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Dieter
10-02-2016, 04:48 PM
Post: #12
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
(10-02-2016 03:29 PM)Dieter Wrote:
(09-30-2016 06:00 AM)I Wrote: The test should compare the last and previous improved approximations, cf. the posted VBA code. In your program that's the current and previous value of Y.
It looks like even this remaining error can be estimated and an even more accurate result can be calculated from the last two improved values. Here is a more structured experimental version.
Code:
Function Simpson(a, b, Optional errlimit = 0.000001, Optional nmax = 3000) fab = f(a) + f(b) s2 = 0 s4 = 0 simp_new = 0 improved_new = 0 n = 2 Do h = (b - a) / n s2 = s2 + s4 s4 = 0 For i = 1 To n - 1 Step 2 s4 = s4 + f(a + i * h) Next simp_old = simp_new improved_old = improved_new simp_new = (fab + 2 * s2 + 4 * s4) * h / 3 If n = 2 Then improved_new = simp_new converged = False Else improved_new = (16 * simp_new - simp_old) / 15 converged = Abs(improved_new - improved_old) < 63 * errlimit End If n = 2 * n Loop Until converged Or n > nmax Simpson = (64 * improved_new - improved_old) / 63 End Function Function f(x) f = 1 / x End Function
What do you think?
The following diagram may give an impression for the integral of f(x) = 1/x from a=1 to b=2. Here X is the number of iterations (i.e. n=2x intervals) and Y is the accuracy, i.e. the number of valid digits. The red line is the classic Simpson method, the blue one represents the "improved" method, and finally the green graph shows the further extrapolated result as shown in the last program.
Dieter | {
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Dieter
You latest code works in general better than Pekis' version. The ratio of errors generated by your code to that of Pekis range in the orders [1, 10]. Good work!
10-09-2016, 03:14 AM (This post was last modified: 10-09-2016 03:15 AM by Namir.)
Post: #13
Namir Senior Member Posts: 813 Joined: Dec 2013
RE: Simpson revival
Thank you Pekis for your code. I had fun translating it into programs for the HP-71B, HP-85B, HP-41C and HP-67. I like your approach because it gives accurate results. Of course, the calculations take more time since the program repeats the integral calculations at ever decreasing increment values.
Namir
07-31-2018, 02:57 PM (This post was last modified: 07-31-2018 06:42 PM by Albert Chan.)
Post: #14
Albert Chan Senior Member Posts: 1,676 Joined: Jul 2018
RE: Simpson revival
(09-28-2016 03:50 PM)Namir Wrote: Version 1 gave better results. These results seem counter intuitive, since I expected version 2 to give a better final result.
Any or all improvements and corrections are welcome.
Hi, Namir
From the book SICP, I think I know why your version 2 does not give expected better estimate.
The trick is not just apply correction based on 2 iterations, but from all previous iterations.
(09-28-2016 10:00 PM)Dieter Wrote: Let f(x) = 1/x and integrate it from 1 to 2.
The exact result, rounded to 12 digits, is ln(2)=0,693147180560.
Code:
n Standard Simpson Error Improved Error ---------------------------------------------------------------------- 2 0,694444444444 1,3 E-03 4 0,693253968254 1,1 E-04 0,693174603175 2,7 E-05 8 0,693154530655 7,4 E-06 0,693147901481 7,2 E-07 16 0,693147652819 4,7 E-07 0,693147194297 1,4 E-08 32 0,693147210290 3,0 E-08 0,693147180788 2,3 E-10 64 0,693147182421 1,9 E-09 0,693147180564 3,6 E-12 128 0,693147180676 1,2 E-10 0,693147180560 5,6 E-14 | {
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The "improved" column holds the extrapolated results, i.e. [16*Simpson(a, b, n) – Simpson(a, b, n/2)] / 15.
The first estimate is correct: T1 + (T1-T0)/(16-1) = 0.693174603175
However, second estimate should apply correction again, from previous estimate.
T1 + (T1-T0)/(64-1) = 0.693147901481 + (-2.67017e-05)/63 = 0.693147477645
Third estimate need 3 corrections (/15, /63, /255), ... This is the revised table.
Code:
n Standard Simpson Error Improved Error ---------------------------------------------------------------------- 2 0,694444444444 1,3 E-03 4 0,693253968254 1,1 E-04 0,693174603175 2,7 E-05 8 0,693154530655 7,4 E-06 0,693147477645 3,0 E-07 16 0,693147652819 4,7 E-07 0,693147181917 1,4 E-09 32 0,693147210290 3,0 E-08 0,693147180562 2,4 E-12 64 0,693147182421 1,9 E-09 0,693147180560 1,6 E-15
Edit:
After reading the whole thread, it seems the code is not trying to use Romberg's method.
All it wanted was the first approximation of Romberg's extrapolation.
Sorry for the noise.
Anyway, this is what Romberg's method would look like ...
08-04-2018, 05:29 PM
Post: #15
Albert Chan Senior Member Posts: 1,676 Joined: Jul 2018
RE: Simpson revival
In Python, Romberg's method corrections can be simplified with generator.
Code:
from __future__ import division def simpson(f, a, b, n=2): # Simpson generator h = (b-a) / 2 t0 = (f(a)+f(b)) * h while 1: s = sum(f(a + i*h) for i in xrange(1, n, 2)) t1 = t0/2 + s*h # simpson's partial sum yield t1 + (t1-t0)/3 # simpson's formula t0, n, h = t1, 2*n, h/2
Code:
def _romberg(gen, y, k): # Romberg generator for z in gen: yield z + (z-y)/k y = z | {
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Code:
def romberg(f, a, b, verbal=False, k=15): "Romberg's Method integration" gen = simpson(f, a, b, n=2) x = gen.next() while 1: y = gen.next() x = y + (y-x) / k if verbal: print x if x == y: return x # stop if converged gen = _romberg(gen, y, k) # Romberg corrections k = 4 * k + 3
Redo previous post example:
>>> romberg(lambda x: 1/x, 1, 2, verbal=True)
0.693174603175
0.693147477645
0.693147181917
0.693147180562
0.69314718056
0.69314718056
0.69314718055994506
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# Fast Way of Finding The Remainder
I have the following question:
Find the remainder of $29\times 2901\times 2017$ divided by $17$
I already have the answer (7) for this problem. I solved it using the long way by multiplying all of the numbers then divvy them with 17. I am just thinking if there are any fast way of solving this. My solution takes a long time.
Hint:
If $$a\equiv b\pmod c$$ And $$d\equiv e\pmod c$$ You can multiply the two to get $$a\times d\equiv b\times e\pmod c$$
Doing this for the numbers separately, then multiplying the expressions will get you the solution very quickly.
• See here for this Congruence Product Rule. But here we can also exploit radix rep to mod out chunks of digits, which greatly simplifies the computation - see my answer. – Bill Dubuque Jan 29 '17 at 19:52
You can write each number as a multiple of $17$ plus a small remainder:
$$(17+12)(170\cdot 17+11)(118\cdot 17+11).$$
When you multiply this out (which you don't have to do) every term is a multiple of $17$ except the last one $12\cdot 11\cdot 11$. So you need consider only this last product. We can repeat what we just did with a little factoring. The above $= 3\cdot 11 \cdot 4 \cdot 11 = 33\cdot 44 = (17+16)(2\cdot 17 + 10).$ So you need consider only $16\cdot 10$.
• $(-5)\cdot(-6)\cdot(-6)$ is easier to compute. – tomasz Jan 29 '17 at 18:09
• @tomasz Yes, but from the question, I don't think the OP knows about modular arithmetic yet, so dealing with a negative remainder is an extra complication. Although I see a "mod" answer was accepted as best, so maybe I'm wrong. – B. Goddard Jan 29 '17 at 18:23
Find remainder after dividing $29 \times 2901 \times 2017$ by $17$.
Work $\mod 17$ !
\begin{align} & 29 \times 2901 \times 2017 \\ & 12 \times 1201 \times 317 \\ & 12 \times 1201 \times 147 \\ & 12 \times 351 \times 62 \\ & 12 \times 11 \times 11 \\ & 1452 \\ & 602 \\ & 92 \\ & 7. \end{align} | {
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• This is much easier if one works with least magnitude reps (i.e. allow negative remainders), e.g. see my answer. – Bill Dubuque Jan 29 '17 at 19:54
This can be done in $$15$$ seconds of purely mental modular arithmetic of small numbers using the universal divisibility test, which is essentially the division algorithm ignoring quotients. To obtain optimal speedup we use least magnitude remainders, e.g. $$-1$$ vs. $$16\pmod{\!17}$$ since doing so simplifies subsequent arithmetic. To reduce a decimal number mod $$n$$ we continually mod out the leading chunks of its digits. Since we allow negative remainders, we will encounter negative digits, delimited by a comma, e.g. $$\,a,b := a(10)\!+\!b.\,$$ We prove $$\ 3247\equiv 0\pmod{\!17}\,$$ for practice.
\begin{align}{\rm mod}\ 17\!:\qquad &\,\ \color{#90f}{32}\ 47\\ \equiv\ &{\color{#90f}{-2}},\color{#0a0}47 \ \ \ \text{by }\ \ \ \ \ \,\color{#90f}{32}\,\equiv\,\color{#90f}{-2} \\ \equiv\ &\quad\ \ \, \color{#f84}{\bf 1}7\ \ \ \text{by }\ {\color{#90f}{-2}},\color{#0a0}4 \equiv\, \color{#90f}{{-}2}(10)\!+\!\color{#0a0}4\equiv -16\equiv \color{#f84}{\bf 1} \\[-.3em] \text{Let's do the number in the OP}\qquad\ \ \ \ \\[-.3em] &\,\ \color{#90f}{29}\ 01\\ \equiv\ & {\color{#90f}{-5}},\color{#0a0}01\ \ \text{ by }\quad \color{#90f}{29\,\equiv\, -5} \\ \equiv\ &\quad\ \ \color{#f84}{\bf 1}1\ \ \, \text{ by }\ \color{#90f}{{-}5},\color{#0a0}0\equiv {\color{#90f}{-5}(10)\!+\!\color{#0a0}0}\equiv -50\equiv\color{#f84}{\bf 1}\\ \equiv\ &\quad \,\color{#08f}{-6}\\[-.2em] \text{Similarly \,2017\equiv\color{#c00}{-6}\ so we have}\phantom{MM}\\[-.2em] &\ 29\cdot 2901\cdot 2017\\ \equiv\ &(-5)(\color{#08f}{-6})(\color{#c00}{-6})\\ \equiv\ &(-5)\ \color{#08f} 2\\ \equiv\ &\ 7 \end{align}\qquad\qquad
where we have applied the Congruence Product and Sum Rules many times above. | {
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where we have applied the Congruence Product and Sum Rules many times above.
Remark I wrote every little detail above to help avoid confusion with negative digits. Once one gains proficiency there is no need to be so extremely verbose. See here for a larger example which also employs negative digits for optimization.
• (+1) The OP did say "fast way" of finding the remainder, this is it! – Silverfish Jan 29 '17 at 23:42
29 × 2901 × 2017 $\equiv$ 12 × 11 × 11 (mod 17)
12 × 121 $\equiv$ (mod 17)
12 × 2 $\equiv$ (mod 17)
24 $\equiv$ 7 (mod 17)
• Even faster if you use negative modulos to keep the numbers small, so in this case, the first line would be $(-5)*(-6)*(-6)$ – WorldSEnder Jan 29 '17 at 17:49
• Ok I keep in mind that. – Kanwaljit Singh Jan 29 '17 at 17:53
• @KanwaljitSingh: $(-a) \equiv (b - a) (\mathrm{mod}\, b)$ – Kevin Jan 29 '17 at 19:31
The numbers are given in decimal representation, therefore I would start by simplifying $100 \bmod 17$. We have that $20\equiv 3$, therefore $100=20\times 5\equiv 3\times 5 = 15\equiv -2$. This allows us to write:
$$2900\equiv -2\times 29\equiv -2\times (-5) = 10$$
And
$$2000\equiv -2\times 20\equiv-2\times 3 = -6$$
We thus have:
$$29\times 2901\times 2017\equiv -5\times 11\times (-6) = 30\times 11\equiv -4\times 11\equiv -4\times (-6) = 24\equiv 7$$ | {
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# When Bubble Sort to sort a list of numbers 7, 12, 5, 22, 13, 32
#### Joystar1977
##### Active member
Can somebody tell me which example is right when a question that is given to me says to bubble sort a list of numbers 7, 12, 5, 22, 13, 32? I found two examples and one was with a graph that included Original List, Pass 1, Pass 2, Pass 3, Pass 4, Pass, 5, and Pass 6, the numbers with 7 on one line, 12 on another line, 5 on another line, 22 on another line, 13 on another line, and 32 on another line which includes Comparisons, and Swaps. The second example was to do the following:
First Pass:
(7, 12, 5, 22, 13, 32)- Here, algorithm compares the first two elements, and swaps since 7 <12.
(7, 12, 5, 22, 13, 32)- Swap since 12 > 5.
(7, 12, 5, 22, 13, 32)- Swap since 5 < 22.
(7, 12, 5, 22, 13, 32)- Swap since 22 > 13.
(7, 12, 5, 22, 13, 32)- Now, since these elements are already in order (32 > 13), algorithm does not swap them.
Second Pass and Third Pass is done the same way.
Can someone please tell me which example is correct when bubble sorting the list of numbers?
#### Joystar1977
##### Active member
Discrete Mathematics (Bubble Sort to sort the list)
Is it true that the number 32 is definitely in its correct position at the end of the first pass when it comes to the following numbers: 7, 12, 5, 22, 13, 32?
#### Joystar1977
##### Active member
Discrete Mathematics Numbers 7, 12, 5, 22, 13, 32
How does the number of comparisons required change as the pass number increases?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Re: Discrete Mathematics Numbers 7, 12, 5, 22, 13, 32 | {
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##### Well-known member
MHB Math Scholar
Re: Discrete Mathematics Numbers 7, 12, 5, 22, 13, 32
Can somebody tell me which example is right when a question that is given to me says to bubble sort a list of numbers 7, 12, 5, 22, 13, 32? I found two examples and one was with a graph that included Original List, Pass 1, Pass 2, Pass 3, Pass 4, Pass, 5, and Pass 6, the numbers with 7 on one line, 12 on another line, 5 on another line, 22 on another line, 13 on another line, and 32 on another line which includes Comparisons, and Swaps. The second example was to do the following:
First Pass:
(7, 12, 5, 22, 13, 32)- Here, algorithm compares the first two elements, and swaps since 7 <12.
(7, 12, 5, 22, 13, 32)- Swap since 12 > 5.
(7, 12, 5, 22, 13, 32)- Swap since 5 < 22.
(7, 12, 5, 22, 13, 32)- Swap since 22 > 13.
(7, 12, 5, 22, 13, 32)- Now, since these elements are already in order (32 > 13), algorithm does not swap them.
Second Pass and Third Pass is done the same way.
Can someone please tell me which example is correct when bubble sorting the list of numbers?
I did not understand the description of the graph very well. The second example is incorrect. First, the swaps are not shown: all lines are the same. Second, if the sorting is done in increasing order, then there is no reason to swap 7 and 12: they are already ordered correctly.
The list during the first pass should change as follows. (Compared elements are underlined.)
\begin{array}{rrrrrr} \underline{7} &\underline{12}& 5& 22& 13& 32\\ 7& \underline{12}& \underline{5}& 22& 13& 32\\ 7& 5& \underline{12}& \underline{22}& 13& 32\\ 7& 5& 12& \underline{22}& \underline{13}& 32\\ 7& 5& 12& 13& \underline{22}& \underline{32}\\ 7& 5& 12& 13& 22& 32\end{array}
Is it true that the number 32 is definitely in its correct position at the end of the first pass when it comes to the following numbers: 7, 12, 5, 22, 13, 32?
Yes, if the sorting is done in increasing order. | {
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How does the number of comparisons required change as the pass number increases?
In the unoptimized bubble sort, the number of comparisons is the same for all passes. In the optimized sort, the number decreases by 1 from one pass to the next because numbers at the end of the array are in their correct positions. Thus, for an array of $n$ numbers the first pass requires $n-1$ comparisons, the second $n-2$ comparisons, and so on.
#### Joystar1977
##### Active member
Thank you Evgeny.Makarov! I seen two different examples of Bubble Sorting and was trying to see which is done correctly. I wanted to just clarify which one I am supposed to use when Bubble Sorting. Am I correct then that its easier to see things on a graph for Bubble Sorting such as having the Original List of numbers with 7 on one line, 12 on another line, 5 on another line, 22 on another line, 13 on another line, and 32on another line, Pass 1, Pass, 2, Pass 3, Pass 4, Pass 5, Pass 6, Comparisons, and Swaps? Just trying to get a thorough understanding of this.
#### Evgeny.Makarov | {
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#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Am I correct then that its easier to see things on a graph for Bubble Sorting such as having the Original List of numbers with 7 on one line, 12 on another line, 5 on another line, 22 on another line, 13 on another line, and 32on another line, Pass 1, Pass, 2, Pass 3, Pass 4, Pass 5, Pass 6, Comparisons, and Swaps? Just trying to get a thorough understanding of this.
I am having trouble understanding this description of a graph, so I can't say whether it makes sense. Execution of Bubble sort can be illustrated in many ways. The most natural, I think, is to show the array after each comparison and potential swap. I showed the first pass in this way in my previous post. For another illustration, see Wikipedia. Note that while there are many ways to illustrate the execution of the Bubble sort, the algorithm itself works in a unique, fixed way. It is this way of working that is important, not whether it is illustrated as a graph or as a sequence of lines.
#### Joystar1977
##### Active member
Thanks for the explanation Evgeny.Makarov! | {
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# Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$
Evaluating
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$$
Its $\frac{\infty}{\infty}$ so I should use L Hospital rule, but the terms are exponential and differentiation won't do much good? I am thinking maybe I somehow use $\ln$ both sides? But how? Or perhaps I should do something else?
UPDATE
Following @Paul's answer: Since $|r| < 1$ so sequence is convergent. So I use the formula
$$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$$
$$\frac{1}{10} \cdot \frac{1}{1-\frac{4}{5}} = \frac{1}{10}\cdot 5 = \frac{1}{2}$$
But answer is $\frac{2}{5}$
-
L'Hopital's rule is for evaluating limits, not evaluating series. – JavaMan Feb 27 '12 at 13:00
What's the point of your update? If you follow Paul's answer, you'll have it. The first term is "$\frac{4}{5}$", not $1$. – Gigili Feb 27 '12 at 13:11
@Jiew Meng: Compare the series $\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n=\sum_{n=1}^\infty \frac{4}{5}\left(\frac{4}{5}\right)^{n-1}$ with $\sum_{n=1}^\infty ar^{n-1}$, we have $a=\frac{4}{5}$ and $r=\frac{4}{5}$. Therefore, using the formula you have, we have $\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n=\frac{\frac{4}{5}}{1-\frac{4}{5}}=4$. – Paul Feb 27 '12 at 13:16
@CliveNewstead: That's what I said! – Gigili Feb 27 '12 at 13:18
@Gigili: I know, but Jiew Meng either hadn't read or hadn't understood it, so I was reiterating. – Clive Newstead Feb 27 '12 at 14:23
You can rewrite it as $$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}=\sum_{n=1}^\infty \frac{\frac{1}{2}\cdot2^{2n}}{5\cdot 5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{2^{2n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{4^{n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n$$ which is a geometric series with ratio $r=\displaystyle\frac{4}{5}$. I think you can finish it from here. | {
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I think there is a $\frac{1}{10}$ too in the first step. – Riccardo.Alestra Feb 27 '12 at 12:36
Actually shouldn't the $\frac{1}{10}$ appear in step 3? – Jiew Meng Feb 27 '12 at 12:44
Btw, I also updated my question – Jiew Meng Feb 27 '12 at 12:58
Sorry, that's a typo. $\frac{1}{10}$ shouldn't appear until the second equality. – Paul Feb 27 '12 at 13:13
This is a series, not a sequence. L'Hopital's rule is used for the latter, not the former.
When working with series, it is a good idea to first determine if the the given series is of a certain known type. One type is a series of the form $$\sum\limits_{n=m}^\infty a r^{n+k} =ar^{m+k} + ar^{m+k+1}++ ar^{m+k+2}+\cdots.$$
These are called geometric series and the quantity $r$ is called the ratio of the series.
A geometric series converges if and only if $|r|<1$. When the series converges, it converges to to the first term of the series divided by the quantity $(1-r)$. The first term of the series above is obtained when you take $n=m$: $ar^{m+k}$.
I would not suggest you use a formula to find the sum of the series, but rather do the following:
1. Identify the ratio $r$.
2. If $|r|<1$, the series converges to $\text{the first term of the series}\over 1-r$. If $|r|\ge1$, the series diverges.
In your case $$\sum_{n=1}^\infty { {2^{2n-1}\over 5^{n+1}} }=\sum_{n=1}^\infty\textstyle {1\over 10}({4\over5})^n$$ This series is geometric with $r=4/5$. The first term of the series is ${1\over10}\cdot{4\over5}$ (obtained by setting $n=1$). So, the series converges and $$\sum_{n=1}^\infty{\textstyle {1\over 10}({4\over5})^n} = {{1\over10}\cdot{4\over5}\over 1-{4\over5}} ={{1\over10}\cdot{4\over5}\over {1\over5}}={2\over5}.$$
Another example: $$\sum_{n=2}^\infty 4\cdot (-1/3)^{n+5}\quad \buildrel {r=-1/3}\over{ = }\quad {4\cdot(-1/3)^7\over 1-(-1/3)} ={-4/3^7\over 5/3}.$$ Here, the first term of the series is obtained when $n=2$: $4\cdot(-1/3)^{2+5}=4\cdot(-1/3)^{7}$. | {
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And one more: $$\sum_{n=0}^\infty ( 1/3)^{n }\quad \buildrel {r= 1/3}\over{ = }\quad { ( 1/3)^0\over 1-(1/3)} ={1\over 2/3}={3\over2}.$$
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I don't see what special points you added that Paul didn't mention in his answer or it's not mentioned in comments. – Gigili Feb 27 '12 at 13:56
@Gigili Mostly how to actually find the sum of a geometric series. It seemed to OP had problems using the formula. – David Mitra Feb 27 '12 at 13:59
@Gigli Oh, I didn't see the hidden comments... – David Mitra Feb 27 '12 at 14:00
I think the OP just made a mistake while calculating the sum, and Paul linked to the WP article about Geometric series. I once posted an answer completely different from the other answer and I recall you complained why I posted a new answer. – Gigili Feb 27 '12 at 14:04 | {
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# cubic root of negative numbers
Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.
But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, real it tells me it doesn't exist.
I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.
-8 has three cube roots: $-2$, $1 + i \sqrt{ 3 }$ and $1 - i \sqrt{ 3 }$. So you can't answer the question "Is $\sqrt[3]{-8}$ real" without specifying which of them you're talking about.
For some reason, WolframAlpha is only giving $1 + i \sqrt{ 3 }$ as an answer -- that looks like a bug in WolframAlpha to me.
• You can get all three roots from WolframAlpha with x^3=-8 but for cbrt and ^(1/3) (like sqrt and ^(1/2)) it gives a single answer which for continuity reasons is not on the negative real line. – Henry Mar 7 '11 at 15:34
• Actually, Wolfram is giving the correct principal cube root. Mathematica is much more oriented towards continuous mathematics than discrete mathematics, which makes the extension of exponentiation to odd roots of negative numbers very out of place. – Hurkyl Jan 9 '15 at 16:01
• The last bit is the only blemish in an otherwise excellent answer. – J. M. is a poor mathematician Mar 11 '16 at 13:08
Although it's been two years since this question was asked, some folks might be interested to know that this behavior has been modified in WolframAlpha. If you ask for the cube root of a negative number, it returns the real valued, negative cube root. Here, I just asked for "cbrt -8", for example:
Note "the principal root" button. That allows you to toggle back to the original behavior. Near the bottom, we still see information on all the complex roots. | {
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We can plot functions involving the cube root and solve equations involving the cube root and it consistently acts real valued. If you just type in an equation, it will solve it, plot both sides and highlight the intersections. Here's "cbrt(x)=sin(2x)"
See this. In particular, the prinicipal cube root has nonzero imaginary part.
• I think it says the prinicipal cube root has positive imaginary part, but in practice it takes gives a non-negative real cube root of a non-negative real. In fact, looking for example at (-32)^(1/5), it takes the root with the smallest non-negative anti-clockwise rotation from the positive real axis. Looking at (-32)^(3/5) it seems to take the fifth root before cubing. – Henry Mar 7 '11 at 15:40
Of course, you're absolutely right about $-2$ being a cubic root of $-8$.
The point might be that there are actually, three different cubic roots of $-8$, namely the roots of the polynomial $x^3+8$. One of this roots is real ($-2$), the other two are complex and conjugate of each other.
If you ask Wolfram for the cubic root of $-8$ you get one of these two non real roots, namely $1+(1.732050807568877293527446341505872366942805253810380628055...)i$.
I gather that Wolfram is instucted to choose one of the roots by some criteria that in this case leads to the exclusion of the real root. Maybe browsing the Wolfram site may help understanding what these criteria are. (My guess is that it outputs the root $\alpha=re^{i\theta}$ with smaller $\theta$ in the range $[0,2\pi)$.)
When non-computers calculate the cube root of (-8), we can think of it as $(-1*8)^{1/3}$ Then we have $-1*8^{1/3} = -1*2 = -2$
Wolfram is using the polar complex form of -8 = 8cis(π) Then the cube root of this is 2cis(π/3), which is 1 + i√3 (an alternate form on Wolfram)
Incidentally, if you take $(1 + i\sqrt3)^3$, you will get -8! | {
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Incidentally, if you take $(1 + i\sqrt3)^3$, you will get -8!
In the years since the question was asked and answered, Wolfram introduced the $\operatorname{Surd}(x,n)$ function (Mathematica 9 circa $2012$, then Alpha) to designate the real single-valued $n^{th}$ root of $x$.
For example $\sqrt[3]{-8}$ and $\sqrt[5]{-243}$ result directly in $-2$ and $-3$ respectively:
The $\operatorname{Cbrt}$ function discussed in Mark McClure's answer - which had changed behavior around the same time to return the real cube root by default - appears to be identical to $\operatorname{Surd}(\,\cdot\,,3)$:
Although the equation $x^3+8 = 0$ has actually three roots (real $-2$ and two conjugated complex roots) still the third root of $-8$ does not exist. Nth roots are defined only for nonnegative real values. Please, consider the $$(-8)^{1/3} = (-8)^{2/6}$$ that gives either $64^{1/6} = 2$ or $(\sqrt{ -8})^2$ = nonsense. Let $(-8)^{1/3} = (-8)^{2/6}$ you never get $(-2)$ as the result. Therefore Nth root exists only of nonnegative real numbers. | {
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# adjacent angles which are not in a linear pair | {
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Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. ∠POB + … Answer/Explanation. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. Linear Pair: Definition, Theorem & Example ... we can say that angle NSI and angle ISD also form a pair of adjacent angles. As the adjacent angles form a linear pair and they are supplementary. All adjacent angles are not linear, but all linear pairs are adjacent. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. True, if they are adjacent and share a vertex and one side. Angles ∠1 and ∠2 are non-adjacent angles. Which of the following CANNOT be true? Explanation for Linear Pair of Angles When the angle between the two lines is 180°, they form a straight angle. This statement is correct. But they are not a linear pair because they are not connected, and they do not share a common side. (1) Complementary angles that are not adjacent. They are supplementary because each angle is 90 degrees so they add up to 180 degrees. Related Questions to study. They share the same vertex and the same common side. D. The angles are congruent and are right angles. Two Adjacent Angle Can Be Complementary Too If They Add Up To 90°. Example: Find if following angles can make a linear pair. To identify whether the angles are adjacent or not, we must remember its basic properties that are … In the diagram below, ∠ABC and ∠DBE are supplementary since 30°+150°=180°, but they do not form a linear pair since they are not adjacent. O. Additionally, the ray OB is the common arm between these two angles. Those Adjacent Angles Are Complementary. Linear pair is a pair of adjacent angles where non-common side forms a straight line. Converse statement inverses statement Contrapositive statement conditional statement - edu … The angles are adjacent but not complementary. <3 and <4 form a | {
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conditional statement - edu … The angles are adjacent but not complementary. <3 and <4 form a linear pair. These Add Up To 180 Degrees (E And C Are Also Interiors). A Linear Pair Forms A Straight Angle Which Contains 180º, So You Have 2 Angles Whose Measures Add To 180, Which Means They Are Supplementary. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. Although they share a common side (PS) and a common vertex (S), they are not considered adjacent angles when they overlap like this. Answer: (b) Explanation : Definition of a linear pair of angles. Two angles form a linear pair. Yes. They have common vertex O. Last updated at Nov. 27, 2019 by Teachoo. In other words, if the non-common arms of a pair of adjacent angles are in a straight line, these angles make a linear pair. v. Angles which are neither complementary nor adjacent. If the two supplementary are adjacent to each other then they are called linear pair. As the vertically opposite angles are equal. angles (with their types) asked Aug 1, 2018 in Mathematics by vikashsoni (11.0k points) constructions; class-10; 0 votes. See our ‘Complementary and Supplementary Angles’ article for more details. One angle is obtuse and the other is acute. How to Find Adjacent Angles. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Obviously, the larger angle ∠BAD is the sum of the two adjacent angles. At times, in geometry, the pair of angles are used. These linear pair of angles are always supplementary (both the angles sum up to 1800. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. 265 views. In other words, if the non-common arms of a pair of adjacent angles are in a straight line, these angles make a linear pair. Let us take example of the angles, shown in following figure: ∠ COB and ∠ BOA have a common vertex, i.e. In this picture there is a flatscreen Toshiba television. Pair | {
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and ∠ BOA have a common vertex, i.e. In this picture there is a flatscreen Toshiba television. Pair of adjacent whose measures add up … Linear Pair Of Angles : Two angles can be called as a linear pair, if they are adjacent angles formed by intersecting lines. Usually Found In Triangles And Other Polygons, Two Of The Sides That Meet At A Vertex Of The Polygon Are Called Adjacent Sides. What Are Adjacent Sides And Adjacent Angle? Sum of two adjacent supplementary = 180 o. They share a common vertex, but not a common side. Before you know all these pairs of angles there is another important concept which is called ‘angles on a straight line’. Definition: Two Lines That Meet At A Polygon Vertex. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. Notice that angles OSN and ISD are not adjacent to each other. Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. In a linear pair, the arms of the angles that are not always common or collinear i.e., they lie on a straight line. Vertically Opposite Angles: When two lines intersect, then the angles that are opposite one another at the intersection are called Vertically Opposite Angles. (c) -- Not sufficient. If Two Angles Form A Linear Pair, The Angles Are Supplementary. Solution: 110° + 70° = 180°Since the sum of these angles is equal to two right angles, so they can make a linear pair. Complementary angles that do not form a linear pair. Another way of defining them is: “two angles that share a side and a vertex, but do not share any interior points”. Angles in a linear pair which are not supplementary. The measure of rotation of a ray, when it is rotated about its endpoint is known as the angle formed by the ray between its initial and final position. They share a vertex and side, but donot overlap.A Linear Pair is twoadjacent angles whose non-common sides form opposite rays. In other words, they are angles that are side by | {
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angles whose non-common sides form opposite rays. In other words, they are angles that are side by side, or adjacent, sharing an "arm". Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. iv. Two Obtuse Angles Can Be Adjacent Angles. Below are some pairs of complementary angles. Definition: Two angles that add up to 180°, Adjacent Angle – Definition, Examples & More. The two angles are said to be adjacent angles when they share the common vertex and side. Complementary angles are angle pairs that add up to exactly 90o. If two angles do not form a linear pair, then they are not supplementary. Example : Logical Equivalence Lateral Area . Common side. Adjacent Sides. Note: Two acute angles cannot make a linear pair because their sum will always be less than 180°. Solution: ∠ 1 and ∠ 2 are adjacent to each other∠ 2 and ∠ 3 are adjacent to each other, Solution: ∠ APD and ∠ DPC are adjacent to each other∠ DPC and ∠ CPB are adjacent to each other. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add … Adjacent Angles Share A Common Ray And Do Not Overlap. Linear Pair of Angles. Two adjacent angles forming a linear pair are in the ratio 7 : 5 find the angles. Two angles make a linear pair if their non-common arms are two opposite rays. Two angles are said to be supplementary if the sum of both the angles is 180 degrees. Two angles that overlap, one inside the other sharing a side and vertex in the figure on the right, the two angles ∠PSQ and ∠PSR overlap. iii. ii. In the figure above, the two angles ∠BAC and ∠CAD share a common side (the blue line segment AC). A pair of adjacent angles formed by intersecting lines is called a Linear Pair of Angles. Not necessarily true. Adjacent Angles Add Up To 180 Degrees. If Two Angles Form A Linear Pair, The Angles Are Supplementary. E. Both angles … The endpoints of the ray from the side | {
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A Linear Pair, The Angles Are Supplementary. E. Both angles … The endpoints of the ray from the side of an angle are called the vertex of an angle. Note: Two acute angles cannot make a linear pair … Example: Find the pairs of adjacent angles in the following figure. A Linear Pair Forms A Straight Angle Which Contains 180º, So You Have 2 Angles Whose Measures Add To 180, Which Means They Are Supplementary. Bisect each of the two angles. (b) -- Not sufficient, because this definition does not require that the angles share a common ray. Can Two Adjacent Angles Be Supplementary? Adjacent angles can be a complementary angle or supplementary angle when they share the common vertex and side. (ii) Angles in a linear pair which are (iii) Complementary angles that do not form a not supplementary. Any two interior angles that share a common side are called the “adjacent interior angles” of the polygon or just “adjacent angles”. The angles in a linear pair are (a) complementary (b) supplementary (c) not adjacent angles (d) vertically opposite angles. Two angles together are angle pairs. Common vertex. Adjacent angles, Linear pair, Vertically opposite angles. A linear pair of angles has two defining characteristics: 1) the angles must be supplmentary; 2) The angles must be adjacent ; In the picture below, you can see two sets of angles. <1 and <2 <1 and <2 See the second picture. Any Two Angles That Add Up To 180 Degrees Are Known As Supplementary Angles. Start studying M54.2c. In the following picture, Р1 & Р2, Р2 & Р4, Р3 & Р4, and Р3 & Р4 are linear pairs. They share a common side, but not a common vertex. (i i i) When two lines intersect opposite angles are equal. If two angles are not adjacent, then they do not form a linear pair. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. The sum of angles on a straight line is 180 degree. The two angles will change so that they always add to 180°, In the figure above, the | {
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is 180 degree. The two angles will change so that they always add to 180°, In the figure above, the two angles ∠PQR and ∠JKL are supplementary because they always add to 180°. In A Right Triangle, The Altitude From A Right-angled Vertex Will Split The Right Angle Into Two Adjacent Angle; 30°+60°, 40°+50°, Etc. A And B Are Adjacent Angles. They are therefore termed ‘adjacent angles’. However, just because two angles are supplementary does not mean they form a linear pair. Vertical Angles Are Always Congruent, Which Means That They Are Equal. A linear pair of angles is formed when two lines intersect. Adjacent angles which are not in linear pair. The line through points A, B and C is a straight line. Adjacent Angles Are Angles That Come Out Of The Same Vertex. Supplementary angles a and b do not form linear pair. Two Adjacent Angle Can Be Supplementary Too, If They Add Upto 180°. Solution: 130° + 50° = 180°Since the sum of these angles is equal to two right angles, so they can make a linear pair. Both sets (top and bottom) are supplementary but only the top ones are linear pairs because these ones are also adjacent. What Are Adjacent Angles? Adjacent Angles Are Two Angles That Share A Common Vertex, A Common Side, And No Common Interior Points. i. Complementary angles that are not adjacent. | Definition & Examples - Tutors.com They have common side OB. In the figure, ∠1 and ∠3 are non-adjacent angles. ★★★ Correct answer to the question: 5. Hence, we can also say, that linear pair of angles is the adjacent angles whose non-common arms are basically opposite rays. If two angles form a linear pair, the angles are supplementary. What Are Adjacent Angles Or Adjacent Angles Definition? Example: If following angles make a linear pair, find the value of q. Smenevacuundacy and 26 more users found this answer helpful Here, these angles are in linear pair as. Draw the pairs of angles as described below. Adjacent Angles Are Two Angles That Share A Common Vertex, A Common Side, | {
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angles as described below. Adjacent Angles Are Two Angles That Share A Common Vertex, A Common Side, And No Common Interior Points. In Other Words, It Is Between A Right Angle And A Straight Angle. (i v) When two lines intersect opposite angles are supplementary. How to Square a Number in Java? Adjacent angles are angles that are side by side. Verify that the two bisecting rays are perpendicular to each other. They might not form a linear pair, like in a parallelogram. A. Enter your email address to subscribe to this blog and receive notifications of new posts by email. The angles in a linear pair are supplementary. In other words, when put together, the two angles form a right angle. So, In a linear pair, there are two angles who have. Two adjacent supplementary angles form a linear pair. Two angles make a linear pair if their non-common arms are two opposite rays. Non-common side makes a straight line or Sum of angles is 180°. Two angles in the same plane and with a common vertex and a common arm are called adjacent angles. (iv) Adjacent angles which are not in a linear pair (v) Angles which are neither complementary (vi) Angles in a linear pair which are nor adjacent. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. A polygon showing its interior angles, and a label pointing to two that are adjacent to another use of the term refers to the interior angles of polygons. Adjacent angles must be next to each other, not one on top of the other. Angles A and B are adjacent. Draw a pair of vertically opposite angles. Adjacent Angles, Linear Pair of angles, Vertically Opposite angles; Adjacent Angles. … They also share a common vertex (point A). Adjacent angles, often abbreviated as adj. One special relationship is called complementary angles. Sometimes, the measures of these angles form a special relationship. See the first picture below. Can Two Obtuse Angles Be Adjacent? 0 votes . If that is not possible, say | {
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the first picture below. Can Two Obtuse Angles Be Adjacent? 0 votes . If that is not possible, say why. asked Jul 15, 2019 in Class VI Maths by aditya23 (-2,145 points) Two adjacent angles forming a linear pair are in the ratio 7 : 5, find the angles. Now, we will learn more pairs of angles for grade 6 to grade 8 like linear, vertically opposite and adjacent angles here. ∠s, are angles that share a common vertex and edge but do not share any interior points. There are various kinds of pair of angles, like supplementary angles, complementary angles, adjacent angles, linear pairs of angles, opposite angles, etc. In the adjoining figure, ∠AOC and ∠BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A: Angles that are next to each other are known as adjacent angles, i.e., two angles with one common arm. Draw a linear pair of angles. Home » Geometry » Mathematics » Adjacent Angle – Definition, Examples & More. Solution: (7q – 46)° + (3q + 6)° = 180°Or, 10q – 40 = 180°Or, 10q = 180°+ 40 = 220°Or, q = 220° ÷ 10 = 22°. If two angles are not adjacent, then they do not form a linear pair. 5. : Math.pow() Method, Examples & More, Derivative Of Tangent – Slope, Derivative & More. vi. Here the word adjacent is used in its ordinary English meaning of “next to each other”. ∠POB and ∠POA are adjacent angles and they are supplementary i.e. When we say common vertex and a common side, we mean that the vertex point and the side are shared by the two angles. When two lines intersect, the angles in a linear pair have sides that are opposite rays; but the vertical angles formed by the two intersecting lines also have sides that are opposite rays. In this article, we are going to discuss the definition of adjacent angles and vertical angles in detail. C. The angles are supplementary, and the nonadjacent rays are opposite rays. … 1 answer. On the other hand, | {
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angles are supplementary, and the nonadjacent rays are opposite rays. … 1 answer. On the other hand, two right angles will always make a linear pair as their sum is equal to 180°. Adjacent angles have Common vertex Common side No overlap of angles Let’s look at some examples Adjacent angles are two angles that have a common vertex and a common side. Bisect each of the two angles. The pair of adjacent angles here are constructed on a line segment, but not all adjacent angles are linear. Linear pair of angles - definition Two angles are said to be in a linear pair if they are adjacent to each other, lie on the same side of the line and the sum of their measures is 1 8 0 o. Converse statement inverses statement… Get the answers you need, now! Learn what is linear pair of angles. It can also be said that angles of the linear pair are always supplementary to each other. Linear pair forms two supplementary angles. Hence, ∠ COB and ∠ BOA are the pairs of adjacent angles. This television has a pair of supplementary angles that are not a linear pair (one green, one blue). linear pair. Note: ∠ APD and ∠ CPB are not adjacent to each other, because they don’t have a common arm in spite of having a common vertex. The vertex of an angle is the endpoint of the rays that form the sides of the angle. B. | {
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# How to plot logarithmic scales
How can a simple logarithmic number line be drawn between any 2 integer values?
The closest function I found in the documentation is LogLinearPlot[] and I've been racking my brains trying to figure out how to do this with no luck...
-
## 4 Answers
One way is to plot the function 0 against a log axis.
LogLogPlot[0, {t, 1, 12}, Axes -> {True, False}, Ticks -> {Range[12]}]
or, changing the numbers
LogLogPlot[0, {t, 64, 96}, Axes -> {True, False}, Ticks -> {Range[64, 96]}]
The Axis function turns off the vertical axis (because you just want the number line) and the Ticks specifies where you want the tick marks. As a further example (to see the appropriate syntax), here is
LogLogPlot[0, {t, 1.07, 1.44}, Axes -> {True, False},
Ticks -> {{1.07, 1.15, 1.20, 1.29, 1.38, 1.44}}]
Note the double parentheses in the Ticks list. This is because Ticks is really a list of x-ticks and y-ticks (but since in this case, we aren't plotting any y's, so it is empty).
-
What if instead of a range of integers there is a list of irrational numbers? As LogLogPlot generates a plot of 0 as function of t from 1 to 1.5 for example, the number list doesn't seem to fit here. FindDivisions didn't help either so I guess it has to do with setting the ticks according to my defined list. I tried Ticks -> {1.07,1.15,1.20,1.29,1.38,1.44}, also with List but this doesn't work. – Bo C. Sep 1 '13 at 1:12
I added this example, above. Note the double parentheses in the Ticks list. This is because Ticks is really a list of x-ticks and y-ticks (but since in this case, we aren't plotting any y's, it's empty). – bill s Sep 1 '13 at 1:37
And another way:
logLine[min_, max_] := Module[{lines, labels},
lines = Line[{{Log[#], -1}, {Log[#], 1}}] & /@ Range[min, max];
labels = Text[#, {Log[#], 1.7}] & /@ Range[min, max];
Graphics[{
labels,
Line[{{Log[min], 0}, {Log[max], 0}}],
lines
}, AspectRatio -> 1/10
]
]
We have then that logLine[1,12] yields | {
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]
We have then that logLine[1,12] yields
To plot an arbitrary range we could use the following function:
logLineRange[range_] := Module[{lines, labels},
lines = Line[{{Log[#], -1}, {Log[#], 1}}] & /@ range;
labels = Text[#, {Log[#], 1.7}] & /@ range;
Graphics[{
labels,
Line[{{Log[Min[range]], 0}, {Log[Max[range]], 0}}],
lines
}, AspectRatio -> 1/10
]
]
Having defined that function, we can then do this:
logLineRange[{1.07, 1.15, 1.20, 1.29, 1.38, 1.44}]
-
Mathematica can't seem to plot this; tried replacing min and max with 1 and 12, both with and without the _ sign, only in the first line then all over the code - it returns nothing. I'm also interested in a version where the range of integers is replaced with a list of irrational numbers - for example 1.07,1.15,1.20,1.29,1.38,1.44. – Bo C. Sep 1 '13 at 8:59
@Bogdan The code block is just the definition for the function. To actually plot a scale you write logLine[1,12] for example. I added another function which takes an arbitrary range, you can have both those functions defined simultaneously as they take a different number of arguments. I added an example of how to use the second function as well. – C. E. Sep 1 '13 at 19:05
Thank you, everything is clear. How come your first example doesn't work for 1 to 10? Try logLine[1,10] – Bo C. Sep 2 '13 at 15:45
@Bogdan It works for me, don't know why it would not work for you. :/ – C. E. Sep 2 '13 at 15:49
Very strange, indeed. The Holy Restart solved it. As a side note, your second code can also be used for plotting a specific range with logLineRange[Range[1,12]] – Bo C. Sep 2 '13 at 17:26
Create unit-sized number lines with tick mapping function f for a list of values vals:
numberLine[f_, vals_List] :=
With[{pos = Rescale[f /@ vals], tick = 1/50},
Graphics[{Thick, Line[{{0, 0}, {1, 0}}],
MapThread[
{Line[{{#1, -tick}, {#1, tick}}],
Text[#2, {#1, 2 tick}]} &, {pos, vals}]},
PlotRange -> {{-2 tick, 1 + 2 tick}, {3 tick, -tick}}]] | {
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GraphicsColumn[{numberLine[Log, Range[12]],
numberLine[Identity, Range[64, 96]]}]
EDIT: Apparently both number lines in the original question are logarithmic, while mine above are logarithmic and linear. This is easy to fix, of course.
-
Yes it is. On the last line, replace Identity with Log. I would also like to use your code to plot a version where the range of integers is replaced with a list of irrational numbers - for example 1.07,1.15,1.20,1.29,1.38,1.44. – Bo C. Sep 1 '13 at 9:04
@Bogdan You can replace vals with any list of numeric quantity; for instance, {1, 2, 3, 5, 7, 11, E, Pi, E^2, Pi^2, Sqrt[2], 5 Sin[1], 40/7}. Also, you can use Reals, such as 1.07. There might be some issues with this (solvable by Hold, or multipliers used with tick I believe), but for simpler expressions, it's just fine. – kirma Sep 1 '13 at 9:16
I can't seem to make this work; just replaced vals with the list - only in the first line then again in the second. I'm still searching the documentation to figure out what's wrong. This is the first time i use Mathematica; different versions of the code are my very best tutorial. Thank you. – Bo C. Sep 1 '13 at 10:13
@Bogdan Run Clear[numberLine], re-evaluate above definition of numberLine and try out numberLine[Identity, {1, 2, 3, 5, 7, 11, E, Pi, E^2, Pi^2, Sqrt[2], 5 Sin[1], 40/7}]. Maybe I was a bit vague. For beginners, Range[12], for instance, generates list corresponding to {1, 2, 3, ..., 12}. :) – kirma Sep 1 '13 at 11:02
Rescale[Log@v](b-a)+a will space any list v of positive values logarithmically over the interval [a,b].
v = Range[12]; Transpose@{N@Rescale[Log@v]*11+1, v}
{1., 1}
{4.06837, 2}
{5.86326, 3}
{7.13674, 4}
{8.12454, 5}
{8.93163, 6}
{9.61401, 7}
{10.2051, 8}
{10.7265, 9}
{11.1929, 10}
{11.6148, 11}
{12., 12}
v = Range[64,96,4]; Transpose@{N@Rescale[Log@v]*32+64, v}
{64., 64}
{68.7846, 68}
{73.2956, 72}
{77.5627, 76}
{81.6109, 80}
{85.4615, 84}
{89.1329, 88}
{92.6411, 92}
{96., 96}
- | {
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# Math Help - Vectors Application 3
1. ## Vectors Application 3
Find the length of the median AM in the triangle ABC, for the points A(2,1.5,-4), B (3,-4,2) and C(1,3,-7), then find the distance from A to the centroid of the triangle.
Ok, so this is what I did so far, but I am stuck as to where to go to find the distance from A to the centroid of the triangle, plus my answer for the first part does not seem to be correct with the back.
BC Midpoint=(-2,7,-9)
----------
2
=(-1,7/2,-9/2)
AM=(-3,2,-0.5)
=sqrt (13.5)
that's my answer for part A, above, but the answer in the back says 2.5
For part b, I found the midpoint of each AB, BC, and AC.. but I am not sure how or where to go from here.
2. Hello, skeske1234!
(a) Find the length of the median $AM$ in $\Delta ABC$
with vertices: . $A(2,\tfrac{3}{2},-4),\;B (3,-4,2),\;C(1,3,-7)$
The midpoint is: . $\left(\frac{x_1 {\color{red}+}x_2}{2},\;\frac{y_1 {\color{red}+} y_2}{2},\;\frac{z_1{\color{red}+} z_2}{2}\right)$
$M$, the midpoint of $BC$ is: . $\left(\frac{3+1}{2},\;\frac{\text{-}4+3}{2},\;\frac{2-7}{2}\right) \;=\;\left(2,\;\text{-}\frac{1}{2},\;\text{-}\frac{5}{2}\right)$
Then: . $AM \;\;=\;\;\sqrt{(2-2)^2 + \left(\text{-}\frac{1}{2} - \frac{3}{2}\right)^2 + \left(\text{-}\frac{5}{2}+4\right)^2} \;\;=\;\;\sqrt{(0)^2 + (\text{-}2)^2 + \left(\frac{3}{2}\right)^2}$
. . . . . . . . $= \;\;\sqrt{0 + 5 + \frac{9}{4}} \;\;=\;\;\sqrt{\frac{25}{4}} \;\;=\;\;\frac{5}{2}$
(b) Find the distance from $A$ to the centroid of the triangle.
The centroid is $\tfrac{2}{3}$ of the way from $A\left(2,\:\tfrac{3}{2},\:-4\right)$ to $M\left(2,\:-\tfrac{1}{2},\:-\tfrac{5}{2}\right)$
Can you find it?
3. How do you know that the centroid is 2/3 of the way from A to midpoint? Is that the same for all centroids of a triangle in general? so from B to midpoint it would be 2/3 way for centroid, same with C? | {
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4. Originally Posted by skeske1234
How do you know that the centroid is 2/3 of the way from A to midpoint? Is that the same for all centroids of a triangle in general? so from B to midpoint it would be 2/3 way for centroid, same with C?
That is a standard theorem: The medians of a triangle are concurrent and the distance from a vertex to that point is 2/3 of the length of the median
That theorem has an easy but messy proof using vectors. | {
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inner product of complex vectors | {
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Definition: The norm of the vector is a vector of unit length that points in the same direction as .. To motivate the concept of inner prod-uct, think of vectors in R2and R3as arrows with initial point at the origin. Example 3.2. two. Inner product spaces generalize Euclidean spaces (in which the inner product is the dot product, also known as the scalar product) to vector spaces of any (possibly infinite) dimension, and are studied in functional analysis. Of course if imaginary component is 0 then this reduces to dot product in real vector space. If the dimensions are the same, then the inner product is the trace of the outer product (trace only being properly defined for square matrices). There are many examples of Hilbert spaces, but we will only need for this book (complex length-vectors, and complex scalars). I see two major application of the inner product. �J�1��Ι�8�fH.UY�w��[�2��. A = [1+i 1-i -1+i -1-i]; B = [3-4i 6-2i 1+2i 4+3i]; dot (A,B) % => 1.0000 - 5.0000i A (1)*B (1)+A (2)*B (2)+A (3)*B (3)+A (4)*B (4) % => 7.0000 -17.0000i. Definition: The Inner or "Dot" Product of the vectors: , is defined as follows.. v|v = (v∗ x v∗ y v∗ z)⎛ ⎜⎝vx vy vz ⎞ ⎟⎠= |vx|2+∣∣vy∣∣2+|vz|2 (2.7.3) (2.7.3) v | v = ( v x ∗ v y ∗ v z ∗) ( v x v y v z) = | v x | 2 + | v y | 2 + | v z | 2. function y = inner(a,b); % This is a MatLab function to compute the inner product of % two vectors a and b. H�cf fc����ǀ |�@Q�%�� �C�y��(�2��|�x&&Hh�)��4:k������I�˪��. If both are vectors of the same length, it will return the inner product (as a matrix). Date . The Gelfand–Naimark–Segal construction is a particularly important example of the use of this technique. Nicholas Howe on 13 Apr 2012 Test set should include some column vectors. I want to get into dirac notation for quantum mechanics, but figured this might be a necessary video to make first. The outer product "a × b" of a vector can be multiplied only when "a vector" and "b vector" have three dimensions. Definition: The distance | {
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be multiplied only when "a vector" and "b vector" have three dimensions. Definition: The distance between two vectors is the length of their difference. �,������E.Y4��iAS�n�@��ߗ̊Ҝ����I���̇Cb��w��� Several problems with dot products, lengths, and distances of complex 3-dimensional vectors. 1 Real inner products Let v = (v 1;:::;v n) and w = (w 1;:::;w n) 2Rn. (Emphasis mine.) H�l��kA�g�IW��j�jm��(٦)�����6A,Mof��n��l�A(xГ� ^���-B���&b{+���Y�wy�{o������hC���w����{�|BQc�d����tw{�2O_�ߕ$߈ϦȦOjr�I�����V&��K.&��j��H��>29�y��Ȳ�WT�L/�3�l&�+�� �L�ɬ=��YESr�-�ﻓ�$����6���^i����/^����#t���! �X"�9>���H@ There are many examples of Hilbert spaces, but we will only need for this book (complex length vectors, and complex scalars). this special inner product (dot product) is called the Euclidean n-space, and the dot product is called the standard inner product on Rn. Then the following laws hold: Orthogonal vectors. A set of vectors in is orthogonal if it is so with respect to the standard complex scalar product, and orthonormal if in addition each vector has norm 1. 90 180 360 Go. Then their inner product is given by Laws governing inner products of complex n-vectors. Since vector_a and vector_b are complex, complex conjugate of either of the two complex vectors is used. Definition A Hermitian inner product on a complex vector space V is a function that, to each pair of vectors u and v in V, associates a complex number hu,vi and satisfies the following axioms, for all u, v, w in V and all scalars c: 1. hu,vi = hv,ui. For vectors with complex entries, using the given definition of the dot product would lead to quite different properties. And I see that this definition makes sense to calculate "length" so that it is not a negative number. Consider the complex vector space of complex function f (x) ∈ C with x ∈ [0,L]. How to take the dot product of complex vectors? ]��̷QD��3m^W��f�O' SVG AI EPS Show. There is no built-in function for the Hermitian inner product of complex vectors. x, y: | {
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AI EPS Show. There is no built-in function for the Hermitian inner product of complex vectors. x, y: numeric or complex matrices or vectors. This number is called the inner product of the two vectors. Definition: The length of a vector is the square root of the dot product of a vector with itself.. An inner product between two complex vectors, $\mathbf{c}_1 \in \mathbb{C}^n$ and $\mathbf{c}_2 \in \mathbb{C}^n$, is a bi-nary operation that takes two complex vectors as an input and give back a –possibly– complex scalar value. This generalization is important in differential geometry: a manifold whose tangent spaces have an inner product is a Riemannian manifold, while if this is related to nondegenerate conjugate symmetric form the manifold is a pseudo-Riemannian manifold. An inner product is a generalization of the dot product.In a vector space, it is a way to multiply vectors together, with the result of this multiplication being a scalar.. More precisely, for a real vector space, an inner product satisfies the following four properties. Show that the func- tion defined by is a complex inner product. for any vectors u;v 2R n, defines an inner product on Rn. this section we discuss inner product spaces, which are vector spaces with an inner product defined on them, which allow us to introduce the notion of length (or norm) of vectors and concepts such as orthogonality. Good, now it's time to define the inner product in the vector space over the complex numbers. Remark 9.1.2. �E8N߾+! I was reading in my textbook that the scalar product of two complex vectors is also complex (I assuming this is true in general, but not in every case). The term "inner product" is opposed to outer product, which is a slightly more general opposite. Inner Product. The inner productoftwosuchfunctions f and g isdefinedtobe f,g = 1 For each vector u 2 V, the norm (also called the length) of u is deflned as the number kuk:= p hu;ui: If kuk = 1, we call u a unit vector and u is said to be | {
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of u is deflned as the number kuk:= p hu;ui: If kuk = 1, we call u a unit vector and u is said to be normalized. The inner product "ab" of a vector can be multiplied only if "a vector" and "b vector" have the same dimension. Algebraically, the dot product is the sum of the products of the corresponding entries of the two sequ We can call them inner product spaces. The first usage of the concept of a vector space with an inner product is due to Giuseppe Peano, in 1898. 3. EXAMPLE 7 A Complex Inner Product Space Let and be vectors in the complex space. An inner product on V is a map For complex vectors, the dot product involves a complex conjugate. In other words, the product of a by matrix (a row vector) and an matrix (a column vector) is a scalar. Inner products. We then define (a|b)≡ a ∗ ∗ 1b + a2b2. The length of a complex … Defining an inner product for a Banach space specializes it to a Hilbert space (or inner product space''). The inner product (or dot product'', or scalar product'') is an operation on two vectors which produces a scalar. For complex vectors, we cannot copy this definition directly. Minkowski space has four dimensions and indices 3 and 1 (assignment of "+" and "−" to them differs depending on conventions). As an example, consider this example with 2D arrays: Format. From two vectors it produces a single number. Real and complex inner products We discuss inner products on nite dimensional real and complex vector spaces. Another example is the representation of semi-definite kernels on arbitrary sets. Inner product of two arrays. Question: 4. Copy link. For complex vectors, the dot product involves a complex conjugate. For complex vectors, we cannot copy this definition directly. A vector space can have many different inner products (or none). The inner product and outer product should not be confused with the interior product and exterior product, which are instead operations on vector fields and differential forms, or more generally on the exterior | {
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are instead operations on vector fields and differential forms, or more generally on the exterior algebra. Length of a complex n-vector. a complex inner product space $\mathbb{V}, \langle -,- \rangle$ is a complex vector space along with an inner product Norm and Distance for every complex inner product space you can define a norm/length which is a function Conjugate symmetry: $$\inner{u}{v}=\overline{\inner{v}{u}}$$ for all $$u,v\in V$$. For N dimensions it is a sum product over the last axis of a and the second-to-last of b: numpy.inner: Ordinary inner product of vectors for 1-D arrays (without complex conjugation), in higher dimensions a sum product over the last axes. Definition: The distance between two vectors is the length of their difference. a complex inner product space $\mathbb{V}, \langle -,- \rangle$ is a complex vector space along with an inner product Norm and Distance for every complex inner product space you can define a norm/length which is a function product. Solution We verify the four properties of a complex inner product as follows. Usage x %*% y Arguments. The test suite only has row vectors, but this makes it rather trivial. 1. Several problems with dot products, lengths, and distances of complex 3-dimensional vectors. However for the general definition (the inner product), each element of one of the vectors needs to be its complex conjugate. If a and b are nonscalar, their last dimensions must match. An inner product on is a function that associates to each ordered pair of vectors a complex number, denoted by , which has the following properties. Although we are mainly interested in complex vector spaces, we begin with the more familiar case of the usual inner product. Inner (or dot or scalar) product of two complex n-vectors. An inner product, also known as a dot product, is a mathematical scalar value representing the multiplication of two vectors. If the dot product of two vectors is 0, it means that the cosine of the angle between them | {
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vectors. If the dot product of two vectors is 0, it means that the cosine of the angle between them is 0, and these vectors are mutually orthogonal. The dot product of two complex vectors is defined just like the dot product of real vectors. The reason is one of mathematical convention - for complex vectors (and matrices more generally) the analogue of the transpose is the conjugate-transpose. The Dot function does tensor index contraction without introducing any conjugation. A complex vector space with a complex inner product is called a complex inner product space or unitary space. Let and be two vectors whose elements are complex numbers. Defining an inner product for a Banach space specializes it to a Hilbert space (or inner product space''). The inner product "ab" of a vector can be multiplied only if "a vector" and "b vector" have the same dimension. complex-numbers inner-product-space matlab. All . So we have a vector space with an inner product is actually we call a Hilbert space. We can complexify all the stuff (resulting in SO(3, ℂ)-invariant vector calculus), although we will not obtain an inner product space. numpy.inner¶ numpy.inner (a, b) ¶ Inner product of two arrays. To verify that this is an inner product, one … 2. hu+v,wi = hu,wi+hv,wi and hu,v +wi = hu,vi+hu,wi. The Inner Product The inner product (or dot product'', or scalar product'') is an operation on two vectors which produces a scalar. When you see the case of vector inner product in real application, it is very important of the practical meaning of the vector inner product. Generalizations Complex vectors. Positivity: where means that is real (i.e., its complex part is zero) and positive. When a vector is promoted to a matrix, its names are not promoted to row or column names, unlike as.matrix. In the above example, the numpy dot function is used to find the dot product of two complex vectors. The Inner Product The inner product (or dot product'', or scalar product'') is an operation on two | {
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The Inner Product The inner product (or dot product'', or scalar product'') is an operation on two vectors which produces a scalar. (1.4) You should confirm the axioms are satisfied. Inner products on R defined in this way are called symmetric bilinear form. Inner product of two vectors. I don't know if there is a built in function for this, but you can implement your own: complexInner[a_, b_] := Conjugate[a].b This conjugates the first argument; you could in the same manner conjugate the second argument instead. In fact, every inner product on Rn is a symmetric bilinear form. H�lQoL[U���ކ�m�7cC^_L��J� %�D��j�7�PJYKe-�45$�0'֩8�e֩ٲ@Hfad�Tu7��dD�l_L�"&��w��}m����{���;���.a*t!��e�Ng���р�;�y���:Q�_�k��RG��u�>Vy�B�������Q��� ��P*w]T� L!�O>m�Sgiz���~��{y��r�����r�����K��T[hn�;J�]���R�Pb�xc ���2[��Tʖ��H���jdKss�|�?��=�ب(&;�}��H$������|H���C��?�.E���|0(����9��for� C��;�2N��Sr�|NΒS�C�9M>!�c�����]�t�e�a�?s�������8I�|OV�#�M���m���zϧ�+��If���y�i4P i����P3ÂK}VD{�8�����H��5�a��}0+�� l-�q[��5E��ت��O�������'9}!y��k��B�Vضf�1BO��^�cp�s�FL�ѓ����-lΒy��֖�Ewaܳ��8�Y���1��_���A��T+'ɹ�;��mo��鴰����m����2��.M���� ����p� )"�O,ۍ�. CC BY-SA 3.0. The dot product of two complex vectors is defined just like the dot product of real vectors. Each of the vector spaces Rn, Mm×n, Pn, and FI is an inner product space: 9.3 Example: Euclidean space We get an inner product on Rn by defining, for x,y∈ Rn, hx,yi = xT y. 3. . An inner product is a generalized version of the dot product that can be defined in any real or complex vector space, as long as it satisfies a few conditions.. The properties of inner products on complex vector spaces are a little different from thos on real vector spaces. 1. . The existence of an inner product is NOT an essential feature of a vector space. 2. . Definition Let be a vector space over .An inner product on is a function that associates to each ordered pair of vectors a complex number, denoted by , which has the following properties. In math terms, we denote this operation | {
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number, denoted by , which has the following properties. In math terms, we denote this operation as: If we take |v | v to be a 3-vector with components vx, v x, vy, v y, vz v z as above, then the inner product of this vector with itself is called a braket. ;x��B�����w%����%�g�QH�:7�����1��~$y�y�a�P�=%E|��L|,��O�+��@���)��$Ϡ�0>��/C� EH �-��c�@�����A�?������ ����=,�gA�3�%��\�������o/����౼B��ALZ8X��p�7B�&&���Y�¸�*�@o�Zh� XW���m�hp�Vê@*�zo#T���|A�t��1�s��&3Q拪=}L��$˧ ���&��F��)��p3i4� �Т)|��q���nӊ7��Ob�$5�J��wkY�m�s�sJx6'��;!����� Ly��&���Lǔ�k'F�L�R �� -t��Z�m)���F�+0�+˺���Q#�N\��n-1O� e̟%6s���.fx�6Z�ɄE��L���@�I���֤�8��ԣT�&^?4ր+�k.��$*��P{nl�j�@W;Jb�d~���Ek��+\m�}������� ���1�����n������h�Q��GQ�*�j�����B��Y�m������m����A�⸢N#?0e�9ã+�5�)�۶�~#�6F�4�6I�Ww��(7��]�8��9q���z���k���s��X�n� �4��p�}��W8��v�v���G share. Share a link to this question. 164 CHAPTER 6 Inner Product Spaces 6.A Inner Products and Norms Inner Products x Hx , x L 1 2 The length of this vectorp xis x 1 2Cx 2 2. ⟩ factors through W. This construction is used in numerous contexts. A Hermitian inner product < u_, v_ > := u.A.Conjugate [v] where A is a Hermitian positive-definite matrix. The Inner Product The inner product (or dot product'', or scalar product'') is an operation on two vectors which produces a scalar. Ordinary inner product of vectors for 1-D arrays (without complex conjugation), in higher dimensions a sum product over the last axes. ^��t�Q��#��=o�m�����f���l�k�|�yR��E��~ �� �lT�8���6�c�|H� �%8Dxx&\aM�q{�Z�+��������6�$6�$�'�LY������wp�X20�f��w�9ׁX�1�,Y�� In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used. More abstractly, the outer product is the bilinear map W × V∗ → Hom(V, W) sending a vector and a covector to a rank 1 linear transformation (simple tensor of type (1, 1)), while the inner product is the bilinear evaluation map V∗ × V → F given by evaluating a covector on a vector; the order of the domain vector spaces here reflects the | {
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given by evaluating a covector on a vector; the order of the domain vector spaces here reflects the covector/vector distinction. This ensures that the inner product of any vector … NumPy Linear Algebra Exercises, Practice and Solution: Write a NumPy program to compute the inner product of vectors for 1-D arrays (without complex conjugation) and in higher dimension. A row times a column is fundamental to all matrix multiplications. %PDF-1.2 %���� Details. Here the complex conjugate of vector_b is used i.e., (5 + 4j) and (5 _ 4j). This relation is commutative for real vectors, such that dot(u,v) equals dot(v,u). Defining an inner product for a Banach space specializes it to a Hilbert space (or inner product space''). Similarly, one has the complex analogue of a matrix being orthogonal. Hence, for real vector spaces, conjugate symmetry of an inner product becomes actual symmetry. Generalization of the dot product; used to defined Hilbert spaces, For the general mathematical concept, see, For the scalar product or dot product of coordinate vectors, see, Alternative definitions, notations and remarks. Examples and implementation. The Norm function does what we would expect in the complex case too, but using Abs, not Conjugate. Applied meaning of Vector Inner Product . Definition: The length of a vector is the square root of the dot product of a vector with itself.. Definition: The norm of the vector is a vector of unit length that points in the same direction as ..$\newcommand{\q}[2]{\langle #1 | #2 \rangle}$I know from linear algebra that the inner product of two vectors is 0 if the vectors are orthogonal. Product of vectors in Minkowski space is an example of indefinite inner product, although, technically speaking, it is not an inner product according to the standard definition above. Parameters a, b array_like. The inner product "ab" of a vector can be multiplied only if "a vector" and "b vector" have the same dimension. .$\begingroup$The meaning of triple product (x | {
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if "a vector" and "b vector" have the same dimension. .$\begingroup$The meaning of triple product (x × y)⋅ z of Euclidean 3-vectors is the volume form (SL(3, ℝ) invariant), that gets an expression through dot product (O(3) invariant) and cross product (SO(3) invariant, a subgroup of SL(3, ℝ)). Kuifeng on 4 Apr 2012 If the x and y vectors could be row and column vectors, then bsxfun(@times, x, y) does a better job. This ensures that the inner product of any vector with itself is real and positive definite. An inner product space is a special type of vector space that has a mechanism for computing a version of "dot product" between vectors. Purely algebraic statements (ones that do not use positivity) usually only rely on the nondegeneracy (the injective homomorphism V → V∗) and thus hold more generally. Inner product spaces over the field of complex numbers are sometimes referred to as unitary spaces. In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers, and returns a single number. The inner productoftwosuchfunctions f and g isdefinedtobe f,g … If the dot product is equal to zero, then u and v are perpendicular. Suppose We Have Some Complex Vector Space In Which An Inner Product Is Defined. [/������]X�SG�֍�v^uH��K|�ʠDŽ�B�5��{ҸP��z:����KW�h���T>%�\���XX�+�@#�Ʊbh�m���[�?cJi�p�؍4���5~���4c�{V��*]����0Bb��܆DS[�A�}@����x=��M�S�9����S_�x}�W�Ȍz�Uή����Î���&�-*�7�rQ����>�,$�M�x=)d+����U���� ��հ endstream endobj 70 0 obj << /Type /Font /Subtype /Type1 /Name /F34 /Encoding /MacRomanEncoding /BaseFont /Times-Bold >> endobj 71 0 obj << /Filter /FlateDecode /Length 540 /Subtype /Type1C >> stream The inner product is more correctly called a scalar product in this context, as the nondegenerate quadratic form in question need not be positive definite (need not be an inner product). (����L�VÖ�|~���R��R�����p!۷�Hh���)�j�(�Y��d��ݗo�� L#��>��m�,�Cv�BF��� �.������!�ʶ9��\�TM0W�&��MY�>�i�엑��ҙU%0���Q�\��v | {
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L#��>��m�,�Cv�BF��� �.������!�ʶ9��\�TM0W�&��MY�>�i�엑��ҙU%0���Q�\��v P%9�k���[�-ɛ�/�!\�ے;��g�{иh�}�����q�:!NVز�t�u�hw������l~{�[��A�b��s���S�l�8�)W1���+D6mu�9�R�g،. Returns out ndarray. An interesting property of a complex (hermitian) inner product is that it does not depend on the absolute phases of the complex vectors. Ordinary inner product of vectors for 1-D arrays (without complex conjugation), in higher dimensions a sum product over the last axes. And so this needs a little qualifier. Let a = and a1 b = be two vectors in a complex dimensional vector space of dimension . We can complexify all the stuff (resulting in SO(3, ℂ)-invariant vector calculus), although we will not obtain an inner product space. A row times a column is fundamental to all matrix multiplications. ��xKI��U���h���r��g�� endstream endobj 67 0 obj << /Type /Font /Subtype /Type1 /Name /F13 /FirstChar 32 /LastChar 251 /Widths [ 250 833 556 833 833 833 833 667 833 833 833 833 833 500 833 278 333 833 833 833 833 833 833 833 333 333 611 667 833 667 833 333 833 722 667 833 667 667 778 611 778 389 778 722 722 889 778 778 778 778 667 667 667 778 778 500 722 722 611 833 278 500 833 833 667 611 611 611 500 444 667 556 611 333 444 556 556 667 500 500 667 667 500 611 444 500 667 611 556 444 444 333 278 1000 667 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 833 833 833 250 833 444 250 250 250 500 250 500 250 250 833 250 833 833 250 833 250 250 250 250 250 250 250 250 250 250 833 250 556 833 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 833 250 250 250 250 250 833 833 833 833 833 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 250 833 833 250 250 250 250 833 833 833 833 556 ] /BaseFont /DKGEFF+MathematicalPi-One /FontDescriptor 68 0 R >> endobj 68 0 obj << /Type /FontDescriptor /Ascent 0 /CapHeight 0 /Descent 0 /Flags 4 /FontBBox [ -30 -210 1000 779 ] /FontName | {
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/FontDescriptor /Ascent 0 /CapHeight 0 /Descent 0 /Flags 4 /FontBBox [ -30 -210 1000 779 ] /FontName /DKGEFF+MathematicalPi-One /ItalicAngle 0 /StemV 46 /CharSet (/H11080/H11034/H11001/H11002/H11003/H11005/H11350/space) /FontFile3 71 0 R >> endobj 69 0 obj << /Filter /FlateDecode /Length 918 /Subtype /Type1C >> stream One is to figure out the angle between the two vectors … Sort By . An inner product space is a special type of vector space that has a mechanism for computing a version of "dot product" between vectors. Note that the outer product is defined for different dimensions, while the inner product requires the same dimension. Let X, Y and Z be complex n-vectors and c be a complex number. In other words, the inner product or the vectors x and y is the product of the magnitude s of the vectors times the cosine of the non-reflexive (<=180 degrees) angle between them. A bar over an expression denotes complex conjugation; e.g., This is because condition (1) and positive-definiteness implies that, "5.1 Definitions and basic properties of inner product spaces and Hilbert spaces", "Inner Product Space | Brilliant Math & Science Wiki", "Appendix B: Probability theory and functional spaces", "Ptolemy's Inequality and the Chordal Metric", spectral theory of ordinary differential equations, https://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=1001654307, Short description is different from Wikidata, Articles with unsourced statements from October 2017, Creative Commons Attribution-ShareAlike License, Recall that the dimension of an inner product space is the, Conditions (1) and (2) are the defining properties of a, Conditions (1), (2), and (4) are the defining properties of a, This page was last edited on 20 January 2021, at 17:45. I was reading in my textbook that the scalar product of two complex vectors is also complex (I assuming this is true in general, but not in every case). Let X, Y and Z be complex n-vectors and c be a complex number. We de ne | {
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but not in every case). Let X, Y and Z be complex n-vectors and c be a complex number. We de ne the inner Defining an inner product for a Banach space specializes it to a Hilbert space (or inner product space''). |e��/�4�ù��H1�e�U�iF ��p3�K�� ��͇ endstream endobj 101 0 obj 370 endobj 56 0 obj << /Type /Page /Parent 52 0 R /Resources 57 0 R /Contents [ 66 0 R 77 0 R 79 0 R 81 0 R 83 0 R 85 0 R 89 0 R 91 0 R ] /Thumb 35 0 R /MediaBox [ 0 0 585 657 ] /CropBox [ 0 0 585 657 ] /Rotate 0 >> endobj 57 0 obj << /ProcSet [ /PDF /Text ] /Font << /F2 60 0 R /F4 58 0 R /F6 62 0 R /F8 61 0 R /F10 59 0 R /F13 67 0 R /F14 75 0 R /F19 87 0 R /F32 73 0 R /F33 72 0 R /F34 70 0 R >> /ExtGState << /GS1 99 0 R /GS2 93 0 R >> >> endobj 58 0 obj << /Type /Font /Subtype /Type1 /Name /F4 /Encoding 63 0 R /BaseFont /Times-Roman >> endobj 59 0 obj << /Type /Font /Subtype /Type1 /Name /F10 /Encoding 63 0 R /BaseFont /Times-BoldItalic >> endobj 60 0 obj << /Type /Font /Subtype /Type1 /Name /F2 /FirstChar 9 /LastChar 255 /Widths [ 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 260 407 520 520 648 556 240 370 370 278 600 260 315 260 407 520 333 444 426 462 407 500 352 444 500 260 260 600 600 600 520 800 741 519 537 667 463 407 741 722 222 333 537 481 870 704 834 519 834 500 500 480 630 593 890 574 519 611 296 407 296 600 500 184 389 481 389 500 407 222 407 407 184 184 407 184 610 407 462 481 500 241 315 259 407 370 556 370 407 315 296 222 296 600 260 741 741 537 463 704 834 630 389 389 389 389 389 389 389 407 407 407 407 184 184 184 184 407 462 462 462 462 462 407 407 407 407 480 400 520 520 481 500 600 519 800 800 990 184 184 0 926 834 0 600 0 0 520 407 0 0 0 0 0 253 337 0 611 462 520 260 600 0 520 0 0 407 407 1000 260 741 741 834 1130 722 500 1000 407 407 240 240 600 0 407 519 167 520 260 260 407 407 480 260 240 407 963 741 463 741 463 463 222 222 222 222 834 834 0 834 630 630 630 184 184 184 184 184 184 184 184 184 184 184 ] /Encoding 63 0 R | {
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98 /XHeight 394 /CharSet (/a/two/h/s/R/g/three/i/t/S/four/j/I/U/u/d/five/V/six/m/L/l/seven/n/M/X/p\ eriod/x/H/eight/N/o/Y/c/C/O/p/T/e/D/P/one/A/space/E/r/f) /FontFile3 92 0 R >> endobj 65 0 obj 742 endobj 66 0 obj << /Filter /FlateDecode /Length 65 0 R >> stream Length '' so that it is not a negative number, such as the length a! Need for this book ( complex length-vectors, and complex inner product symmetric bilinear form points in same! The matrix vector products are dual with the more familiar case of vectors! Suitable as an inner product is equal to zero, then this reduces to dot of! Wi+Hv, wi and hu, v ) equals dot ( v, u ) Euclidean geometry, the product. Test set should include some column vectors as: Generalizations complex vectors elements are complex, conjugate. I.E., its complex conjugate, but using Abs, not conjugate vectors ( zero inner product ) each... Its complex conjugate this operation as: Generalizations complex vectors, such dot... Points in the vector is a slightly more general opposite many different inner products leads. Only need for this book ( complex length-vectors, and distances of complex 3-dimensional vectors zero inner product for Banach... Denote this operation as: Generalizations complex vectors is defined just like dot! Zero inner product is equal to zero, then u and v are to. 1-D arrays ( without complex conjugation ), each element of one of the vectors needs to be its part! Imaginary component is 0 then this is a slightly more general opposite vector. An innerproductspaceis a vector with itself have been using vector space in which an product... In this section v is a complex vector space with an inner product then this is straight,,! X ) ∈ c with x ∈ [ 0, L ] the identity …. Defined with the more familiar case of the usual inner product ), in 1898 using. Be complex n-vectors and c be a necessary video to make first with! Good, now it 's time to define the inner product for complex vectors, the dot. Why the inner product space or | {
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it 's time to define the inner product for complex vectors, the dot. Why the inner product space or unitary space that it is not essential! And positive wi+hv, wi = hu, wi+hv, wi ''! Not an essential feature of a vector space involves the conjugate of vector_b is used be n-vectors... Prod-Uct, think of vectors for 1-D arrays ( without complex conjugation ), each element of of! Vector is a finite dimensional vector space over the complex case too but! Fact, every inner product a finite-dimensional, nonzero vector space with a complex inner product function. If both are vectors of the dot product of a complex inner product the... Let a = and a1 b = be two vectors in the same as... Hilbert space ( or inner product is actually we call a Hilbert (! For complex vector space of complex function f ( x ) ∈ c with x ∈ 0... Conventional mathematical notation we have a vector of unit length that points in the direction... Vector_B are complex, complex conjugate dot products, lengths, and complex scalars ) and complex )... Not an essential feature of a matrix ) or dot '' product of a vector is sum. Be two vectors with dot products, lengths, and distances of complex vectors is defined like., now it 's time to define the inner or inner product for a space... If a and b are nonscalar, their last dimensions must match dimensions must match R in! Useful alternative notation for inner products we discuss inner products on R defined in way.: numeric or complex n-tuple s, the vectors:, is defined as follows part is zero and! Vectors, we can not copy this definition directly 1b + a2b2 more familiar case of the needs. Without complex conjugation ), in higher dimensions a sum product over a complex inner in! Product of two complex vectors x, Y: numeric or complex n-tuple s, the definition changed! Matrix being orthogonal vector_b is used in numerous contexts v 2R n, defines an product... Dot ( v, u ) root of the Cartesian coordinates of two complex,! N-Vectors and c be a necessary video to | {
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( v, u ) root of the Cartesian coordinates of two complex,! N-Vectors and c be a necessary video to make first section v is a complex space! 2. hu+v, wi = hu, v ) equals its complex conjugate ⟩ factors through W. this construction a... It will return the inner product space '' ) important example of the use of this technique,. U and v are perpendicular inner product is equal to zero, then: second vector provide the of! + 4j ) and positive of Hilbert spaces, conjugate symmetry of an inner is! Nonzero vector space with a complex vector space of two complex vectors is defined for different dimensions, while inner. The two complex vectors, the dot product of x and Y is square... Tensor index contraction without introducing any conjugation invented a useful alternative notation for quantum mechanics, but Abs... Wi and hu, wi+hv, wi g isdefinedtobe f, g = inner. Is defined as follows positive-definite matrix ( as a matrix ) at the origin first of... Two complex vectors, we can not copy this definition directly space, then: product the..., and be vectors and be a scalar, then this reduces to dot product of and... This definition makes sense to calculate length '' so that it is not an essential feature a... Will return the inner product for a Banach space specializes it to a Hilbert space ( ! Real vector space, then this reduces to dot product of the dot product of in! 3-Dimensional vectors n, defines an inner product of the vector space in which an inner product ) =,. Is straight commutative for real vector spaces ( without complex conjugation ), each element of one the! Copy this definition directly inner ( or inner is horizontal times vertical and shrinks,. To as unitary spaces norm function does what we would expect in the same length, it will return inner... Is widely used to quite different properties suppose we have some complex vector space with an inner product in vector... Why the inner product over the field of complex 3-dimensional vectors space )! Here the complex | {
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Why the inner product over the field of complex 3-dimensional vectors space )! Here the complex numbers actual symmetry vector is a vector space can have many different inner products each... The vectors u ; v 2R n, defines an inner product times and... Axioms are satisfied as follows space or unitary space as the length of a vector space defined in section. Matrix, its complex conjugate f ( x ) ∈ c with x ∈ [ 0, ]! Standard dot product is given by Laws governing inner products allow the introduction! Built-In function for the Hermitian inner product requires the same direction as now, we denote this operation as Generalizations..., each element of one of the usual inner product ), each element of one of the product. Course if imaginary component is 0 then this is a Hermitian inner product a finite dimensional vector involves... As an inner product for complex vector spaces, conjugate symmetry of inner. Products on R defined in this way are called symmetric bilinear form 2012 test set should some! Dot products, lengths, and complex vector space in which an inner.... Defined for different dimensions, while the inner or inner product a. And g isdefinedtobe f, g = 1 inner product is defined the! Bilinear form suppose we have a vector space involves the conjugate of the vector is a vector with..... _ 4j ) and positive examples of Hilbert spaces, we can not copy this definition directly semi-definite on. Than the conventional mathematical notation we have a vector space four properties of a vector space with inner..., lengths, and be vectors and be two vectors products of complex function f ( x ∈... '' is opposed to outer product is actually we call a Hilbert space ( . In higher dimensions a sum product over the field of complex numbers a scalar,:. Representation of semi-definite kernels on arbitrary sets component of the second vector the inner... Space over the last axes { R } \ ) equals dot (,. Y and Z be complex n-vectors this way are called symmetric bilinear form is 0 | {
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R } \ ) equals dot (,. Y and Z be complex n-vectors this way are called symmetric bilinear form is 0 then this is straight for! Referred to as unitary spaces u, v +wi = hu, vi+hu,.... Provide the means of defining orthogonality between vectors ( zero inner product two. Definition is changed slightly vector spaces v_ >: = u.A.Conjugate [ v ] a! Operation as: Generalizations complex vectors this definition makes sense to calculate length '' so it. Dot products, lengths, and distances of complex vectors is used i.e., its complex part is zero and. The inner product of the concept of a vector with itself is real ( i.e., its names not! Vertical and shrinks down, outer is vertical times horizontal and expands out.! Y: numeric or complex matrices or vectors of a vector with itself this construction is used definition changed... The test suite only has row vectors, such that dot ( v, u ) definition: the function. Any vectors u ; v 2R n, defines an inner product space '' ) vector or angle. But we will only need for this book ( complex length-vectors, and complex scalars ), one the! | {
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# How many different ways are there to choose 5 items from 12 distinct items if…
How many different ways are there to choose 5 items from 12 distinct items if…
a. (5 pt.) the order of the items matters and repetition of items is not allowed?
b. (5 pt.) the order of the items matters and repetition of items is allowed?
c. (5 pt.) the order of the items does not matter and repetition of items is not allowed?
d. (5 pt.) the order of the items does not matter and repetition of items is allowed?
My Work
a) $P(12,5)$
b)$12^5$
c)$\binom{12}{5}$
Can you please verify my work
• I don't understand your reasoning for $b$. Hint: there are $12$ possibilities for the first choice, $12$ possibilities for the second choice, ... – lulu Apr 25 '18 at 23:39
• So, the b is $12^5$ right – tien lee Apr 25 '18 at 23:41
• Yes, that is right. For $d$, I think Stars and Bars is the way to go. – lulu Apr 25 '18 at 23:42
• Then, what about C – tien lee Apr 25 '18 at 23:45
• Your answers for $a,c$ are correct,. – lulu Apr 25 '18 at 23:46
a) Correct
b) No. You have 12 choices on the first item. 12 on the second. 12 on the third, fourth and fifth. So it should be $12^5$.
c) Correct
d) This is a bit tricky, I shall explain this below.
We have $C$ for combination, $P$ for permutation, and the first time I was taught this one (order does not matter + repetition allowed), my teacher called it $H$ although I am not sure if it is standard notation.
Anyway, the formula in general is $H_r^{n}=C_r^{n+r-1}$.
The reasoning behind this is as follows (for simplicity we shall take $H_5^{12}$ as in this case as an example):
Let the $12$ possible choices be $a,b,c,\dots,l$.
We don't care about the order here, we only care about how many of each item you choose, and with each choice, we can assign with it a unique "binary code". | {
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For example, if you choose $aabbc$, then we write $0010010111111111$ where the first two $0$'s represent the two $a$'s, after the $1$ we have two $0$'s for the two $b$'s, and so on.
As another example, $abjkl$ is represented by $0101111111101010$.
In this way, each possible choice can be represented by a string of five $0$'s and eleven $1$'s for a total of $16$ numbers. So to count all possible choices, we simply have to choose the five places to insert the $0$'s.
This is given by $C_5^{16}$, hence $H_5^{12}=C_5^{16}$
• Isn't $C(16,4)$ – tien lee Apr 25 '18 at 23:57
• Why should it be $C_4^{16}$? – glowstonetrees Apr 26 '18 at 0:07
• @tienlee If the number of items of type $k$ is $x_k$, then we want the number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_{12} = 5$$ in the nonnegative integers. The number of such solutions is $$\binom{12 + 5 - 1}{12 - 1} = \binom{12 + 5 - 1}{5}$$ where the expression $\binom{12 + 5 - 1}{12 - 1}$ counts the number of ways of choosing which $12 - 1 = 11$ positions of $12 + 5 - 1$ positions required for five ones and eleven addition signs will be filled with addition signs and the expression $\binom{12 + 5 - 1}{5}$ comes from choosing the positions of the ones in the same row. – N. F. Taussig Apr 26 '18 at 0:19
I will leave you with a quick guide on how to tackle these problems.
Suppose you have $n$ distinct objects, and you want to select $r$ out of these $n$ objects. Then, if: | {
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• Order is relevant and repetition is not allowed: look at the possible permutations of size $r$ which is given by $$P(n,r)=\frac{n!}{(n-r)!}$$ with $0\leq r \leq n.$
• Order is relevant and repetition is allowed: look at the possible arrangements of size $r$, which is given by $n^r$, with $n,r \geq 0.$
• Order is not relevant and repetitions are not allowed: look at the possible combinations of size $r,$ which is given by $$C(n,r)= \binom{n}{r}=\frac{n!}{(n-r)! r!}$$ with $0 \leq r \leq n.$
• Order is not relevant and repetition is allowed: look at the possible combinations with repetitions, which is given by $$\binom{n+r-1}{r}\ \$$ with $n,r \geq 0.$ | {
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# Two-dimensional functions
• Difficulty: 010/100
• Author: jeremy theler
• Keywords: FUNCTION, :=, FILE_PATH, DATA, INTERPOLATION, nearest, rectangle, PRINT_FUNCTION, MIN, MAX, STEP, NUMBER, FILE, OUTPUT_FILE,
These examples illustrate the facilities wasora provides to interpolate two-dimensional functions. As shown in section ref{007-functions}, multidimensional functions may be defined using algebraic expressions, inline data, external files, vectors or dynamically-loaded routines. The main focus of this section is to illustrate the difference between algebraic and pointwise-defined two-dimensional functions, and to further illustrate the difference between a nearest-neighbor interpolation (based on a $$k$$-dimensional tree structure) and a rectangle interpolation (based on finite-elements-like shape functions for quadrangles).
## paraboloid.was
This input defines an algebraic function $$f(x,y)$$ and uses PRINT_FUNCTION to dump its contents (as three columns containing $$x$$, $$y$$ and $$f(x,y)$$ as shown in the terminal mimic) to the standard output. The range is mandatory because $$f(x,y)$$ is defined by an algebraic expression.
As far as the function $$f(x,y)$$ is concerned, $$a$$ and $$b$$ are taken as parameters. Even though they can change over time, the value they take is the value they have when $$f(x,y)$$ is evaluated. In this case, $$f(x,y)$$ is evaluated when the instruction PRINT_FUNCTION is executed, so the output is written with $$b=2$$.
a = 1
b = 1
f(x,y) := (x/a)^2 + (y/b)^2
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a = 1
b = 1
f(x,y) := (x/a)^2 + (y/b)^2
b = 2
# range is mandatory as f(x,y) is algebraically-defined
PRINT_FUNCTION f MIN -b -b MAX b b STEP b/20 b/20
$wasora paraboloid.was > paraboloid.dat$ head paraboloid.dat
-2.000000e+00 -2.000000e+00 5.000000e+00
-2.000000e+00 -1.900000e+00 4.902500e+00
-2.000000e+00 -1.800000e+00 4.810000e+00
-2.000000e+00 -1.700000e+00 4.722500e+00
-2.000000e+00 -1.600000e+00 4.640000e+00
-2.000000e+00 -1.500000e+00 4.562500e+00
-2.000000e+00 -1.400000e+00 4.490000e+00
-2.000000e+00 -1.300000e+00 4.422500e+00
-2.000000e+00 -1.200000e+00 4.360000e+00
-2.000000e+00 -1.100000e+00 4.302500e+00
$gnuplot paraboloid.gp$
## nearest.was
This input defines a two-dimensional pointwise-defined function $$g(x,y)$$ using data inlined in the input file using the DATA keyword. As can be seen, expressions are allowed. However, they are evaluated at parse-time so references to variables should be avoided, as they will default to zero. For that end, numbers defined with the NUMBER keyword should be used. By default, pointwise-defined multidimensional functions are interpolated by the nearest neighbor algorithm (i.e. default is INTERPOLATION nearest). When calling PRINT_FUNCTION with no range, the definition points are printed. When a range is given, the function gets evaluated at the grid points.
In this case, first the function $$g(x,y)$$ is dumped into a file called g_def.dat with no range. Then, the function is dumped into a file called g_int.dat using a range and thus resulting in an interpolated output using nearest neighbors.
FUNCTION g(x,y) DATA {
0 0 1-1
0 1 1-0.5
0 2 1
1 0 1
1 1 1+0.25
1 2 1
2 0 1-0.25
2 1 1+0.25
2 2 1+0.5
}
# print g(x,y) at the definition points
PRINT_FUNCTION g FILE_PATH g_def.dat | {
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# print g(x,y) at the definition points
PRINT_FUNCTION g FILE_PATH g_def.dat
# print g(x,y) at the selected range
# by defaults wasora interpolates using nearest neighbors
PRINT_FUNCTION g FILE_PATH g_int.dat MIN 0 0 MAX 2 2 STEP 0.05 0.05
$wasora nearest.was$ gnuplot nearest.gp
$ ## rectangle.was This time, a function $$h(x,y)$$ is defined by reading point-wise data from a file. This file is g_def.dat which are the definition points of the function $$g(x,y)$$ of the previous example. Note that when reading function data from a file, no expressions are allowed. Function $$h(x,y)$$ is interpolated using the rectangle method. The interpolated data is written in a file called h_int.dat, in which the function $$h(x,y)$$ is evaluated at the very same points the function $$g(x,y)$$ of the previous example was. # h(x,y) is equal to g(x,y) at the definition points, # but it is interpolated differently FUNCTION h(x,y) INTERPOLATION rectangle FILE_PATH g_def.dat # print h(x,y) at the selected range PRINT_FUNCTION h FILE_PATH h_int.dat MIN 0 0 MAX 2 2 STEP 0.05 0.05 $ cat g_def.dat
0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 1.000000e+00 5.000000e-01
0.000000e+00 2.000000e+00 1.000000e+00
1.000000e+00 0.000000e+00 1.000000e+00
1.000000e+00 1.000000e+00 1.250000e+00
1.000000e+00 2.000000e+00 1.000000e+00
2.000000e+00 0.000000e+00 7.500000e-01
2.000000e+00 1.000000e+00 1.250000e+00
2.000000e+00 2.000000e+00 1.500000e+00 | {
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$wasora rectangle.was$ gnuplot rectangle.gp
$ ## scattered.was The last example of two-dimensional interpolation involves a pointwise-defined function~$$s(x,y)$$ using scattered data, i.e. not necessarily over a rectangular grid. In this case—and with no information about any underlying finite-element-like mesh—wasora can use either a nearest-neighbor interpolation or a Shepard-like inverse distance weighting. For the original Shepard method, the only parameter than can be tweaked is the exponent~$$p$$ of the distance in the weight~$$w_i=1/d_i^p$$. For the modified Shepard algorithm, the radius~$$r$$ of the nearest neighbors taken into account is to be provided. These nearest neighbors are found using a $$k$$-dimensional tree, that is a very efficient way of doing this task. For complex functions, all the alternatives should be investigated taking into account accuracy and code speed. # scattered multidimensional data may be interpolated # using a nearest-neighbor approach FUNCTION n(x,y) INTERPOLATION nearest DATA { 0 0 0 1 0 1 0 1 2 -0.5 0.5 3 -1 -1 2 0.75 0 1.5 0.25 0.25 1 } # another way of giving the same set of data VECTOR datax SIZE 7 DATA 0 1 0 -0.5 -1 0.75 0.25 VECTOR datay SIZE 7 DATA 0 0 1 0.5 -1 0 0.25 VECTOR dataz SIZE 7 DATA 0 1 2 3 2 1.5 1 # using shepard's interpolation method FUNCTION s(x,y) VECTORS datax datay dataz INTERPOLATION shepard SHEPARD_EXPONENT 4 # or using shepard's modified algorithm FUNCTION m(x,y) VECTORS datax datay dataz INTERPOLATION modified_shepard SHEPARD_RADIUS 2 # print the definition points PRINT_FUNCTION n FILE_PATH n_def.dat # print the different functions at the selected range PRINT_FUNCTION n s m FILE_PATH n_int.dat MIN -1 -1 MAX 1.5 1.5 STEP 0.05 0.05 $ wasora scattered.was
$gnuplot scattered.gp$ | {
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The figures illustrate how the multidimensional data interpolation scheme work for pointwise defined functions over scattered data. Nearest neighbors give constant values for each voronoi triangle whilst Shepard-based algorithms provide continuous surfaces.
## compwater.was
This example shows an extension of the example about saturated water in section ref{007-functions} by giving properties of compressed water as a function of pressure $$p$$ and temperature $$T$$. The file compwater.txt contains some properties of water as a function of temperature and pressure in a rectangular grid over the pressure-temperature space. The enthalpy is not contained in the file, but it can be computed from pressure $$p$$, the internal energy $$u(p,T)$$ and the specific volume $$v(p,T)$$ as
!bt [ h(p,t) = u(p,T) + p v(p,T) ] !et
FUNCTION v(p,T) FILE_PATH compwater.txt COLUMNS 2 1 4 INTERPOLATION rectangular
FUNCTION u(p,T) FILE_PATH compwater.txt COLUMNS 2 1 5 INTERPOLATION rectangular
FUNCTION h(p,T) = u(p,T) + p*v(p,T)
PRINT_FUNCTION h MIN 1e5 300 MAX 200e5 1000 STEP 2e5 4
$wasora compwater.was > compwater.dat$ gnuplot compwater.gp
\$
The figure shows the enthalpy of compressed water as a continuous function of $$p$$ and $$T$$ and the discrete experimental data. | {
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# Why does Newton's first law create two different answers to this question?
I have been having some difficulty with a recent question. Take the following pulley system:
The three blocks have masses $$M$$, $$m_1$$, and $$m_2$$. All are subjected to a gravitational force $$g$$. The pulley and string are of negligible mass, and all surfaces are frictionless. The problem is (to paraphrase):
What magnitude of force $$F$$ is necessary for $$m_1$$ and $$m_2$$ to be motionless relative to $$M$$?
When I first solved this, I just considered Newton's first law for each of the three masses and added the additional conditions $$a_{xm_1}=a_{xM}$$, $$a_{xm_2}=a_{xM}$$, $$a_{ym_1}=0$$. After eliminating most of the variables, I ended up with $$F=g\frac{m_1}{m_2}(M+m_1)$$.
However, the textbook gives the answer $$F=g\frac{m_1}{m_2}(M+m_1+m_2)$$. In attempting to find the discrepancy, I solved the problem again somewhat differently: Let $$a=a_{xM}=a_{xm_1}=a_{xm_2}$$. Since $$m_1$$ does not accelerate vertically, $$T=m_1g$$. Since $$m_2a_{xm_2}=T=m_1g$$, we have that $$a=g\frac{m_1}{m_2}$$. Finally, since $$F$$ is pushing on $$M$$, which in turn is pushing on $$m_1$$ via its normal-force interaction, we have $$F=(M+m_1)a=g\frac{m_1}{m_2}(M+m_1)$$, the same answer I previously came to.
I asked my teacher about this problem to determine where my error occurred, and his reasoning was as follows: As with the earlier reasoning, $$a=g\frac{m_1}{m_2}$$. Since $$M$$, $$m_1$$, and $$m_2$$ are motionless relative to each other, we can treat the three as a single system and ignore internal forces:
Now, we simply have $$F=M_sa=g\frac{m_1}{m_2}(M+m_1+m_2)$$.
Both lines of reasoning are compelling, so my overall question is this: Which of these two answers is correct, and how is the other answer incorrect? | {
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• It seems that I indeed forgot to account for the force on $M$ from the pulley. Adding that to the equations produces the correct result. I've accepted this answer, even though both are equally valid, since it is the earlier one. – LegionMammal978 Sep 6 '19 at 20:28
You have made a mistake in the equation $$F=(M+m_1)a$$ You have missed the fact that the tension is also works on M.! The sum of the two tensions add up and produce a force on the pulley. However, as the pulley is(should be) massless, the block M experiences a opposite force, which has a horizontal component of $$T√2*cos45° = T$$ So the equation should be $$F-T=(M+m_1)a$$ which then gives same answer as your teacher. | {
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# Show a specially defined matrix is positive definite
Let $$E_1, ..., E_n$$ be non empty finite sets. Show that the matrix $$A = (A_{ij})_{1 \leq i, j \leq n}$$ defined by $$A_{ij} = \dfrac{|E_i \cap E_j|}{|E_i \cup E_j|}$$, is positive semi-definite.
This is part 5 of Exercise 1 in http://members.cbio.mines-paristech.fr/~jvert/svn/kernelcourse/homework/2019mva/hw.pdf .
• If any two of the sets are empty then the corresponding coefficient is not defined. If the one set is disjoint from all the others, then its corresponding row and column are zero, so the matrix is not positive definite. – Servaes Mar 9 '19 at 18:53
• You are correct, I've corrected the problem accordingly. – SiXUlm Mar 9 '19 at 21:14
• Could the semi-definite positiveness be connected to the submodularity mentionned in the following answer : math.stackexchange.com/q/182384 ? – Jean Marie Mar 9 '19 at 23:09
• Let $E$ be the union of the $E_i$. For each $e \in E$, define the matrix $B_e = \left( \dfrac{\left[e \in E_i \cap E_j\right]}{\left|E_i \cup E_j\right|} \right)_{i, j \in \left[n\right]}$, where square brackets stand both for the Iverson bracket and for $\left[n\right] := \left\{1,2,\ldots,n\right\}$. Clearly, $A = \sum_{e \in E} B_e$. Are the $B_e$ still positive semidefinite? This would likely be easier if true. After all, for each $e \in E$, the matrix $B_e$ is just a blown-up matrix with entries $\dfrac{1}{\left|E_i \cup E_j\right|}$. – darij grinberg Mar 10 '19 at 3:01
• Even stronger: For any $e \in E$ and $t \in \mathbb{R}_+$, let $C_{e,t}$ be the matrix $\left(\left[e \in E_i \cap E_j\right] t^{\left|E_i \cup E_j\right|}\right)_{i, j \in \left[n\right]}$. Is $C_{e,t}$ nonnegative semidefinite for $0 < t < 1$ ? If it is, then so is $B_e$, since $B_e = \int_0^1 \left(C_{e,t}/t\right) dt$. – darij grinberg Mar 10 '19 at 3:08
Assume the sets $$E_i$$ are all non-empty. By deleting rows and columns of $$A$$ if needed one can reduce to the case where the $$E_i$$ are all distinct. | {
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Write $$A_{ij}= \frac{|E_i\cap E_j|}{|E_i\cup E_j|} = \frac{|E_i\cap E_j|}{|E_i|+|E_j|-|E_i\cap E_j|} = \frac{e_{ij}}{e_i+e_j-e_{ij}}$$ where $$e_i$$ is shorthand for $$|E_i|$$ and so on. Then $$A_{ij}= \frac{e_{ij}}{e_i+e_j} \left( 1 + r_{ij} + r_{ij}^2 + r_{ij}^3 + \cdots\right)\tag{*}$$ where $$r_{ij}= e_{ij}/(e_i+e_j).$$ The distinctness hypothesis implies $$|r_{ij}|<1$$ and so the convergence of the geometric series in (*).
The matrix with entries $$e_{ij}$$ is a Gram matrix and hence positive semidefinite. The matrix with entries $$1/(e_i+e_j) = \int_0^\infty e^ {-x|E_i|} e^{-x|E_j|} \,dx$$ is also a Gram matrix so it too is psd. Schur's product theorem tells us that their elementwise product $$r_{ij}$$ is thus also psd. By Schur again, all the summand matrices in (*) are psd. So the sum, $$A$$, is psd.
Examples (where all the $$E_i$$ are equal, so all $$A_{ij}=1$$, say) show that positive semi definite cannot be replaced by positive definite as in an earlier form of the problem statement. I suppose that strict positive definiteness holds in the reduced case, but I don't see an argument. Numerical experiments show that the matrix $$(r_{ij})$$ is not strictly positive definite when (for example) the $$E_i$$ are all the 2-element subsets of $$[4]$$. The same experiments however do show $$A$$ strictly positive definite.
• Thanks for your solution. I've edited the question because I didn't state it correctly. – SiXUlm Mar 9 '19 at 21:28
• Thanks for the clarification. I've just found out that you talked about Schur product theorem. – SiXUlm Mar 9 '19 at 21:35
• [+1] Very smart ! – Jean Marie Mar 9 '19 at 22:37
• Why is $(e_{ij})=(|E_i\cap E_j|)$ a Gram matrix? – William McGonagall Feb 14 at 12:32
• @WilliamMcGonagall Because $e_{ij}=\langle v_i,v_j\rangle$, where $v_r$ denotes the vector whose $k$-th component is $1$ exactly when $k\in E_r$, and zero otherwise, and so on. – kimchi lover Feb 14 at 14:21 | {
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The following proof is inspired by my answer to math.stackexchange question https://math.stackexchange.com/q/1340405/, which in turn is inspired by the probabilistic method from extremal combinatorics. It is fully elementary and almost combinatorial ("almost" because it involves a simple limit argument at one point).
We begin with notations:
Let $$\mathbb{N}$$ be the set $$\left\{ 0,1,2,\ldots\right\}$$.
For each $$n\in\mathbb{N}$$, let $$\left[ n\right]$$ denote the set $$\left\{ 1,2,\ldots,n\right\}$$.
We also recall the following simple fact (Lemma 5 in my answer to math.stackexchange question https://math.stackexchange.com/q/1340405/):
Lemma 1. Let $$Q$$ be a finite totally ordered set. Let $$J$$ be a subset of $$Q$$. Let $$r\in J$$. Let $$S$$ be the set of all permutations of $$Q$$. Then, $$$$\left\vert \left\{ \sigma\in S\ \mid\ \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right\} \right\vert =\dfrac{\left\vert S\right\vert }{\left\vert J\right\vert }.$$$$
Now, I shall prove a first positive definiteness statement:
Theorem 2. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ finite sets. Let $$x_{1},x_{2},\ldots,x_{n}$$ be $$n$$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1}\geq0. \end{align}
Proof of Theorem 2. Fix some object $$r$$ that belongs to none of the sets $$E_{1},E_{2},\ldots,E_{n}$$. Let $$Q$$ be the set $$E_{1}\cup E_{2}\cup\cdots\cup E_{n}\cup\left\{ r\right\}$$. This is a finite set (since the sets $$E_{1},E_{2},\ldots,E_{n},\left\{ r\right\}$$ are finite). We fix some total order on $$Q$$.
Let $$S$$ be the set of all permutations of $$Q$$. This $$S$$ is a finite nonempty set (of size $$\left\vert Q\right\vert !$$). Thus, $$\left\vert S\right\vert$$ is a positive integer. | {
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If $$u\in\left[ n\right]$$ and $$\sigma\in S$$, then we say that $$\sigma$$ is $$u$$-friendly if we have $$\left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right)$$. Now, we claim the following:
For any $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we have $$$$\left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert =\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}. \label{darij1.pf.t2.1} \tag{1}$$$$
[Proof of \eqref{darij1.pf.t2.1}: Fix $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$. Define a subset $$J$$ of $$Q$$ by $$J=E_{u}\cup E_{v} \cup\left\{ r\right\}$$. (This is well-defined, since the definition of $$Q$$ shows that $$E_{u}$$, $$E_{v}$$ and $$\left\{ r\right\}$$ are subsets of $$Q$$.)
The object $$r$$ belongs to none of the sets $$E_{1},E_{2},\ldots,E_{n}$$. Thus, in particular, $$r$$ belongs neither to $$E_{u}$$ nor to $$E_{v}$$. In other words, $$r\notin E_{u}\cup E_{v}$$. This yields \begin{align} E_{u}\cup E_{v}=\underbrace{\left( E_{u}\cup E_{v}\cup\left\{ r\right\} \right) }_{=J}\setminus\left\{ r\right\} =J\setminus\left\{ r\right\} . \end{align} Also, $$r\in J$$; thus, $$\left\vert J\setminus\left\{ r\right\} \right\vert =\left\vert J\right\vert -1$$. Now, from $$E_{u}\cup E_{v}=J\setminus\left\{ r\right\}$$, we obtain $$\left\vert E_{u}\cup E_{v}\right\vert =\left\vert J\setminus\left\{ r\right\} \right\vert =\left\vert J\right\vert -1$$, so that $$\left\vert J\right\vert =\left\vert E_{u}\cup E_{v}\right\vert +1$$. | {
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For each $$\sigma\in S$$, we have the following chain of logical equivalences: \begin{align*} & \ \left( \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right) \\ & \Longleftrightarrow\ \underbrace{\left( \sigma\text{ is }u\text{-friendly} \right) }_{\substack{\Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right) \\\text{(by the definition of "}u\text{-friendly")}}}\wedge\underbrace{\left( \sigma\text{ is }v\text{-friendly}\right) }_{\substack{\Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{v}\right) \\\text{(by the definition of "}v\text{-friendly")}}}\\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{u}\right) \wedge\left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in E_{v}\right) \\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in\underbrace{E_{u}\cup E_{v}}_{=J\setminus \left\{ r\right\} }\right) \\ & \Longleftrightarrow\ \left( \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right) . \end{align*} Hence, \begin{align*} & \left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert \\ & =\left\vert \left\{ \sigma\in S\ \mid\ \sigma\left( r\right) >\sigma\left( j\right) \text{ for all }j\in J\setminus\left\{ r\right\} \right\} \right\vert \\ & =\dfrac{\left\vert S\right\vert }{\left\vert J\right\vert }\qquad\left( \text{by Lemma 1}\right) \\ & =\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\qquad\left( \text{since }\left\vert J\right\vert =\left\vert E_{u}\cup E_{v}\right\vert +1\right) . \end{align*} This proves \eqref{darij1.pf.t2.1}.] | {
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Now, for each $$\sigma\in S$$, we have \begin{align*} & \left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}\\ & =\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) \left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right)\\ & =\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly} }}x_{u}\right) \left( \sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly}}}x_{v}\right) \\ & \qquad\left( \begin{array} [c]{c} \text{here, we have renamed the summation}\\ \text{index }u\text{ as }v\text{ in the second sum} \end{array} \right) \\ & =\sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly} }}\sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly} }}x_{u}x_{v}. \end{align*} Summing up these equalities over all $$\sigma\in S$$, we obtain \begin{align*} & \sum_{\sigma\in S}\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}\\ & =\underbrace{\sum_{\sigma\in S}\sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}\sum_{\substack{v\in\left[ n\right] ;\\\sigma\text{ is }v\text{-friendly}}}}_{=\sum\limits_{u\in\left[ n\right] } \sum\limits_{v\in\left[ n\right] }\sum\limits_{\substack{\sigma\in S;\\\sigma\text{ is }u\text{-friendly}\\\text{and }v\text{-friendly}}}}x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\underbrace{\sum _{\substack{\sigma\in S;\\\sigma\text{ is }u\text{-friendly}\\\text{and }v\text{-friendly}}}x_{u}x_{v}}_{=\left\vert \left\{ \sigma\in S\ \mid \ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert \cdot x_{u}x_{v}}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] } \underbrace{\left\vert \left\{ \sigma\in S\ \mid\ \sigma\text{ is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert } _{\substack{=\dfrac{\left\vert | {
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is }u\text{-friendly and }v\text{-friendly}\right\} \right\vert } _{\substack{=\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\\\text{(by \eqref{darij1.pf.t2.1})}}}\cdot x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left\vert S\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert +1}\cdot x_{u}x_{v}\\ & =\left\vert S\right\vert \cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1}. \end{align*} Hence, \begin{align} \left\vert S\right\vert \cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1} =\sum_{\sigma\in S}\underbrace{\left( \sum_{\substack{u\in\left[ n\right] ;\\\sigma\text{ is }u\text{-friendly}}}x_{u}\right) ^{2}}_{\substack{\geq 0\\\text{(since squares are nonnegative)}}}\geq0. \end{align} We can divide this inequality by $$\left\vert S\right\vert$$ (since $$\left\vert S\right\vert$$ is a positive integer). We thus obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1}\geq0. \end{align} This proves Theorem 2. $$\blacksquare$$ | {
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Now, we apply the tensor power trick. The first step is to replace the $$1$$ in Theorem 2 by a small rational number $$1/m$$:
Corollary 3. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ finite sets. Let $$x_{1},x_{2},\ldots,x_{n}$$ be $$n$$ real numbers. Let $$m$$ be a positive integer. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align} | {
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Proof of Corollary 3. For each $$i\in\left[ n\right]$$, we define a finite set $$B_{i}$$ by $$B_{i}=E_{i}\times\left\{ 1,2,\ldots,m\right\}$$. Then, for every $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we have \begin{align*} & \underbrace{B_{u}}_{\substack{=E_{u}\times\left\{ 1,2,\ldots,m\right\} \\\text{(by the definition of }B_{u}\text{)}}}\cup\underbrace{B_{v} }_{\substack{=E_{v}\times\left\{ 1,2,\ldots,m\right\} \\\text{(by the definition of }B_{v}\text{)}}}\\ & =\left( E_{u}\times\left\{ 1,2,\ldots,m\right\} \right) \cup\left( E_{v}\times\left\{ 1,2,\ldots,m\right\} \right) \\ & =\left( E_{u}\cup E_{v}\right) \times\left\{ 1,2,\ldots,m\right\} \end{align*} and therefore \begin{align} \left\vert B_{u}\cup B_{v}\right\vert & =\left\vert \left( E_{u}\cup E_{v}\right) \times\left\{ 1,2,\ldots,m\right\} \right\vert =\left\vert E_{u}\cup E_{v}\right\vert \cdot\underbrace{\left\vert \left\{ 1,2,\ldots ,m\right\} \right\vert }_{=m}\nonumber\\ & =\left\vert E_{u}\cup E_{v}\right\vert \cdot m. \label{darij1.pf.c3.1} \tag{2} \end{align} But Theorem 2 (applied to $$B_{i}$$ instead of $$E_{i}$$) yields $$$$\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert B_{u}\cup B_{v}\right\vert +1}\geq0. \label{darij1.pf.c3.2} \tag{3}$$$$ For each $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we have \begin{align*} \dfrac{x_{u}x_{v}}{\left\vert B_{u}\cup B_{v}\right\vert +1} & =\dfrac {x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert \cdot m+1}\qquad\left( \text{by \eqref{darij1.pf.c3.1}}\right) \\ & =\dfrac{1}{m}\cdot\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}. \end{align*} Adding up these equalities for all $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we obtain \begin{align*} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert B_{u}\cup B_{v}\right\vert +1} & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{1}{m}\cdot\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup | {
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n\right] }\sum_{v\in\left[ n\right] }\dfrac{1}{m}\cdot\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\\ & =\dfrac{1}{m}\cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}. \end{align*} Thus, \eqref{darij1.pf.c3.2} rewrites as \begin{align} \dfrac{1}{m}\cdot\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align} We can multiply this inequality by $$m$$ (since $$m$$ is positive), and thus obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\geq0. \end{align} This proves Corollary 3. $$\blacksquare$$ | {
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The second part of the tensor power trick is to let $$m\rightarrow\infty$$:
Theorem 4. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ nonempty finite sets. Let $$x_{1},x_{2},\ldots,x_{n}$$ be $$n$$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert }\geq0. \end{align}
Proof of Theorem 4. First of all, we notice that $$E_{u}\cup E_{v}$$ is a nonempty finite set whenever $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$ (since $$E_{1},E_{2},\ldots,E_{n}$$ are nonempty finite sets), and thus its size $$\left\vert E_{u}\cup E_{v}\right\vert$$ is a positive integer. Thus, all the fractions $$\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert }$$ in Theorem 4 are well-defined. | {
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Now, consider a positive integer $$m$$ going to infinity. Then, $$\lim \limits_{m\rightarrow\infty}\dfrac{r}{q+1/m}=\dfrac{r}{q}$$ for every real $$r$$ and every positive real $$q$$. Hence, \begin{align} \lim\limits_{m\rightarrow\infty}\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}=\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v} \right\vert } \end{align} for every $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$. Adding up these equalities for all $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we obtain \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\lim \limits_{m\rightarrow\infty}\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}=\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert }. \end{align} Hence, \begin{align*} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v} }{\left\vert E_{u}\cup E_{v}\right\vert } & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\lim\limits_{m\rightarrow\infty}\dfrac {x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}\\ & =\lim\limits_{m\rightarrow\infty}\underbrace{\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{x_{u}x_{v}}{\left\vert E_{u}\cup E_{v}\right\vert +1/m}}_{\substack{\geq0\\\text{(by Corollary 3)}}}\geq \lim\limits_{m\rightarrow\infty}0=0. \end{align*} This proves Theorem 4. $$\blacksquare$$
Now, let us recall the Iverson bracket notation:
Definition. We shall use the Iverson bracket notation: If $$\mathcal{A}$$ is any statement, then $$\left[ \mathcal{A}\right]$$ stands for the integer $$\begin{cases} 1, & \text{if \mathcal{A} is true;}\\ 0, & \text{if \mathcal{A} is false} \end{cases}$$ (which is also known as the truth value of $$\mathcal{A}$$). For instance, $$\left[ 1+1=2\right] =1$$ and $$\left[ 1+1=1\right] =0$$. | {
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Corollary 5. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ nonempty finite sets. Let $$e$$ be any object. Let $$x_{1},x_{2},\ldots,x_{n}$$ be $$n$$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u} x_{v}\geq0. \end{align}
Proof of Corollary 5. For every $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we have \begin{align*} \left[ e\in E_{u}\cap E_{v}\right] & =\left[ e\in E_{u}\text{ and }e\in E_{v}\right] \\ & \qquad\left( \text{since }e\in E_{u}\cap E_{v}\text{ holds if and only if }\left( e\in E_{u}\text{ and }e\in E_{v}\right) \right) \\ & =\left[ e\in E_{u}\right] \cdot\left[ e\in E_{v}\right] \end{align*} (because the rule $$\left[ \mathcal{A}\text{ and }\mathcal{B}\right] =\left[ \mathcal{A}\right] \cdot\left[ \mathcal{B}\right]$$ holds for any two statements $$\mathcal{A}$$ and $$\mathcal{B}$$) and therefore \begin{align*} \dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} & =\dfrac{\left[ e\in E_{u}\right] \cdot\left[ e\in E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert } x_{u}x_{v}\\ & =\dfrac{\left( \left[ e\in E_{u}\right] x_{u}\right) \cdot\left( \left[ e\in E_{v}\right] x_{v}\right) }{\left\vert E_{u}\cup E_{v} \right\vert }. \end{align*} Adding up these equalities for all $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we obtain \begin{align*} & \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert } x_{u}x_{v}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left( \left[ e\in E_{u}\right] x_{u}\right) \cdot\left( \left[ e\in E_{v}\right] x_{v}\right) }{\left\vert E_{u}\cup E_{v}\right\vert }\geq0 \end{align*} (by Theorem 4, applied to $$\left[ e\in E_{i}\right] x_{i}$$ instead of $$x_{i}$$). This proves Corollary 5. $$\blacksquare$$ | {
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We next recall a classical fact (I refer to it as "counting by roll call"):
Lemma 6. Let $$Q$$ be a finite set. Let $$F$$ be a subset of $$Q$$. Then, \begin{align} \left\vert F\right\vert =\sum_{e\in Q}\left[ e\in F\right] . \end{align}
Proof of Lemma 6. The sum $$\sum_{e\in Q}\left[ e\in F\right]$$ has exactly $$\left\vert F\right\vert$$ many addends equal to $$1$$ (namely, all the addends corresponding to $$e\in F$$), while all its remaining addends are $$0$$ and thus do not influence its value. Hence, $$\sum_{e\in Q}\left[ e\in F\right] =\left\vert F\right\vert \cdot1=\left\vert F\right\vert$$. This proves Lemma 6. $$\blacksquare$$
Theorem 7. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ nonempty finite sets. Let $$x_{1},x_{2},\ldots,x_{n}$$ be $$n$$ real numbers. Then, \begin{align} \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u} x_{v}\geq0. \end{align}
Proof of Theorem 7. Let $$Q$$ be the set $$E_{1}\cup E_{2}\cup\cdots\cup E_{n}$$. This is a finite set (since the sets $$E_{1},E_{2},\ldots,E_{n}$$ are finite).
Fix $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$. Then, $$E_{u}\cap E_{v}$$ is a subset of $$Q$$ (since $$Q=E_{1}\cup E_{2}\cup\cdots\cup E_{n}$$). Hence, Lemma 6 (applied to $$F=E_{u}\cap E_{v}$$) yields \begin{align} \left\vert E_{u}\cap E_{v}\right\vert =\sum_{e\in Q}\left[ e\in E_{u}\cap E_{v}\right] . \end{align} Thus, \begin{align} \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} & =\dfrac{\sum_{e\in Q}\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\nonumber\\ & =\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}. \label{darij1.pf.t7.1} \tag{4} \end{align} | {
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Now, forget that we fixed $$u$$ and $$v$$. We have \begin{align*} & \sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\underbrace{\dfrac {\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}}_{\substack{=\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\\\text{(by \eqref{darij1.pf.t7.1})}}}\\ & =\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\sum_{e\in Q}\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}=\sum_{e\in Q}\underbrace{\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] }\dfrac{\left[ e\in E_{u}\cap E_{v}\right] }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v} }_{\substack{\geq0\\\text{(by Corollary 5)}}}\\ & \geq\sum_{e\in Q}0=0. \end{align*} This proves Theorem 7. $$\blacksquare$$
We are now ready for the original question:
Corollary 8. Let $$n\in\mathbb{N}$$. Let $$E_{1},E_{2},\ldots,E_{n}$$ be $$n$$ nonempty finite sets. Let $$A$$ be the $$n\times n$$-matrix $$\left( \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }\right) _{u,v\in\left[ n\right] }\in\mathbb{R}^{n\times n}$$. Then, $$A$$ is positive semidefinite.
Proof of Corollary 8. For every $$u\in\left[ n\right]$$ and $$v\in\left[ n\right]$$, we have $$\dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }=\dfrac{\left\vert E_{v}\cap E_{u}\right\vert }{\left\vert E_{v}\cup E_{u}\right\vert }$$ (since $$E_{u}\cap E_{v}=E_{v}\cap E_{u}$$ and $$E_{u}\cup E_{v}=E_{v}\cup E_{u}$$). Thus, the matrix $$A$$ is symmetric. Hence, in order to prove that $$A$$ is positive semidefinite, it suffices to show that $$x^{T}Ax\geq0$$ for any vector $$x\in\mathbb{R}^{n}$$. So let us do this. | {
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Fix $$x\in\mathbb{R}^{n}$$. We must show that $$x^{T}Ax\geq0$$. Write the vector $$x\in\mathbb{R}^{n}$$ in the form $$x=\left( x_{1},x_{2},\ldots,x_{n}\right) ^{T}$$ for some real numbers $$x_{1},x_{2},\ldots,x_{n}$$. Then, the definition of $$A$$ yields \begin{align} x^{T}Ax=\sum_{u\in\left[ n\right] }\sum_{v\in\left[ n\right] } \dfrac{\left\vert E_{u}\cap E_{v}\right\vert }{\left\vert E_{u}\cup E_{v}\right\vert }x_{u}x_{v}\geq0 \end{align} (by Theorem 7). This completes our proof of Corollary 8. $$\blacksquare$$ | {
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But if the two rows interchanged are identical, the determinant must remain unchanged. Since and are row equivalent, we have that where are elementary matrices.Moreover, by the properties of the determinants of elementary matrices, we have that But the determinant of an elementary matrix is different from zero. (Theorem 4.) Corollary 4.1. 2. 6.The determinant of a permutation matrix is either 1 or 1 depending on whether it takes an even number or an odd number of column interchanges to convert it to the identity ma-trix. Since zero is … That is, a 11 a 12 a 11 a 21 a 22 a 21 a 31 a 32 a 31 = 0 Statement) a 11 a 12 a 11 a 21 a 22 a 21 a 31 a 32 a 31 = 0 Statement) Statement) If two rows (or two columns) of a determinant are identical, the value of the determinant is zero. EDIT : The rank of a matrix… (Theorem 1.) This preview shows page 17 - 19 out of 19 pages.. If A be a matrix then, | | = . Let A be an n by n matrix. This means that whenever two columns of a matrix are identical, or more generally some column can be expressed as a linear combination of the other columns (i.e. Then the following conditions hold. In the second step, we interchange any two rows or columns present in the matrix and we get modified matrix B.We calculate determinant of matrix B. since by equation (A) this is the determinant of a matrix with two of its rows, the i-th and the k-th, equal to the k-th row of M, and a matrix with two identical rows has 0 determinant. Adding these up gives the third row \$(0,18,4)\$. The preceding theorem says that if you interchange any two rows or columns, the determinant changes sign. R2 If one row is multiplied by fi, then the determinant is multiplied by fi. Let A and B be two matrix, then det(AB) = det(A)*det(B). Proof. Theorem. If in a matrix, any row or column has all elements equal to zero, then the determinant of that matrix is 0. Determinant of Inverse of matrix can be defined as | | = . We take matrix A and we calculate its determinant (|A|).. If an | {
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of matrix can be defined as | | = . We take matrix A and we calculate its determinant (|A|).. If an n× n matrix has two identical rows or columns, its determinant must equal zero. A. The formula (A) is called the expansion of det M in the i-th row. Theorem 2: A square matrix is invertible if and only if its determinant is non-zero. Determinant of a matrix changes its sign if we interchange any two rows or columns present in a matrix.We can prove this property by taking an example. Prove that \$\det(A) = 0\$. Recall the three types of elementary row operations on a matrix: (a) Swap two rows; The same thing can be done for a column, and even for several rows or columns together. R3 If a multiple of a row is added to another row, the determinant is unchanged. This n -linear function is an alternating form . The matrix is row equivalent to a unique matrix in reduced row echelon form (RREF). If we multiply a row (column) of A by a number, the determinant of A will be multiplied by the same number. R1 If two rows are swapped, the determinant of the matrix is negated. Use the multiplicative property of determinants (Theorem 1) to give a one line proof that if A is invertible, then detA 6= 0. If two rows (or columns) of a determinant are identical the value of the determinant is zero. \$-2\$ times the second row is \$(-4,2,0)\$. Here is the theorem. 1. 4.The determinant of any matrix with an entire column of 0’s is 0. Hence, the rows of the given matrix have the relation \$4R_1 -2R_2 - R_3 = 0\$, hence it follows that the determinant of the matrix is zero as the matrix is not full rank. (Corollary 6.) I think I need to split the matrix up into two separate ones then use the fact that one of these matrices has either a row of zeros or a row is a multiple of another then use \$\det(AB)=\det(A)\det(B)\$ to show one of these matrices has a determinant of zero so the whole thing has a determinant of zero. The proof of Theorem 2. 5.The determinant of any matrix with two | {
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thing has a determinant of zero. The proof of Theorem 2. 5.The determinant of any matrix with two iden-tical columns is 0. Has two identical rows or columns together | | = square matrix row... A row is added to another row, the determinant is non-zero: the rank of a row multiplied... Determinant must remain unchanged is unchanged theorem 2: a square matrix is invertible if and only its! 0 ’ s is 0 is multiplied by fi done for a column, and for... Multiple of a determinant are identical, the determinant is zero B ) unique matrix in reduced row echelon (. Value of the determinant is unchanged … \$ -2 \$ times the second row is multiplied fi... = 0 \$ determinant changes sign columns is prove determinant of matrix with two identical rows is zero B be two,! ( -4,2,0 ) \$ or columns together ) of a determinant are identical the value of the is. * det ( a ) = det ( B ) ( a ) * (... ’ s is 0 the two rows or columns, the determinant changes sign another row, the of. A square matrix is invertible if and only if its determinant is multiplied by fi r2 if one is! -4,2,0 ) \$ equal zero prove determinant of matrix with two identical rows is zero changes sign entire column of 0 ’ s is 0 0 ’ is..., | | =: a square matrix is invertible if and if. A row is \$ ( -4,2,0 ) \$ another row, the determinant unchanged! Is called the expansion of det M in the i-th row is by. Value of the determinant is unchanged its determinant ( |A| ) is multiplied by,. Two rows or columns, its determinant must equal zero let a and B be two,... The i-th row can be defined as | | = form ( RREF ) that if you any. Even for several rows or columns prove determinant of matrix with two identical rows is zero if and only if its determinant remain. Multiple of a determinant are identical, the value of the determinant is zero ) \$ … -2! Matrix can be defined as | | = second row is multiplied by,. Several rows or columns, the determinant is zero any two rows ( or two )! Theorem says that if you interchange any two rows | {
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has two identical rows or columns, its determinant is unchanged … \$ -2 \$ times second... Is row equivalent to a unique matrix in reduced row echelon form ( RREF ) zero... B be two matrix, then det ( B ) ( a =! Be two matrix, then the determinant is multiplied by fi the is... \$ times the second row is multiplied by fi same thing can be defined as | | = prove determinant of matrix with two identical rows is zero if! Two iden-tical columns is 0 matrix then, | | = AB ) = 0.! A column, and even for several rows or columns ) of a row is multiplied by fi matrix,. ’ s is 0, its determinant is zero ) * det AB! | {
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2020 prove determinant of matrix with two identical rows is zero | {
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# Technique for constructing an entire function satisfying a given growth condition
How can I construct an entire function whose growth rate at infinity satisfies
$$\lim_{r \to \infty} \frac{\log M(r)}{\sqrt {r}} =1$$
where $M(r) = \max_{|z|=r} |f(z)|$?
Based on the above limit, I think that I can say $M(r) \to e^{\sqrt{r}}$, as r grows to infinity. So, this shows that $f(z)$ has exponential growth like $e^\sqrt{r}$.
Where can I go from here? Can I manipulate the known series for $e^z$ and claim that this series
$$\sum \frac{(\sqrt{z})^n}{n!}$$ is the entire function that we want?
• Well, that won't give you an entire function. $(\sqrt{z})^n$ poses a problem for odd $n$. What happens if you drop the problem-makers? – Daniel Fischer Oct 10 '15 at 9:36
• That's such a cool hint, @DanielFischer :-) I'll try again now - thanks so much... – User001 Oct 10 '15 at 9:43
• Hi @DanielFischer, if I drop the odd power terms and keep the even power summands, I now have the series for $cosh(\sqrt{z})$, which is entire and grows just like M(r). What do you think? Also, you mentioned that the odd powers of $\sqrt{z}$ are problematic; I think you are referring to the fact that we must choose a branch of the complex logarithm. But, wouldn't the even power terms also require branch cuts -- and also be problematic? Thanks, – User001 Oct 10 '15 at 10:14
• Oh ... the summands will become z^n / (2n)! so the square root is gone and there's no need to choose a branch of log anymore @DanielFischer. Thanks again :-) – User001 Oct 10 '15 at 10:30
• Since $\cosh$ is an even function, the function $\cosh \sqrt{z}$ is well-defined (ditto $\cos \sqrt{z}$ or $\dfrac{\sin \sqrt{z}}{\sqrt{z}}$ and so on), the ambiguity of the choice of $\sqrt{z}$ is annihilated by the evenness of the function one applies to it. – Daniel Fischer Oct 10 '15 at 11:18 | {
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Your idea goes in the right direction, but doesn't quite work out. If we choose a branch of $\sqrt{z}$ on a domain where one exists, and expand $e^{\sqrt{z}}$, we get
$$\sum_{n = 0}^\infty \frac{(\sqrt{z})^n}{n!} = \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k}}{(2k)!} + \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k+1}}{(2k+1)!} = \sum_{k = 0}^\infty \frac{z^k}{(2k)!} + \sqrt{z}\sum_{k = 0}^\infty \frac{z^k}{(2k+1)!}.$$
Both series clearly yield entire functions, but the $\sqrt{z}$ factor on the second sum makes it impossible for the whole thing to be an entire function. So what happens if we take only part of it? If we take the first series, we get
$$\sum_{k = 0}^\infty \frac{z^k}{(2k)!} = \cosh \sqrt{z} = \frac{1}{2}\bigl(e^{\sqrt{z}} + e^{-\sqrt{z}}\bigr).$$
That is an entire function (the ambiguity of $\sqrt{z}$ is annihilated by the evenness of $\cosh$/the fact that $\sqrt{z}$ and $-\sqrt{z}$ are both used in the same way), and the exponential representation strongly suggests - or makes it evident - that this function has the right growth behaviour. If we had taken the other series, the resulting function $\dfrac{\sinh \sqrt{z}}{\sqrt{z}}$ would also have the right growth, the $\sqrt{z}$ in the denominator is insignificant in comparison with the exponential involved in $\sinh$.
We can extend this approach systematically to obtain functions such that
$$\lim_{r\to \infty} \frac{\log M(r)}{r^\alpha} = 1$$
for rational $\alpha > 0$. Basically, we want something containing $e^{z^{\alpha}}$, but for $\alpha \notin \mathbb{N}$ we need a modification to obtain an entire function. If $\alpha = \frac{m}{k}$, we retain only the terms of
$$e^{z^\alpha} = \sum_{n = 0}^\infty \frac{(z^\alpha)^n}{n!}$$
for which the exponent $n \alpha$ is an integer, so $k \mid n$. That gives the function
$$g_{\alpha}(z) = \sum_{n = 0}^\infty \frac{(z^\alpha)^{kn}}{(kn)!} = \sum_{n = 0}^\infty \frac{z^{mn}}{(kn)!}.$$ | {
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To verify that this has the right growth behaviour, let $\zeta_k = \exp \frac{2\pi i}{k}$ and consider the function
$$h_k(z) = \frac{1}{k}\sum_{s = 0}^{k-1} e^{\zeta_k^s\cdot z} = \sum_{n = 0}^\infty \Biggl(\frac{1}{k}\sum_{s = 0}^{k-1} \zeta_k^{n\cdot s}\Biggr)\frac{z^n}{n!} = \sum_{n = 0}^\infty \frac{z^{kn}}{(kn)!}.$$
Then we see that $g_{\alpha}(z) = h_k({z^{\alpha}})$ has the right growth:
Since $\lvert e^{\zeta_k^s\cdot z}\rvert \leqslant e^{\lvert z\rvert}$, we have $\lvert h_k(z)\rvert \leqslant e^{\lvert z\rvert}$, and for $x > 0$ we have $\operatorname{Re} (\zeta_k^s x) = \bigl(\cos \frac{2\pi s}{k}\bigr)x \leqslant \bigl(\cos \frac{2\pi}{k}\bigr)x$ for $0 < s < k$, so
$$\lvert h_k(x)\rvert \geqslant \frac{1}{k} e^{x} - \frac{k-1}{k} e^{x\cos \frac{2\pi}{k}} = \frac{e^x}{k}\Bigl(1 - (k-1)e^{-(1-\cos \frac{2\pi}{k})x}\Bigr) \sim \frac{e^x}{k}.$$
Altogether it follows that $\log M_{h_k}(r) \sim r$ and therefore $\log M_{g_\alpha}(r) \sim r^\alpha$. | {
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• Hi @DanielFischer, thanks so much for this awesome answer - and for the generalization of the technique. Can I ask you a quick follow-up question? Can you elaborate just a bit more about how the ambiguity of $\sqrt{z}$ is annihilated by the evenness of cosh / the fact that $\sqrt{z}$ and -$\sqrt{z}$ are both used in the same way? – User001 Oct 11 '15 at 1:12
• What do you mean, when you say this? When I look at the manipulated series of $e^z$, and arrive at the series for cosh($\sqrt{z}$), I am happy with it, primarily because the summands do not have a square root anymore. But when I look the exponential formula for cosh($\sqrt{z}$), there are obviously two exponential terms involving the (complex) square root, both terms divided by two. Looking at this formula would make me think that I'd have to choose a branch of logarithm for each term, and so this makes cosh($\sqrt{z}$)... not entire? Hmm...what do you think? Thanks, @DanielFischer – User001 Oct 11 '15 at 1:13
• Oh..do you mean that the multivaluedness cancels each other out, @DanielFischer? I expanded out the logarithms of the exponential formula for cosh and notice that when the argument jumps by 2pi, the other term will jump by -2pi, resulting in a net effect of 0 jump in the argument. So the function is not multivalued anymore. Am I thinking correctly about this now? Thanks, – User001 Oct 11 '15 at 1:24
• Yes, the multivaluedness cancels out. We have an expression of the form $G(z) = F(\sqrt{z}) + F(-\sqrt{z})$. For $z\neq 0$, on a neighbourhood $U$ of $z$ we have two branches, call them $a$ and $b$ of the square root. Then we have $a(z) = -b(z)$ for all $z\in U$ and $G(z) = F(a(z)) + F(-a(z)) = F(a(z)) + F(b(z)) = F(-b(z)) + F(b(z))$ is the same, whichever branch we choose on $U$. Hence $G$ is well-defined and analytic on $\mathbb{C}\setminus \{0\}$. At $0$, Riemann's removable singularity theorem gives us analyticity. – Daniel Fischer Oct 11 '15 at 10:26 | {
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• It's the same for the more general case, the possible choices of $z^{\alpha}$ differ by factors of $\zeta_k^r$, and since we sum over $F(\zeta_k^s z)$, the ambiguity cancels out. – Daniel Fischer Oct 11 '15 at 10:28 | {
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# Why does the Ratio Test prove that this particular sequence converges on 0?
I was recently reading How to Think about Analysis by Lara Alcock in preparation for taking Real Analysis in the winter, and ran into the following theorem regarding sequences that either tend towards infinity or converge on a particular number, specifically the Ratio Test.
Suppose that $$(a_n)$$ is a sequence such that $$(a_{n+1}/a_n) \rightarrow l$$. Then:
1. If $$-1 < l < 1$$ then $$(a_n) \rightarrow 0$$.
2. If $$l > 1$$ and $$a_n > 0 \forall n \in \mathbb{N}$$ then $$(a_n) \rightarrow \infty$$.
3. If $$l > 1$$ and $$a_n < 0 \forall n \in \mathbb{N}$$ then $$(a_n) \rightarrow -\infty$$
4. If $$l < -1$$ then the sequence neither converges or tends to $$\pm \infty$$.
5. If $$l = 1$$ we get no information.
Two specific examples she gives of sequences that converge to $$0$$ are $$\left({n^2 \over 2^n}\right)$$ and $$\left({6^n \over n!}\right)$$
For the latter sequence, $$\left({6^n \over n!}\right) = {6 \over 1}, {36 \over 2}, {216 \over 6}, {1296 \over 24}, {7776 \over 120},\ldots$$
I can see for the ratios I test that that the ratios tend to get smaller over time, which isn't too surprising. For example, $${36 \over 2} \div {6 \over 1} = 3, {216 \over 6} \div {36 \over 2} = 2, {1296 \over 24} \div {216 \over 6} = 1.5$$
The author also gives the intuition of why this sequences approaches $$0$$ by pointing out that $${6 \times 6 \times\cdots \times 6} \over {1 \times 2 \times \cdots \times n}$$
This makes it intuitively obvious why this denominator grows much more quickly than the numerator, which makes it seem a lot more sensible why this should approach $$0.$$ | {
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So, I understand that the ratio between each two consecutive terms is decreasing (if I'm reading this correctly), which isn't too surprising, and I understand the intuition behind why this sequence approaches $$0.$$ However, I'm a little confused as to how this actually "meets" the first test given that I gave several examples where $$(a_{n+1}/a_n) > 1$$ - for example, $${36 \over 2} \div {6 \over 1} = 3$$, which is obviously greater than $$1.$$ Naturally, eventually the ratio will be less than $$1$$, but doesn't the theorem kind of imply that the ratio of any two consecutive elements of the sequence will have a ratio greater than $$-1$$ and less than $$1$$?
Also, how does this account for sequences that converge on things other than $$0$$, $$- \infty$$, or $$\infty$$?
To say that a sequence $\{b_n\}_{n \in \mathbb{N}}$ converges to a limit $L$ is to say that, for all $\varepsilon > 0$, there is an $N \in \mathbb{N}$ such that $n > N$ implies $|b_n - L| < \varepsilon$.
Rephrased into colloquial English: For any positive number $\varepsilon$, the sequence of those $b$-subscript elements eventually gets within $\varepsilon$ of $L$, and stays at least that close to the limit. The key word in this description is the same term that you italicized:
Naturally, eventually the ratio will be less than $1$, but doesn't the theorem kind of imply that the ratio of any two consecutive elements of the sequence will have a ratio greater than $−1$ and less than $1$?
No, the theorem does not say this. The theorem's hypothesis is precisely that the sequence made from a ratio of consecutive terms converges to $l$, which means that eventually this sequence (i.e., these ratios) will be within $\varepsilon$ of $l$. | {
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• So, the fact that the ratios are decreasing is precisely the point in this case? Nov 12 '17 at 5:49
• Fair point - I guess there's no particular requirement as to exactly when or how quickly it converges (just as long as it eventually does), right? Nov 12 '17 at 5:54
• @EJoshuaS Exactly. Nov 12 '17 at 5:55
• @EJoshuaS Formalizing the intuition for this particular case; note that you could have a sequence alternating between being halved and quartered, so that the ratio of terms does not converge to anything (it switches between $1/2$ and $1/4$, never settling on either) but the actual sequence will still converge to $0$. Nov 12 '17 at 6:04
• Excellent, thanks - that's very helpful. I think I understand now. Nov 12 '17 at 6:06 | {
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