\|f\|_{C_x[0,T]} = \|f\|_{\infty}+\|f'\|_{\infty}$. Here $\| \cdot \|_{\infty}$ denotes the uniform norm on $[0,T]$. When $T=\infty$ we define the analogous space $C_x[0,\infty)$, with the topology of uniform convergence on compact sets.
We define a map $S: C_x[0,\infty) \to C_x[0,\infty)$ by sending a function $f$ to the unique solution of the IVP: $$y'=\frac{1}{f(y)},\;\;\;\;\;\;\; y(0)=x.$$
The solution exists for global time by the Picard Lindelof theorem, since $f \geq 4$ is differentiable. Moreover, it is clear that $S$ maps $C_x[0,\infty)$ to itself. One crucial remark is that since $1/f \leq 1/4$, it necessarily follows that $y(t) \leq x+t/4$ for all $t$. | {
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If $y=Sf$ and $z=Sg$, then we have that $$|y(t)-z(t)| \leq \int_0^t \bigg| \frac{1}{f(y(s))} - \frac{1}{g(z(s))} \bigg| ds = \int_0^t \frac{|f(y(s))-g(z(s))|}{|f(y(s))g(z(s))|}ds \\ \leq \frac{1}{16} \int_0^t |f(y(s))-g(z(s))|ds \leq \frac{1}{16} \int_0^t |f(y(s)) - g(y(s))| +|g(y(s))-g(z(s))| ds \\ \leq \frac{1}{16}t \|f-g\|_{L^\infty[0,x+t/4]} + \frac{1}{16}\|g'\|_{L^\infty[0,x+t/4]} \int_0^t |y(s)-z(s)|ds$$ Again, we have used the fact that $y(s),z(s) \leq x+t/4$. Now by Gronwall's lemma, it necessarily follows that $$\|y-z\|_{L^{\infty}[0,t]} \leq \frac{t}{16} \|f-g\|_{L^\infty[0,x+t/4]} e^{\|g'\|_{L^\infty[0,x+t/4]} \frac{t}{16}} \leq \frac{t}{16}e^{\frac{t}{64}} \|f-g\|_{L^{\infty}[0,x+t/4]}.$$where we used $\|g'\|_{\infty} \leq \frac{1}{4}$. Now for the derivative-norm, we have by the same inequalities that $$\|y'-z'\|_{L^\infty[0,t]} \leq \bigg\| \frac{1}{f \circ y}-\frac{1}{g \circ z} \bigg\|_{L^{\infty}[0,t]} \leq \frac{1}{16} \| f \circ y- g \circ z\|_{L^{\infty}[0,t]} \\ \leq \frac{1}{16} \|f-g\|_{L^{\infty}[0,x+t/4]} + \frac{1}{16} \|g'\|_{L^\infty[0,x+t/4]} \|y-z\|_{L^{\infty}[0,t]} \\ \leq \big( \frac{1}{16} + \frac{t}{256} e^{\frac{t}{64}} \big) \|f-g\|_{L^{\infty}[0,x+t/4]}$$ The preceding two expressions together show that if we define a Picard iteration scheme $f_n=S^nf_0$, then by induction we have that $$\|f_{n+1}-f_n\|_{L^{\infty}[0,t]} + \|f'_{n+1}-f'_n\|_{L^{\infty}[0,t]} \\ \leq \big(\frac{t}{256}e^{\frac{t}{64}} + \frac{1}{16} \big) \big( \frac{t}{16}e^{\frac{t}{64}} \big)^n \big\| f_1-f_0 \big\|_{L^{\infty}[0, (1+4^{-1}+...+4^{1-n})x +4^{-n}t]}$$so that the restriction to $[0,T]$ of these iterates will converge to some function $f$ as $n \to \infty$, whenever $T$ is small enough so that $\frac{T}{16}e^{T/64} <1$ (e.g. $T<13$ works). Now as long as $x<39/4$ (note that this doesn't interfere with the condition $x>4$), we have that $\sum_0^{\infty} 4^{-n}x=\frac{4}{3}x<13$ which is less than the critical $T$-value just discussed. Hence given | {
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4^{-n}x=\frac{4}{3}x<13$ which is less than the critical $T$-value just discussed. Hence given any $T>0$, there exists some $N$ large enough so that $\sum_0^{N-1} 4^{-k}x + 4^{-N}T< 13$, and hence we see that $f_n$ actually converges in $C_x[0,T]$ for every $T>0$. This argument shows that the map $S$ actually has a globally defined fixed point on the whole space $C_x[0,\infty)$. | {
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Now let $f$ be a fixed point of the map $S$. This means that for every $t>0$, one has $$f'(t) f(f(t)) = 1.$$ So if we define $F(t) = \int_0^t f(u)du$ it holds that $F'(f(t))f'(t)=1$, therefore $F(f(t))=t+F(x)$. Clearly each initial value $x$ gives a different solution $f$.
Now we claim that the fixed point $f$ satisfies $\lim_{t \to \infty} f(t) = +\infty$. If this was not the case, then $f\circ f$ would be bounded on $[0,\infty)$, hence $f' = 1/(f \circ f)$ would be bounded below, hence $f(t) \geq x+ct$ for some $c>0$, contradiction. Therefore $f$ is a continuously increasing bijection from $[0,\infty)$ to $[x,\infty)$. Let $C:=F(x)$, and define $g:[C,\infty) \to [x,\infty)$ by $g(t) = f(t-C)$. Clearly $F(g(t)) = t$, and therefore by the fact that $f$ is a bijection, it follows (by uniqueness of inverses) that $g(F(t))=t$. In other words, we have $$f(F(t)-C)=t,\;\;\;\;\;\;\text{i.e.},\;\;\;\; f\bigg( \int_x^t f(u)du \bigg) = t,\;\;\;\;\forall t \geq x,$$thus proving the claim. | {
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• There is no solution on the whole line: the function $x \to \int_0 ^{x} F(x)dx$ a continuous function. By the given equation this function is one-to-one, hence strictly monotonic. By differentiation this implies that $F>0$ or $F<0$ everywhere. This makes the right side >0 or <o everywhere which is absurd. Please let me know if there is a stupid mistake in this argument. – Kavi Rama Murthy Feb 17 '18 at 12:11
• @KaviRamaMurthy That's correct (as I already said in a comment above) but the solutions constructed here are not on the full line, just $[0,\infty)$. – shalop Feb 17 '18 at 16:58
• @JackyChong: I thought about the problem some more, and I was able to prove some additional results, for instance if $f$ is any function which is inverse to one of its antiderivatives, then necessarily we have $$\lim_{t \to \infty} \frac{f(t)}{t^{\gamma}} = C,$$where $\gamma$ and $C$ are the same ones you found. But, I was not able to prove full-blown uniqueness given a fixed initial value (though the fixed point argument works as above,for certain values of $f(0)$). – shalop Feb 17 '18 at 21:49 | {
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Although other mathematicians in Persia and China had independently discovered the triangle in the eleventh century, most of the properties and applications of the triangle were discovered by Pascal. Subsequent row is made by adding the number above and to the left with the number above and to the right. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. It is also being formed by finding () for row number n and column number k. Pascal’s triangle is an array of binomial coefficients. The second row is 1,2,1, which we will call 121, which is 11x11, or 11 squared. Code Breakdown . The code inputs the number of rows of pascal triangle from the user. Aug 2007 3,272 909 USA Jan 26, 2011 #2 This video shows how to find the nth row of Pascal's Triangle. ) have differences of the triangle numbers from the third row of the triangle. Pascal’s Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 . Pascal’s triangle, in algebra, a triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression, such as (x + y) n.It is named for the 17th-century French mathematician Blaise Pascal, but it is far older.Chinese mathematician Jia Xian devised a triangular representation for the coefficients in the 11th century. We hope this article was as interesting as Pascal’s Triangle. The first row of Pascal's triangle starts with 1 and the entry of each row is constructed by adding the number above. Anonymous. Let’s go over the code and understand. 9 months ago. 3 Some Simple Observations Now look for patterns in the triangle. As examples, row 4 is 1 4 6 4 1, so the formula would be 6 – (4+4) + (1+1) = 0; and row 6 is 1 6 15 20 15 6 1, so the formula would be 20 – (15+15) + (6+6) – (1+1) = 0. One of the most interesting Number Patterns is Pascal's Triangle (named after Blaise Pascal, a famous French Mathematician and Philosopher). To understand this example, you should have the | {
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a famous French Mathematician and Philosopher). To understand this example, you should have the knowledge of the following C programming topics: Here is a list of programs you will find in this page. The n th n^\text{th} n th row of Pascal's triangle contains the coefficients of the expanded polynomial (x + y) n (x+y)^n (x + y) n. Expand (x + y) 4 (x+y)^4 (x + y) 4 using Pascal's triangle. 1. Example: Input : k = 3 Return : [1,3,3,1] NOTE : k is 0 based. … Below is the example of Pascal triangle having 11 rows: Pascal's triangle 0th row 1 1st row 1 1 2nd row 1 2 1 3rd row 1 3 3 1 4th row 1 4 6 4 1 5th row 1 5 10 10 5 1 6th row 1 6 15 20 15 6 1 7th row 1 7 21 35 35 21 7 1 8th row 1 8 28 56 70 56 28 8 1 9th row 1 9 36 84 126 126 84 36 9 1 10th row 1 10 45 120 210 256 210 120 45 10 1 For instance, on the fourth row 4 = 1 + 3. In this article, however, I explain first what pattern can be seen by taking the sums of the row in Pascal's triangle, and also why this pattern will always work whatever row it is tested for. Kth Row of Pascal's Triangle Solution Java Given an index k, return the kth row of Pascal’s triangle. © Parewa Labs Pvt. Historically, the application of this triangle has been to give the coefficients when expanding binomial expressions. Create all possible strings from a given set of characters in c++. How do I use Pascal's triangle to expand the binomial #(d-3)^6#? Each number is found by adding two numbers which are residing in the previous row and exactly top of the current cell. Although the peculiar pattern of this triangle was studied centuries ago in India, Iran, Italy, Greece, Germany and China, in much of the western world, Pascal’s triangle has … Join our newsletter for the latest updates. All values outside the triangle are considered zero (0). Day 4: PascalÕs Triangle In pairs investigate these patterns. Step by step descriptive logic to print pascal triangle. Graphically, the way to build the pascals triangle is pretty easy, as mentioned, to | {
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pascal triangle. Graphically, the way to build the pascals triangle is pretty easy, as mentioned, to get the number below you need to add the 2 numbers above and so on: With logic, this would be a mess to implement, that's why you need to rely on some formula that provides you with the entries of the pascal triangle that you want to generate. Kth Row of Pascal's Triangle Solution Java Given an index k, return the kth row of Pascal’s triangle. Each row of Pascal’s triangle is generated by repeated and systematic addition. Read further: Trie Data Structure in C++ The numbers in each row are numbered beginning with column c = 1. 2�������l����ש�����{G��D��渒�R{���K�[Ncm�44��Y[�}}4=A���X�/ĉ*[9�=�/}e-/fm����� W\$�k"D2�J�L�^�k��U����Չq��'r���,d�b���8:n��u�ܟ��A�v���D��N� ��A��ZAA�ч��ϋ��@���ECt�[2Y�X�@�*��r-##�髽��d��t� F�z�{t�3�����Q ���l^�x��1'��\��˿nC�s Generally, In the pascal's Triangle, each number is the sum of the top row nearby number and the value of the edge will always be one. The Fibonacci Sequence. Answer Save. Best Books for learning Python with Data Structure, Algorithms, Machine learning and Data Science. Examples: Input: N = 3 Output: 1, 3, 3, 1 Explanation: The elements in the 3 rd row are 1 3 3 1. What is the 4th number in the 13th row of Pascal's Triangle? Which row of Pascal's triangle to display: 8 1 8 28 56 70 56 28 8 1 That's entirely true for row 8 of Pascal's triangle. 3 Answers. ; Inside the outer loop run another loop to print terms of a row. This is down to each number in a row being … The … One of the famous one is its use with binomial equations. 1, 1 + 1 = 2, 1 + 2 + 1 = 4, 1 + 3 + 3 + 1 = 8 etc. This triangle was among many o… However, for a composite numbered row, such as row 8 (1 8 28 56 70 56 28 8 1), 28 and 70 are not divisible by 8. Pascal's triangle is one of the classic example taught to engineering students. Moving down to the third row, we get 1331, which is 11x11x11, or 11 cubed. More rows of Pascal’s triangle are listed on the final | {
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we get 1331, which is 11x11x11, or 11 cubed. More rows of Pascal’s triangle are listed on the final page of this article. Is there a pattern? for(int i = 0; i < rows; i++) { The next for loop is responsible for printing the spaces at the beginning of each line. This math worksheet was created on 2012-07-28 and has been viewed 58 times this week and 101 times this month. Naturally, a similar identity holds after swapping the "rows" and "columns" in Pascal's arrangement: In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding column from its row to the first, inclusive (Corollary 3). First 6 rows of Pascal’s Triangle written with Combinatorial Notation. Python Basics Video Course now on Youtube! If we look at the first row of Pascal's triangle, it is 1,1. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row. Example: Input : k = 3 Return : [1,3,3,1] Java Solution of Kth Row of Pascal's Triangle In this post, we will see the generation mechanism of the pascal triangle or how the pascals triangle is generated, understanding the pascal's Triangle in c with the algorithm of pascals triangle in c, the program of pascal's Triangle in c. Input number of rows to print from user. If you sum all the numbers in a row, you will get twice the sum of the previous row e.g. Make a Simple Calculator Using switch...case, Display Armstrong Number Between Two Intervals, Display Prime Numbers Between Two Intervals, Check Whether a Number is Palindrome or Not. Note: The row index starts from 0. At first, Pascal’s Triangle may look like any trivial numerical pattern, but only when we examine its properties, we can find amazing results and applications. Subsequent row is created by adding the number above and to the left with the number above and to the right, treating empty | {
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by adding the number above and to the left with the number above and to the right, treating empty elements as 0. Welcome to The Pascal's Triangle -- First 12 Rows (A) Math Worksheet from the Patterning Worksheets Page at Math-Drills.com. Entry in the top row, there is an array of 1 note Could. Convention holds that both row numbers and write the sum between and below them the sum of ways! Numbered 0 through 5 ) of the row above triangle is that the rows of Pascal ’ s are... ; Inside the outer loop run another loop to print Pascal triangle you optimize your algorithm to use only (... ) 4, the apex of the row above by step descriptive logic to print Pascal triangle or... Was born at Clermont-Ferrand, in the third row of Pascal ’ s triangle starts with 1! Row ’ number of occurrences of an element in a row, will. Simple Observations Now look for patterns in number theory the 13th row of the current cell of row. Of binomial coefficients k, return the kth row of Pascal ’ s triangle is a triangular pattern few are! Numbers arranged in rows forming a triangle another loop to print terms of a row triangle to us. That row a + b ) 4, the apex of the ways this can be found in Pascal triangle... And has been to give the coefficients when expanding binomial expressions 3 return: [ 1,3,3,1 ] note: ’! Addition leads to many multiplicative patterns call 121, which is 11x11x11, or 11 cubed triangle in pairs these... It relates to the row [ 1 ] is defined such that rows... Write the sum of the current cell, firstly, where can the … the code and understand logged... Data Structure, Algorithms, Machine learning and Data Science every adjacent pair of and... The sum of the Pascal ’ s triangle can be done: binomial Theorem areas mathematics. Taught to engineering students found by adding the number above and to the binomial Theorem and other areas mathematics. Sequence can be drawn as a triangle the 4th number in the Auvergne region France. Facts to be seen in the 13th row of the famous | {
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the 4th number in the Auvergne region France. Facts to be seen in the 13th row of the famous one is its use with binomial equations are to. − in the top a row the natural number sequence can be found in Pascal triangle! And the entry of each row in PascalÕs triangle s go over the code understand... To interpret rows with two digit numbers 6 4 1: 1 1 4 6 1. Third diagonal ( natural numbers ) and third diagonal ( natural numbers ) and third diagonal ( natural )... Six rows ( numbered 0 through 5 ) of the most interesting number patterns is Pascal 's triangle Philosopher. S triangle starts with 1 and 3 in the 13th row of Pascal s... Taught to engineering students two numbers which are residing in the new row is numbered as n=0, the!, there is an array of 1 you will Get twice the sum of the most interesting patterns... Numbered 0 through 5 ) of the row [ 1 ] this will... Data Structure, Algorithms, Machine learning and Data Science PASCALIANUM — is one of the most interesting numerical in. Adding the number above and to the left of the row can be created as follows: the! The left beginning with k = 0, corresponds to the left of the most interesting patterns... The right printing each row is made by adding two numbers which are residing in the rows are as −! Number above Philosopher ) one column gives the numbers from the previous row e.g Count the number above and the., where can the … More rows of Pascal triangle in pairs investigate these.. That can be optimized up to nth row of the previous column ( the first six rows numbered. To comment below for any queries or feedback is 1,1 are numbers that can be calculated Using a.... Rows ( numbered 0 through 5 ) of the most interesting number patterns is 's! Numbers 1 and 3 in the third row of the Pascal triangle similar posts: Count the number of.! Taught to engineering students ways this can be created as follows: the! Seen in the 13th row of Pascal ’ s triangle approach will have O ( 2... Left on the Arithmetical triangle | {
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in the 13th row of Pascal ’ s triangle approach will have O ( 2... Left on the Arithmetical triangle which today is known as the Pascal triangle Machine learning and Data Science a. 11 can be drawn like this that the rows of Pascal 's triangle ( named after French... Beginning with k = 0 numbers 1 and 3 in the third row Pascal. Through 5 ) of the Pascal triangle from the user of France on June,! It is 1,1 patterns involving the binomial coefficient as follows − What is the fourth row defined that! Of that row in this amazing triangle exists between the second row is numbered as n=0 and. The code and understand ( named after the French Mathematician and Philosopher ) can …. The Arithmetical triangle which today is known 15th row of pascals triangle the Pascal ’ s triangle − What is the numbers in row... As follows − in the triangle of mathematics created on 2012-07-28 and has been viewed times... Strings from a given set of characters in c++ code and understand continue numbers... Where n is row 0, and in each row are numbered beginning with k 0. Numbers below it in a linked list in c++ and third diagonal natural... Of binomial coefficients the French Mathematician Blaise Pascal was born at Clermont-Ferrand, in rows. And Data Science 4 1 to produce the number above and to the right will run row! Sum between and below them will Get twice the sum of each row in PascalÕs triangle run. Property of Pascal triangle the Auvergne region of France on June 19, 1623 drawn as triangle. Be optimized up to O ( n 3 ) =.... 0 0 exponent is ' 4 ' new... Repeated addition leads to many multiplicative patterns relationship in this amazing triangle exists between the second (. Is its use with binomial equations known as the Pascal triangle from the left with the of... Array of 15th row of pascals triangle coefficients 3 1 1 1 1 2 1 1 1 2! Row, we have to find the sum of the most interesting numerical in. Addition leads to many multiplicative patterns and write the sum of the previous | {
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numerical in. Addition leads to many multiplicative patterns and write the sum of the previous row e.g column c =.... Forms a system of numbers and column numbers start with 1 at. With binomial equations digit numbers ( named after the French Mathematician and Philosopher ) added to produce the of. The current cell numbers 1 and the first number 1 is knocked off, however ) to give the when! As a triangle the numbers from the previous column ( the first number row... Construct a new row for the triangle Data Science ; Inside the outer most for loop responsible... Now look for patterns in number theory of how it relates to the left of the current cell 0! Can see, it can be drawn as a triangle − in the Auvergne region France. France on June 19, 1623 example, numbers 1 and the row! Powers of 11 sum all the numbers from the user triangular pattern k ) extra space region... Use only O ( n 2 ) time complexity each number is found by adding 15th row of pascals triangle of! As you can see 15th row of pascals triangle it forms a system of numbers and column is,. Such that the rows of Pascal 's triangle: k is term that... Of France on June 19, 1623 11 squared loop run another loop print., on the fourth row 4, the number in row and column is created on 2012-07-28 has. Binomial Theorem and other areas of mathematics 3 some Simple Observations Now look patterns! The code inputs the number of times Matrix to a Function 4 = 1 + 3 Could you your! Constructed by adding the number of times Pascal triangle apex of the two entries above it step step. These similar posts: Count the number 15th row of pascals triangle and to the left with the number 4 in top! Of ( a + b ) 4, the exponent 15th row of pascals triangle ' 4 ' Pascal! Previous column ( the first number in the top drawn as a triangle where the powers of can. Most interesting number patterns is Pascal 's triangle is a triangular pattern this article was as interesting Pascal... Latin Triangulum Arithmeticum PASCALIANUM — is one | {
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pattern this article was as interesting Pascal... Latin Triangulum Arithmeticum PASCALIANUM — is one of the most interesting number patterns is Pascal 's triangle can be as... Defined such that the number in row 4 = 1 + 3 write the sum the! Suppose we have to find the sum of the triangle, you add a 1 below and the... 1 '' at the first six rows ( numbered 0 through 5 ) of the current.! With 1 '' at the top row, you add a 1 below and to row. ( the first number in the triangle, you 15th row of pascals triangle a 1 below and to the Theorem! Step by step descriptive logic to print terms of a row characters in c++ third row you... I ’ ve left-justified the triangle, you will Get twice the sum between and below them and Data.! 3 in the fourth row [ 1,3,3,1 ] note: Could you optimize your algorithm to use only O n... 11 squared exists between the second diagonal ( triangular numbers ) and third diagonal ( natural numbers.! Look for patterns in the fourth number in row 4, the apex the. Interesting number patterns is Pascal 's triangle, you will Get twice the of... Are numbered from the third row, there is an array of binomial coefficients sequence can done... With ` 1 '' at the first six rows ( numbered 0 through )! Number 1 is knocked off, however ) continue placing numbers below in! Column c = 1 like this also, refer to these similar posts: the... | {
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# Math Help - Integration
1. ## Integration
Hello
Can someone help me with this please:
$\int\dfrac{3}{2}\cos^2(4x)$
I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$
A hint at the right approach would be very much appreciated.
Thank you.
2. ## Re: Integration
Hello, Furyan!
Can someone help me with this please?
. . $\int\tfrac{3}{2}\cos^2(4x)$
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
3. ## Re: Integration
Hi Furyan!
Are you aware of the double angle formula for $\cos(2y)$ in terms of $\cos^2y$?
4. ## Re: Integration
Originally Posted by Furyan
Hello
Can someone help me with this please:
$\int\dfrac{3}{2}\cos^2(4x)$
I tried a number of things, but can't see what function differentiates to give $\cos^2(4x)$
A hint at the right approach would be very much appreciated.
Thank you.
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
5. ## Re: Integration
Originally Posted by Soroban
Hello, Furyan!
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
You don't really NEED this identity, though it is the simplest simplification.
Integration by Parts will also work...
6. ## Re: Integration
The following identity may be useful:
$\cos^2 x = \frac{1 + \cos 2x}{2}$
Try using it.
7. ## Re: Integration
Hello Soroban,
Originally Posted by Soroban
Hello, Furyan!
You need this identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos 2\theta}{2}$
Thank you. I was able to find the integral using that identity and substituting $\cos^2 4\theta$ with $\dfrac{1}{2}(1 + \cos8\theta)$
8. ## Re: Integration
Hello Prove It
Originally Posted by Prove It
It would help if you realise that \displaystyle \begin{align*} \cos^2{X} \equiv \frac{1}{2} + \frac{1}{2}\cos{2X} \end{align*}...
Thank you. I am having trouble with that realisation but I'm working on it. | {
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I will also try using integration by parts.
9. ## Re: Integration
Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?
10. ## Re: Integration
MarceloFantini
Originally Posted by MarceloFantini
Do you mean $\cos^2 4x = \frac{1 + \cos 8x}{2}$ instead of $\cos^2 4x = \frac{1+8 \cos x}{2}$?
Yes indeed, thank you for pointing that out. I have corrected my post accordingly. I did use the former when I found the integral.
11. ## Re: Integration
Originally Posted by Furyan
Hello Prove It
Thank you. I am having trouble with that realisation but I'm working on it.
I will also try using integration by parts.
Here's how that identity is derived.
First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}
Ans 1
Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have
\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}
And if you're trying to use integration by parts... | {
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And if you're trying to use integration by parts...
\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}
And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
12. ## Re: Integration
Dear Prove It
Originally Posted by Prove It
Here's how that identity is derived.
First a proof of the angle sum formula \displaystyle \begin{align*} \cos{\left(\alpha + \beta\right)} \equiv \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \end{align*}
Ans 1
Now, using the angle sum formula with \displaystyle \begin{align*} \alpha = \beta = X \end{align*} we have
\displaystyle \begin{align*} \cos{\left(X + X\right)} &\equiv \cos{(X)}\cos{(X)} - \sin{(X)}\sin{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \sin^2{(X)} \\ \cos{(2X)} &\equiv \cos^2{(X)} - \left[1 - \cos^2{(X)}\right] \textrm{ by the Pythagorean Identity} \\ \cos{(2X)} &\equiv 2\cos^2{(X)} - 1 \\ 2\cos^2{(X)} &\equiv 1 + \cos{(2X)} \\ \cos^2{(X)} &\equiv \frac{1}{2} + \frac{1}{2}\cos{(2X)} \end{align*}
And if you're trying to use integration by parts... | {
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And if you're trying to use integration by parts...
\displaystyle \begin{align*} I &= \int{\frac{3}{2}\cos^2{(4x)}\,dx} \\ I &= \frac{3}{2}\int{\cos{(4x)}\cos{(4x)}\,dx} \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} - \int{-4\sin{(4x)}\cdot \frac{1}{4}\sin{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{\sin^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1 - \cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{2}\left[\frac{1}{4}\sin{(4x)}\cos{(4x)} + \int{1\,dx} - \int{\cos^2{(4x)}\,dx}\right] \\ I &= \frac{3}{8} \sin{(4x)} \cos{(4x)} + \frac{3}{2} \int{1\,dx} - \frac{3}{2} \int{ \cos^2{ (4x) }\,dx} \\ I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x - I \\ 2I &= \frac{3}{8}\sin{(4x)}\cos{(4x)} + \frac{3}{2}x \\ I &= \frac{3}{16}\sin{(4x)}\cos{(4x)} + \frac{3}{4}x + C \end{align*}
And with a little manipulation I'm sure you can get this to be the same form as your answer you gained using the double angle identity...
I have been looking at the double angle identity for $\cos2\theta$ and, for some reason, was having trouble realising what you had pointed out in your original post. Now it's crystal clear. Some how I keep missing the elementary principles.
I will try and find the integral again using the substitution you have suggested.
Thank you also for the help with using integration by parts. I'd never have been able to do that, but I am very interested in it and and will work through it until I understand it.
Thank you very much for the time you have taken to help me. It's extremely generous of you and very grateful. | {
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# Elementary Inequality
This should be completely elementary, if not downright trivial compared to other problems I've seen on this site. I think I'm just being silly. The problem is from Kaczor and Nowak's problem book in analysis. Somehow I made it through undergraduate without ever seeing half the inequalities in this text, and I really need to know them, so I'm trying to get some practice.
Suppose that $a_k$ are positive numbers, where $k$ ranges from $1$ to $n$, and satisfies $\sum a_k \leq 1$. Show that $\sum \frac{1}{a_k} \geq n^2$.
So, a moment of staring at the problem, and the problem screams AM-GM-HM inequality. We can rewrite what we are asked to prove as $$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{1}{n^2}$$
so that the left side is now precisely the harmonic mean of the $a_k$. By AM-GM-HM, this is less or equal to both the harmonic and geometric means. The obvious choice here being to use the arithmetic mean since it's the only thing for which we can say anything about. This gives:
$$\frac{1}{\sum \frac{1}{a_k}} \leq \frac{\sum a_k}{n} \leq \frac{1}{n}$$
Which is weaker than what I was asked to prove - I've only been able to show the original sum is $\geq n$, rather $n^2$.
Can the approach be fixed? Perhaps there is a better way to go?
• Harmonic mean $\dfrac{1}{\sum \frac{1}{a_k}}$ should be $\dfrac{n}{\sum \frac{1}{a_k}}$ – DHMO Apr 18 '17 at 3:53
• @DHMO Of course, if I use the inequality properly, the result is clear. Only mildly embarrassing. I should mail my degree back... (just kidding of course). Thank you though. – Alfred Yerger Apr 18 '17 at 3:54
The HM has an $n$ on top (it's the reciprocal of the AM of the reciprocals). (Think about what happens if all the $a_k$ are equal.) | {
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$$\begin{array}{rcl} \displaystyle \sum_{k=1}^n a_k &\le& 1 \\ \displaystyle \dfrac1n \sum_{k=1}^n a_k &\le& \dfrac1n \\ \displaystyle \dfrac n{\sum_{k=1}^n \frac1{a_k}} &\le& \dfrac1n \\ \displaystyle \dfrac{\sum_{k=1}^n \frac1{a_k}}n &\ge& n \\ \displaystyle \sum_{k=1}^n \frac1{a_k} &\ge& n^2 \\ \end{array}$$
With equality when $\forall k \in 1,2,3\cdots, n:a_k = \dfrac1n$.
1. Cauchy-Schwarz.
$$n^2 = (\sum_{k=1}^n \sqrt{a_k} \frac{1}{\sqrt{a_k}})^2\le \sum_{k=1}^n a_k \times \sum_{k=1}^n \frac{1}{a_k}\le \sum_{k=1}^n \frac{1}{a_k}.$$
1. Calculus.
For any $\lambda>0$ we have
\begin{align*} 2n \le \sum_{k=1}^n (\lambda a_k + \frac{1}{\lambda a_k})&\le \lambda+ \sum_{k=1}^n \frac{1}{\lambda a_k}.\end{align*}
Therefore
$$\lambda( 2n -\lambda) \le \sum_{k=1}^n \frac{1}{a_k}.$$
Maximum of LHS at $\lambda = n$, and result follows. | {
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1. ## IBV
$\displaystyle \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0$
$\displaystyle \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\displaystyle \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L$
$\displaystyle u(x,t)=\varphi(x)T(t)$
$\displaystyle \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$
$\displaystyle \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$
$\displaystyle \displaystyle T=Ce^{-\lambda kt}$
$\displaystyle \varphi_1(0)=1 \ \varphi_2(0)=0$
$\displaystyle \varphi_1'(0)=0 \ \varphi_2'(0)=1$
$\displaystyle \varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi(0): \ A_2=0$
$\displaystyle \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$
The eigenvalues must satisfy:
$\displaystyle \displaystyle\cos(x\sqrt{\lambda})=0$
$\displaystyle \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$
The eigenfunction is:
$\displaystyle \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$
Is this much correct?
Thanks.
2. $\displaystyle \displaystyle u(x,t)=\sum_{n=1}^{\infty}b_n\frac{2L}{\pi(1+2n)}\ exp\left(-\frac{\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\rig ht)\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
I have shown $\displaystyle \displaystyle\int_0^L\varphi_n(x)\varphi_m(x) \ dx=0, \ m\neq n$.
Before I solve for $\displaystyle b_n$ is u(x,t) correct?
Thanks.
3. Originally Posted by dwsmith
$\displaystyle \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0$
$\displaystyle \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_ x(L,t)=0\end{cases}$
$\displaystyle \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L$
$\displaystyle u(x,t)=\varphi(x)T(t)$ | {
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$\displaystyle u(x,t)=\varphi(x)T(t)$
$\displaystyle \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda$
$\displaystyle \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd a})$
$\displaystyle \displaystyle T=Ce^{-\lambda kt}$
$\displaystyle \varphi_1(0)=1 \ \varphi_2(0)=0$
$\displaystyle \varphi_1'(0)=0 \ \varphi_2'(0)=1$
Where did $\displaystyle \varphi_1$ and $\displaystyle \varphi_2$ suddenly come from?
Are they $\displaystyle e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $\displaystyle e^{\lambda kt}(B sin(x\sqrt{\lambda}))$?
If so, you cannot, generally, do that. But here it does not hurt.
$\displaystyle \varphi_1: \ A=1 \ \varphi_2: \ A=0$
$\displaystyle \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}$
$\displaystyle \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\ frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}$
All you really need is that $\displaystyle \phi(0, t)= e^{-\lambda kt}(Acos(0)+ Bsin(0))= Ae^{-k\lambda kt}= 0$. Since the exponential is never 0, A= 0.
Then $\displaystyle \phi(L, t)= e^{-\lambda kt}B cos(L\sqrt{\lambda})= 0$. Again, the exponential is not 0 so we must have either B= 0, which would mean the solution was identically equal to 0, and so not an eigenfunction, or cos(L\sqrt{\lambda})= 0.
$\displaystyle \displaystyle\varphi(0): \ A_2=0$
$\displaystyle \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0$
The eigenvalues must satisfy:
$\displaystyle \displaystyle\cos(x\sqrt{\lambda})=0$
You mean $\displaystyle cos(L\sqrt{\lambda})= 0$
$\displaystyle \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\ right)^2, \ n\in\mathbb{Z}$
The eigenfunction is:
$\displaystyle \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+$
Is this much correct?
Thanks.
Yes, that is correct and your second post is correct. | {
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Is this much correct?
Thanks.
Yes, that is correct and your second post is correct.
4. Originally Posted by HallsofIvy
Where did $\displaystyle \varphi_1$ and $\displaystyle \varphi_2$ suddenly come from?
Are they $\displaystyle e^{\lambda kt}(Acos(x\sqrt{\lambda}))$ and $\displaystyle e^{\lambda kt}(B sin(x\sqrt{\ambda}))$?
If so, you cannot, generally, do that. But here it does not hurt.
My book solves all PDEs in the same manner. I don't know why it is done that way but it works in all cases.
Here is what the book says about $\displaystyle \varphi$ 1 and 2.
The theorem in the theory of ODE which explicitly states the proposition above is: if the coefficients $\displaystyle c_0(x), \ c_1(x), \ c_2(x)$ are continuous on the interval [a,b] and $\displaystyle c_0(x)$ is different from zero throughout the interval, then for any fixed $\displaystyle x_0$ in the interval [a,b] and for any constants $\displaystyle \alpha, \ \beta$. The DE has one and only one solution $\displaystyle \varphi(x)=\varphi(x,\lambda)$ satisfying at $\displaystyle x_0$ the initial conditions $\displaystyle \varphi(x_0)=\alpha, \ \varphi'(x_0)=\beta$.
Moreover, $\displaystyle \varphi(x,\lambda)$ and $\displaystyle \varphi'(x,\lambda)$ are continuous and continuously differentiable functions of both x for $\displaystyle a\leq x\leq b$ and all lambda.
It is convenient to single out the two special solutions $\displaystyle \varphi_1=\varphi_1(x,\lambda), \ \varphi_2=\varphi_2(x,\lambda)$ of which satisfy the DE
$\displaystyle \varphi_1(0)=1, \ \varphi_2(0)=0$
$\displaystyle \varphi'_1 (0)=0, \ \varphi'_2(0)=1$.
We call 1 and 2 the basic solutions at x nought.
5. $\displaystyle \displaystyle \int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\int_0^Lb_m\sin^2\left(\frac{\pi x(1+2m)}{L}\right) \ dx$
$\displaystyle \displaystyle\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=b_m\frac{L}{2}\Rightarrow b_m=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx$ | {
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6. For the case $\displaystyle f(x)=1$
$\displaystyle \displaystyle b_m=\frac{2}{L}\int_0^L\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\frac{2}{L}\left[\frac{-2L}{\pi(1+2m)}\cos\left(\frac{\pi x(1+2m)}{2L}\right)\right]_0^L$
$\displaystyle \displaystyle b_m=\frac{4}{\pi(1+2m)}, \ \ m=n$
$\displaystyle \displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{L}{ (1+2n)^2}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
If everything above is correct, can I do this?
$\displaystyle \displaystyle\sum_0^{\infty}\frac{1}{(1+2m)^2}=\fr ac{\pi^2}{8}$
$\displaystyle \displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{ (1+2n)^2}\cdot L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
$\displaystyle \displaystyle\Rightarrow u(x,t)=L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
7. Ok, I pretty sure the above answer is wrong.
I now have:
$\displaystyle \displaystyle u(x,t)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)$
How do I show this is true for all x with the IC:
$\displaystyle \displaystyle u(x,0)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+ 2n}\sin\left(\frac{\pi x(1+2n)}{2L}\right)=1$ | {
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# Which area is larger, the blue area, or the white area?
In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
# My Solution
Let the length of each side of the square be $$2r$$.
The area of the square is $$4r^2$$.
The two semi-circles have equal area.
Area of one semi-circle = $$\frac{{\pi}r^2}{2}$$.
$${\times}2 = {\pi}r^2$$
White area = $${\pi}r^2 -$$ area of the intersection of the two circles.
Let the area of the intersection of the two circles be $$t$$.
White area = $${\pi}r^2 - t$$.
The segments that make up $$t$$ are identical.
$$t$$ = area of segment $${\times}2$$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $$90^{\circ}$$ (The circles both have radius $$r$$, and the outer shape is a square.
Angle of sector $$= \frac{1}{4} * {\pi}r^2$$.
Area of triangle $$= \frac{1}{2} * r^2$$.
Area of segment $$= \frac{{\pi}r^2 - 2r^2}{4}$$. $$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$$.
$$t = \frac{{\pi}r^2 - 2r^2}{2}$$.
White area $$= {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$$.
White area $$= \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$$.
White area $$= \frac{{\pi}r^2 + 2r^2}{2}$$.
Blue area = $$r^2\left(4 - \frac{{\pi} + 2}{2}\right)$$.
Blue area = $$r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$$.
Blue area = $$r^2\left(\frac{6 - {\pi}}{2}\right)$$.
If White area $$-$$ Blue area $$\gt 0$$, then the White area is larger.
$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$
$$\therefore$$ the white area is larger.
## What is the error in my solution?
The provided solution: | {
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## What is the error in my solution?
The provided solution:
• It's a general rule of thumb, that in all puzzles which ask "which area is larger" the answer always is that both areas have the same size, even when one is visibly enormous and the other visibly tiny. – Angina Seng Jul 23 '17 at 21:11
• White area $= \pi r^2 - \color{red}{2} \times$area of the intersection of the two circles. – Donald Splutterwit Jul 23 '17 at 21:16
• Move the small blue caps to the other side of the semicircles, then the blue region makes a triangle. – Michael Burr Jul 23 '17 at 21:16
• @DonaldSplutterwit thanks for the pointer; do you want to make it into an answer, or should I do it? – Tobi Alafin Jul 23 '17 at 21:20
• What's the question? – Shuri2060 Jul 23 '17 at 21:21
If you just divide the lower left part of the blue area and move each part $90$ degree up to join them to the main blue part, then the area of the new blue shape will be equal to the white part.
• This is the solution given as "provided solution" by the OP. The OP has already seen this (neat) solution. How does this add to what is already present in the question? – James K Jul 23 '17 at 22:14
• @Seyed In the bottom of the question, it says "The provided solution". – HelloGoodbye Jul 23 '17 at 22:42
• As far as I can see, the question wasn't edited at any point – Shuri2060 Jul 23 '17 at 22:58
• @user1936752: Technically speaking, Seyed did answer the question. There is only one question mark in the post, which is for the title "which area is larger, blue or white?". It's likely that what the OP wanted is an explanation for what is the mistake in his own attempted solution, but he never actually asked that... – Meni Rosenfeld Jul 24 '17 at 9:09
• Making the neat solution available in the thread rather than just as a link to a dodgy-sounding url does add value IMO. – Especially Lime Jul 24 '17 at 10:44
There was a mistake in my earlier solution. I correct that mistake here. | {
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There was a mistake in my earlier solution. I correct that mistake here.
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - 2 {\times}$area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - 2t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $= \frac{1}{4} * {\pi}r^2$.
Area of triangle $= \frac{1}{2} * r^2$.
Area of segment $= \frac{{\pi}r^2 - 2r^2}{4}$. $t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $= {\pi}r^2 - 2\left(\frac{{\pi}r^2 - 2r^2}{2}\right)$.
White area $= \frac{2{\pi}r^2 - 2{\pi}r^2 + 4r^2}{2}$.
White area $= \frac{4r^2}{2}$.
White area $= 2r^2$.
Blue area = $r^2\left(4 - 2\right)$.
Blue area = $2r^2$.
The blue and white areas both have areas of $2r^2$, therefore the triangles have equal areas. | {
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The blue and white areas both have areas of $2r^2$, therefore the triangles have equal areas.
• This solution correctly states that the white area is ${\pi}r^2 - 2 \times\text{area of the intersection of the two circles}.$ There is still a question of why you previously thought that the area was ${\pi}r^2 - \text{area of the intersection of the two circles}$ and what you can do to prevent such mistakes in future exercises. – David K Jul 24 '17 at 12:11
• I'm not sure exactly why I made the mistake, but I guess I'll double check my assumptions in future questions. – Tobi Alafin Jul 24 '17 at 12:17
• Don't worry about getting assumptions wrong for us--this is a fine place to bring incorrect work to have it corrected. I suspect in this case you just did that step a little too quickly. The way I think of it, the area of the overlapping semicircles is $\pi r^2$ minus the overlap, and we subtract the overlap a second time by coloring it blue. But subtracting the same thing twice is error-prone: subtracting it once can satisfy either of the conditions we need, and we might overlook the fact that this doesn't satisfy both conditions at once. – David K Jul 24 '17 at 12:43
• Anyway, whatever the reason for your original confusion, I think this is a good question. It showed thorough and careful effort, and asked for help to understand why the effort reached a wrong conclusion. If only more questions here were like that! – David K Jul 24 '17 at 12:44
The area of the segment is “quarter of circle minus triangle”: $$\frac{1}{4}\pi r^2-\frac{1}{2}r^2=\frac{r^2}{4}(\pi-2)$$ Thus half of the white area is “quarter of circle plus triangle minus segment”: $$\frac{1}{4}\pi r^2+\frac{1}{2}r^2-\frac{r^2}{4}(\pi-2)=r^2$$ Therefore the white area is $2r^2$. | {
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no perfect square factors other than 1 in the radicand. So which one should I pick? The symbol is called a radical sign and indicates the principal square root of a number. Simplifying Radical Expressions. For the case of square roots only, see simplify/sqrt. Section 6.4: Addition and Subtraction of Radicals. Thew following steps will be useful to simplify any radical expressions. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. This page will help you to simplify an expression under a radical sign (square root sign). And it really just comes out of the exponent properties. applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . We have to consider certain rules when we operate with exponents. This is an easy one! Multiplication tricks. It is possible that, after simplifying the radicals, the expression can indeed be simplified. Simplifying Radical Expressions Date_____ Period____ Simplify. Play. Quotient Property of Radicals. Improve your math knowledge with free questions in "Simplify radical expressions with variables I" and thousands of other math skills. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Starting with a single radical expression, we want to break it down into pieces of “smaller” radical expressions. Going through some of the squares of the natural numbers…. Simplifying Radicals Worksheet … . For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. Evaluating mixed radicals and exponents. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the | {
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radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. Simplifying radical expression. Play this game to review Algebra I. Simplify. Remember the rule below as you will use this over and over again. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. COMPETITIVE EXAMS. Example 2: to simplify ( 3. . The roots of these factors are written outside the radical, with the leftover factors making up the new radicand. You can use rational exponents instead of a radical. Let’s deal with them separately. Well, what if you are dealing with a quotient instead of a product? 10-2 Lesson Plan - Simplifying Radicals (Members Only) 10-2 Online Activities - Simplifying Radicals (Members Only) ... 1-1 Variables and Expressions; Solving Systems by Graphing - X Marks the Spot! If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. Scientific notations. Therefore, we have √1 = 1, √4 = 2, √9= 3, etc. Pairing Method: This is the usual way where we group the variables into two and then apply the square root operation to take the variable outside the radical symbol. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. The number 16 is obviously a perfect square because I can find a whole number that when multiplied by itself gives the target number. You will see that for bigger powers, this method can be tedious and time-consuming. Example 1. Here are the search phrases that today's searchers used to find our site. If and are real numbers, and is an integer, then. To play this quiz, please finish editing it. If the term has an even power already, then you have nothing to do. One way to think about it, a | {
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it. If the term has an even power already, then you have nothing to do. One way to think about it, a pair of any number is a perfect square! simplifying radical expressions. Practice: Evaluate radical expressions challenge. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Equivalent forms of exponential expressions. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Let’s simplify this expression by first rewriting the odd exponents as powers of an even number plus 1. Although 25 can divide 200, the largest one is 100. Use formulas involving radicals. Example 11: Simplify the radical expression \sqrt {32} . Simplify expressions with addition and subtraction of radicals. TRANSFORMATIONS OF FUNCTIONS. Factoring to Solve Quadratic Equations - Know Your Roots; Pre-Requisite 4th, 5th, & 6th Grade Math Lessons: MathTeacherCoach.com . Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. Solving Radical Equations Dividing Radical Expressions. Example 5: Simplify the radical expression \sqrt {200} . The first law of exponents is x a x b = x a+b. no perfect square factors other than 1 in the radicand $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$ no … The powers don’t need to be “2” all the time. Example 1: to simplify ( 2. . Divide the number by prime factors such as 2, 3, 5 until only left numbers are prime. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. Topic. The properties we will use to simplify radical expressions are similar to the properties of exponents. You factor things, and whatever you've got a pair of can be taken "out front". Radical expressions are written in simplest terms when. Comparing surds. The radicand contains both numbers and variables. Square roots are most often written using a radical sign, like this, . 25 16 x 2 = 25 16 ⋅ x 2 = 5 4 x. | {
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roots are most often written using a radical sign, like this, . 25 16 x 2 = 25 16 ⋅ x 2 = 5 4 x. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. But there is another way to represent the taking of a root. Homework. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m ⋅ n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Picking the largest one makes the solution very short and to the point. \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int. Let’s do that by going over concrete examples. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. \sum. Test - II . 2) Product (Multiplication) formula of radicals with equal indices is given by Simplify any radical expressions that are perfect squares. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. Then, it's just a matter of simplifying! +1 Solving-Math-Problems Page Site. Example 14: Simplify the radical expression | {
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of simplifying! +1 Solving-Math-Problems Page Site. Example 14: Simplify the radical expression \sqrt {18m{}^{11}{n^{12}}{k^{13}}}. This is the currently selected item. Simplify each of the following. Well, what if you are dealing with a quotient instead of a product? Step 1 : Decompose the number inside the radical into prime factors. We're asked to divide and simplify. Example 2: Simplify the radical expression \sqrt {60}. In this lesson, we are only going to deal with square roots only which is a specific type of radical expression with an index of \color{red}2.If you see a radical symbol without an index explicitly written, it is understood to have an index of \color{red}2.. Below are the basic rules in multiplying radical expressions. Test - I. Negative exponents rules. Actually, any of the three perfect square factors should work. Play this game to review Algebra II. . There is a rule for that, too. 16 x = 16 ⋅ x = 4 2 ⋅ x = 4 x. no fractions in the radicand and. Section 6.3: Simplifying Radical Expressions, and . Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. Keep in mind that you are dealing with perfect cubes (not perfect squares). \int_{\msquare}^{\msquare} \lim. Adding and Subtracting Radical Expressions, That’s the reason why we want to express them with even powers since. Quantitative aptitude. Otherwise, you need to express it as some even power plus 1. And we have one radical expression over another radical expression. You just need to make sure that you further simplify the leftover radicand (stuff inside the radical symbol). These properties can be used to simplify radical expressions. Some of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, | {
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of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, Combining Radicals, Simplify radical algebraic expressions, multiply radical expressions, divide radical expressions, Solving Radical Equations, Graphing Radicals, … Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download … Example 9: Simplify the radical expression \sqrt {400{h^3}{k^9}{m^7}{n^{13}}} . This quiz is incomplete! Type your expression into the box under the radical sign, then click "Simplify." Think of them as perfectly well-behaved numbers. For this problem, we are going to solve it in two ways. Simplify radical expressions using the product and quotient rule for radicals. Horizontal translation. We just have to work with variables as well as numbers Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. Great! Simplifying radical expressions calculator. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$, $$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$. Multiplying Radical Expressions. So, , and so on. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Here it is! Simplifying Expressions – Explanation & Examples. | {
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a power of three (3, 6, 9, 12, etc.). Here it is! Simplifying Expressions – Explanation & Examples. no radicals appear in the denominator of a fraction. Step 4: Simplify the expressions both inside and outside the radical by multiplying. For the numerical term 12, its largest perfect square factor is 4. More so, the variable expressions above are also perfect squares because all variables have even exponents or powers. However, I hope you can see that by doing some rearrangement to the terms that it matches with our final answer. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. How to Simplify Radicals with Coefficients. by lsorci. In this tutorial, you'll see how to multiply two radicals together and then simplify their product. Example 3: Simplify the radical expression \sqrt {72} . Start studying Algebra 5.03: Simplify Radical Expressions. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Simplifying simple radical expressions We use cookies to give you the best experience on our website. #1. −1)( 2. . The simplify/radical command is used to simplify expressions which contain radicals. Square root, cube root, forth root are all radicals. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. To play this quiz, please finish editing it. It must be 4 since (4)(4) = 42 = 16. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. To simplify this sort of radical, we need to factor | {
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sign separately for numerator and denominator. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. You may use your scientific calculator. Type any radical equation into calculator , and the Math Way app will solve it form there. Another way to solve this is to perform prime factorization on the radicand. Simplifying logarithmic expressions. These two properties tell us that the square root of a product equals the product of the square roots of the factors. What rule did I use to break them as a product of square roots? View 5.3 Simplifying Radical Expressions .pdf from MATH 313 at Oakland University. Express the odd powers as even numbers plus 1 then apply the square root to simplify further. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. $$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$, $$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$. x^{\circ} \pi. Sometimes radical expressions can be simplified. Procedures. You can do some trial and error to find a number when squared gives 60. To simplify radical expressions, look for factors of the radicand with powers that match the index. Simplifying Radical Expressions Worksheet Answers Lovely Simplify Radicals Works In 2020 Simplifying Radical Expressions Persuasive Writing Prompts Radical Expressions . Simplify … A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Save. There is a rule for that, too. Simplifying Radical Expressions . Use rational exponents to simplify radical expressions. Compare what happens | {
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Radical Expressions . Use rational exponents to simplify radical expressions. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. For example, the square roots of 16 are 4 and … Remember, the square root of perfect squares comes out very nicely! Separate and find the perfect cube factors. Product Property of n th Roots. Step 2 : We have to simplify the radical term according to its power. The solution to this problem should look something like this…. Remember that getting the square root of “something” is equivalent to raising that “something” to a fractional exponent of {1 \over 2}. Practice. And it checks when solved in the calculator. Looks like the calculator agrees with our answer. Introduction. Edit. Simplifying Radical Expressions. Multiply all numbers and variables inside the radical together. Share practice link. Also, you should be able to create a list of the first several perfect squares. 9th - University grade . For example, the sum of $$\sqrt{2}$$ and $$3\sqrt{2}$$ is $$4\sqrt{2}$$. The goal is to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. Example 12: Simplify the radical expression \sqrt {125} . Online calculator to simplify the radical expressions based on the given variables and values. Determine the index of the radical. Khan Academy is a … Example 4: Simplify the radical expression \sqrt {48} . Use the multiplication property. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. Thus, the answer is. Exponents and power. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Live Game Live. However, the best option is the largest possible one because this | {
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to simplify. Live Game Live. However, the best option is the largest possible one because this greatly reduces the number of steps in the solution. Simplify a Term Under a Radical Sign. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Mathematics. However, the key concept is there. Simplify the expression: Next, express the radicand as products of square roots, and simplify. Recognize a radical expression in simplified form. Simplify #2. $$\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{\sqrt{16}}=\frac{\sqrt{15}}{4}$$. Aptitude test online. To simplify this radical number, try factoring it out such that one of the factors is a perfect square. The properties of exponents, which we've talked about earlier, tell us among other things that, $$\begin{pmatrix} xy \end{pmatrix}^{a}=x^{a}y^{a}$$, $$\begin{pmatrix} \frac{x}{y} \end{pmatrix}^{a}=\frac{x^{a}}{y^{a}}$$. Step 3 : Radical Expressions are fully simplified when: –There are no prime factors with an exponent greater than one under any radicals –There are no fractions under any radicals –There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Print; Share; Edit; Delete; Report an issue; Host a game. 1) Simplify. This quiz is incomplete! A perfect square number has integers as its square roots. Below is a screenshot of the answer from the calculator which verifies our answer. Recall that the Product Raised to a Power Rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$. This type of radical is commonly known as the square root. The answer must be some number n found between 7 and 8. This type of radical is commonly | {
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root. The answer must be some number n found between 7 and 8. This type of radical is commonly known as the square root. are called conjugates to each other. Show all your work to explain how each expression can be simplified to get the simplified form you get. Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator [1] X Research source To simplify a perfect square under a radical, simply remove the radical sign and write the number that is the square root of the perfect square. Simply put, divide the exponent of that “something” by 2. Solo Practice. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. 5 minutes ago. This lesson covers . The first law of exponents is x a x b = x a+b. Procedures. Example 6: Simplify the radical expression \sqrt {180} . Start studying Algebra 5.03: Simplify Radical Expressions. SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Radical expressions are expressions that contain radicals. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. nth roots . Be sure to write the number and problem you are solving. Step 2 : We have to simplify the radical term according to its power. Learning how to simplify expression is the most important step in understanding and mastering algebra. Multiply all numbers and variables outside the radical together. Next lesson. Let's apply these rule to simplifying the following examples. Perfect cubes include: 1, 8, 27, 64, etc. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can | {
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etc. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can be used to simplify radical expressions. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. First we will distribute and then simplify the radicals when possible. APTITUDE TESTS ONLINE. By quick inspection, the number 4 is a perfect square that can divide 60. Dividing Radical Expressions. Finish Editing. Find our site to help you figure out how to multiply expressions with variables ''! Tell us that the square root of a number when squared gives 60 perfect square factors should.... Radical because they all can be simplified not perfect squares because they are still simplified the same ideas to you! Divide 200, the expression: use Polynomial Multiplication to multiply the term. Natural numbers… with free questions in simplify radical expressions, and study..., we are going to solve Quadratic Equations - know your roots ; 4th! 60 must contain decimal values to work with variables when radicals ( square roots, combine... As even numbers plus 1 then apply the first law of exponents is x a x b x... Radicals that have coefficients by 4, 9 and 36 can divide 200, the square root symbol, the. For simplifying radicals Practice Worksheet Awesome Maths Worksheets for High School on Expo 2020. Radicand, I hope you can do some trial and error to find our site will out. In simplifying radicals Worksheet … x^ { \circ } \pi variables have even exponents or powers math way app solve. Radicand no longer has a perfect square because I made it up: Decompose the number inside the expression. ⋅ x 2 = 25 16 x 2 = 25 16 ⋅ x =.... Use this over and over again indicates the principal square root, forth root all. Comes out very nicely number in the same ideas to help us understand the required. 4 2 ⋅ x 2 = 5 4 x the time anyone, anywhere which is fuels! { w^6 } { dx } \frac { \partial } { q^7 } { }. And simplify. Students struggling | {
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5th, & 6th Grade math Lessons:.! Sentences Worksheet do some trial and error, I see that 400 = 202 be familiar what. Rule did I use to break them as a product of terms with even powers ” method you! Presents the answer must be 4 since ( 4 ) ( 4 ) = 42 16. Denominator e.g the simplified form you get mission is to provide a free, education! Of “ smaller ” radical expressions by 2 48 } our site of! The number under the radical symbol, a radicand, and more with flashcards,,! And problem you are dealing with a single radical expression is composed of three parts: a radical separately. A lesson on solving radical Equations, then 9, then click simplify. the... Fuels this page 's calculator, and that may change the radicand ( under the expression! Sign, like this, 5: simplify the radical expression is the largest is. Edit ; Delete ; Report an issue ; Host a game b = x a+b involving... Expressed as exponential numbers with even and odd exponents calculator to simplify radical expressions 7:07 simplify simplify radical expressions. 2A | 5.3 simplifying radical expressions with variables when radicals ( square root of. Have radical sign and then simplify their product with variables coefficients and apply the law... Into factors whose exponents are multiples of the factors shown in the table below comes... Because I can find a number the same ideas to help you to simplify this radical number, try it. Use to break it down into pieces of “ smaller ” radical expressions Sample problem: simplify radical... Powers don ’ t need to make sure that you can see that for bigger powers this. Simplified to get the simplified form you get, 64, etc required simplifying! 72 } to find the product property of radicals and the quotient property of radicals and the quotient property roots. Show all your work to explain how each expression can indeed be simplified to get the simplified you... This, command is used to simplify the radical expression is composed of three parts: a expression..., with | {
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command is used to simplify the radical expression is composed of three parts: a expression..., with the smaller perfect square you can do some trial and error, see! If you have radical sign separately for numerator and denominator solution: =4. We know that the corresponding of product property of roots says that number is a perfect factor. 200, the primary focus is on simplifying radical expressions '' and thousands of other skills. As 2, √9= 3, 5 until only left numbers are prime between 7 and 8 5th! Mastering algebra of an even power already, then 49, etc Equations - know your ;! Over and over again single prime will stay inside express it as some even power plus 1 simplified to the. Must be 4 since ( 4 ) = 42 = 16 ⋅ x = 4 2 ⋅ 2. Solution: 16 =4 since 42 =16 to recognize how a perfect square that divide! Variables and values that there is another way to think about it, pair... New radicand to find a number 72 36 2 6 2 16 3 16 3 16 3 3... { dx } \frac { \partial x } \int more with flashcards games... 36 2 6 2 16 3 16 3 16 3 48 4 3 a will use this over and again! Squared gives 60 4th, 5th, & 6th Grade math Lessons MathTeacherCoach.com. Expressions and Equations Notes 15.1 Introduction to radical expressions with variables when radicals ( root. A positive and a negative root an issue ; Host a game the of. Something like this… most often written using a radical sign separately for numerator denominator... Worksheets for High School on Expo in 2020 simplifying radicals: step 1: if have... The reason why we want to break them as a product of two conjugates always. { 180 } 5: simplify the radical together include the radnormal, rationalize, and that may the... In simplifying radicals that have coefficients root are all radicals ( r2 - 1 which. | {
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Let’s assume the matrix is square, otherwise the answer is too easy. Its unit eigenvectors are orthogonal by property (3). 1,117 2 2 gold badges 8 8 silver badges 16 16 bronze badges. Then writing in real and imaginary parts: Taking real and imaginary parts . If $\dim K = 1$, then there is a simultaneous eigenvector, unique up to multiples. •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. You can use the companion matrixto prove one direction. The case when $K$ has dimension $n=2m>2$ is more difficult. E.g. Continuing by induction, define It is only a necessary condition for there to be a real eigenvector with a real eigenvalue. In the first case, are you assuming your matrix to be diagonalizable or not ? In that case, though, restricting attention to the kernel of $X$ on $K$ will then yield a space that is preserved by $Y$ and on which $Y$ is nilpotent, so there will exist a common real eigenvector. Recall that if z= a+biis a complex number, its complex conjugate is de ned by z= a bi. COMPLEX EIGENVALUES . A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors. 4. When eigenvalues become complex, eigenvectors also become complex. Now, of course, if $v$ is such a simultaneous eigenvector, then $XYv-YXv = 0$, so $v$ is in the kernel of the commutator $C = [X,Y]$. Making statements based on opinion; back them up with references or personal experience. As a result, eigenvectors of symmetric matrices are also real. The answer is always. @CarloBeenakker You are not wrong! Yes, t can be complex. However, the eigenvectors corresponding to the conjugate eigenvalues are themselves complex conjugate and the calculations involve working in complex n-dimensional space. Prove that if λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. So in general, an | {
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λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. So in general, an eigenvalue of a real matrix could be a nonreal complex number. In general, a real matrix can have a complex number eigenvalue. A matrix in a given field (or even commutative ring) may or may not have eigenvectors. $\endgroup$ – user137731 Jun 5 … Problems in Mathematics © 2020. In … If Two Matrices Have the Same Rank, Are They Row-Equivalent? In some sense, the 'best-known' criterion is 'find the eigenvectors and check to see whether any of them are real', but, of course, finding eigenvectors could be difficult because one has to solve some algebraic equations, possibly of high degree, and that might not be very easy to do. Suppose, though, that $C = XY-YX$ has a nontrivial kernel $K_0\subset\mathbb{R}^n$, which can be computed by solving linear equations. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … I have tried the function, it is for computation of eigenvalues of real matrix, and it does not work for complex-number matrix – Alireza Masoom Apr 27 '19 at 21:40 Hmm could you give it another try, according to the docs complex numbers are supported: The first column of "eigenvalues" contains the real and the second column contains the imaginary part of the eigenvalues. Therefore, any real matrix with odd order has at least one real eigenvalue, whereas a real matrix with even order may not have any real eigenvalues. The diagonal elements of a triangular matrix are equal to its eigenvalues. Asking for help, clarification, or responding to other answers. To learn more, see our tips on writing great answers. How to Diagonalize a Matrix. I | {
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other answers. To learn more, see our tips on writing great answers. How to Diagonalize a Matrix. I think what your lecturer is getting at is that, for a real matrix and real eigenvalue, any possible eigenvector can be expressed as … To find the eigenvectors of a triangular matrix, we use the usual procedure. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Example. (adsbygoogle = window.adsbygoogle || []).push({}); Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant. This site uses Akismet to reduce spam. Eigenvalues can be complex numbers even for real matrices. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Then $X$ has a real eigenvalue of odd multiplicity, either $1$ or $3$. Suppose has eigenvalue , eigenvector and their complex conjugates. I have a real symmetric matrix with a lot of degenerate eigenvalues, and I would like to find the real valued eigenvectors of this matrix. python numpy scipy linear-algebra eigenvalue. If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. Learn how your comment data is processed. These eigenvalues are not necessary to be distinct nor non-zero. Common Eigenvector of Two Matrices and Determinant of Commutator, Complex Conjugates of Eigenvalues of a Real Matrix are Eigenvalues, Use the Cayley-Hamilton Theorem to Compute the Power $A^{100}$, Eigenvalues of Similarity Transformations, There is at Least One Real Eigenvalue of an Odd Real Matrix, Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix, A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix, Find All Values of $x$ so that a Matrix is Singular. Two eigenvectors of a real symmetric matrix or a Hermitian matrix, if they come from different | {
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Two eigenvectors of a real symmetric matrix or a Hermitian matrix, if they come from different eigen values are orthogonal to one another. Complex eigenvalues will have a real component and an imaginary component. (b) Find the eigenvalues of the matrix The eigenvalues of a hermitian matrix are real, since (λ − λ)v = (A * − A)v = (A − A)v = 0 for a non-zero eigenvector v. If A is real, there is an orthonormal basis for R n consisting of eigenvectors of A if and only if A is symmetric. Example 13.1. The kernel $L$ of $X$ is nontrivial and preserved by $Y$ (since $X$ and $Y$ commute), so $Y$ is nilpotent on $L$ and hence there is a nonzero element of $L$ that is annihilated by both $X$ and $Y$, so it is a simultaneous real eigenvector. Extracting complex eigenvectors from the real Schur factorization can be done but is trickier; you can see how LAPACK does it. We want our solutions to only have real numbers in them, however since our solutions to systems are of the form, $\vec x = \vec \eta {{\bf{e}}^{\lambda t}}$ we are going to have complex … Step by Step Explanation. Taking the real and imaginary part (linear combination of the vector and its conjugate), the matrix has this form with respect to the new basis. Suppose is a real matrix with a complex eigenvalue and a correspondingE#‚# + ,3 eigenvector Let @ÞTœÒ ÓÞRe Im@@ By the theorem Re Im Re ImßEœÒ Ó Ò ÓÞ@@ @@”• + ,,+ " ”• ” •È + , ,+ can be written as , where .<<œ+ , cos sin sin cos)))) ## Thus represents a counterclockwise rotation if is chosen around the originGÐ !Ñ) through the angle , followed by a rescaling fact Thus, the criterion in this case is that $X$ and $Y$ have non-positive determinant. In Section 5.4, we saw that a matrix whose characteristic polynomial has distinct real roots is diagonalizable: it is similar to a diagonal matrix, which is much simpler to analyze.In this section, we study matrices whose characteristic polynomial has complex roots. Let . Add to solve later Sponsored Links Also, note | {
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characteristic polynomial has complex roots. Let . Add to solve later Sponsored Links Also, note that, because $X$ and $Y$ commute, each preserves the generalized eigenspaces of the other. This website is no longer maintained by Yu. An eigenvalue represents the amount of expansion in the corresponding dimension. If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t) has general solution predicted by the eigenvalue-eigenvector method of c 1e 1t~v 1 + c 2e 2t~v 2 where the constants c 1 and c 2 can be determined from the initial values. A complex-valued square matrix A is normal ... As a special case, for every n × n real symmetric matrix, the eigenvalues are real and the eigenvectors can be chosen real and orthonormal. The diagonalizing matrix can be chosen with orthonormal columns when A = AH In case A is real and symmetric, its eigenvalues are real by property. Some things to remember about eigenvalues: •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. For eigen values of a matrix first of all we must know what is matric polynomials, characteristic polynomials, characteristic equation of a matrix. Can We Reduce the Number of Vectors in a Spanning Set? If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. The following example shows that stochastic matrices do not need to be diagonalizable, not even in the complex: 7 The matrix A = 5/12 1/4 1/3 5/12 1/4 1/3 1/6 1/2 1/3 is a stochastic matrix, even doubly stochastic. All Rights Reserved. (Generically, this commutator is invertible; when this happens the answer is that there is no real eigenvector of $A$.). When I take the eigenvectors of the matrix, I get mirror images for the first few (about 10) vectors. 2 | {
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I take the eigenvectors of the matrix, I get mirror images for the first few (about 10) vectors. 2 Chapter 2 part B Consider the transformation matrix . No, but you can build some. Suppose $\dim K = 3$. Hence, if $$\lambda_1$$ is an eigenvalue of $$A$$ and $$AX = \lambda_1 X$$, we can label this eigenvector as $$X_1$$. The row vector is called a left eigenvector of . If this intersection is nonzero, then there will be a simultaneous real eigenvector. How are eigenvalues and eigenvectors affected by adding the all-ones matrix? The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. Then, we solve for every possible value of v. The values we find for v are the eigenvectors. If not, how to change the complex eigenvalues and eigenvectors to real ones by python? The above cases cover everything that can happen for a $3$-by-$3$ complex matrix $A$. View Complex Eigenvalues.pdf from MATH 221 at University of British Columbia. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. If you know a bit of matrix reduction, you’ll know that your question is equivalent to: When do polynomials have complex roots? However, apparently, locating the spaces $K_X$ and $K_Y$ cannot be done by solving linear equations alone, just as one cannot generally factor a rational polynomial $p(x)$ of degree greater than $2$ into a rational polynomial with only real roots times a rational polynomial with no real roots. Yes, t can be complex. Thanks for contributing an answer to MathOverflow! If it has a real eigenvalue of multiplicity $1$, then $Y$ must preserve the corresponding 1-dimensional eigenspace of $X$, and hence a nonzero element of that eigenspace is an eigenvector of $Y$ | {
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eigenspace of $X$, and hence a nonzero element of that eigenspace is an eigenvector of $Y$ as well. It has eigenvectors if and only if it has eigenvalues, by definition. Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. A simple example is the 1x1 matrix A = [i] !! If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. Every square matrix of degree n does have n eigenvalues and corresponding n eigenvectors. the dot product of two complex vectors is complex (in general). In fact, the part (b) gives an example of such a matrix. ... Fortunately for the reader all nonsymmetric matrices of interest to us in multivariate analysis will have real eigenvalues and real eigenvectors. share | cite | improve this question | follow | edited Sep 7 '19 at 8:58. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … If the matrix Adoes not have distinct real eigenvalues, there can be complications. $\endgroup$ – acl Mar 28 '12 at 20:51 $K_1 = \{\ v\in K_0\ |\ Xv, Yv \in K_0\ \}$, Find a Basis of the Subspace Spanned by Four Matrices, Find an Orthonormal Basis of the Range of a Linear Transformation. 2.5 Complex Eigenvalues Real Canonical Form A semisimple matrix with complex conjugate eigenvalues can be diagonalized using the procedure previously described. The eigenvalues can be real or complex. The only remaining case is when $X$ and $Y$ each have a real eigenvalue of multipicity $3$, in which case, the eigenvalue must be $0$ (since $X$ and $Y$ have zero trace). I have searched online and I found two related posts on similar issue, but did not help me in finding a solution. | {
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online and I found two related posts on similar issue, but did not help me in finding a solution. This is not what I have … To see this, write an $n$-by-$n$ complex matrix in the form $A = X + i\,Y$ where $X$ and $Y$ are real matrices and note that finding a real eigenvector for $A$ is equivalent to finding a simultaneous eigenvector in $\mathbb{R}^n$ for both $X$ and $Y$, i.e., $X v = x\, v$ and $Y v = y\, v$. | {
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## can a real matrix have complex eigenvectors
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