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$$\binom{5}{4}=5, \binom{5}{1,2}=\frac{5!}{1!2!2!}=30, \binom{5}{2}=10$$ respectively.
Each column value of $$k$$ can be displayed in $$\binom{3}{k}$$ ways.
All patterns have now been achieved once and once only (by inspection).
The total number is therefore $$5\cdot 3^5 + 30\cdot 3^4 + 10\cdot 3^3 = 1215+2430+270=3915$$.
• Thank you very much for the time and effort! Would it be a bother for you to elaborate on the the relation between partitions of 9 into 5 positive parts, each part less than 4- to the problem? I can't see it :( It's a really interesting solution and I'd like to understand it better!
– omer
Dec 23, 2019 at 12:47
• 5 parts is one for each column, hence positive because each column must be contain at least one filled cell, and each column can have a maximum height of 3, hence less than 4.
– JMP
Dec 23, 2019 at 12:48
• Thank you again! This is a great new way to see this problem!
– omer
Dec 23, 2019 at 12:50 | {
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# Injective Homomorphism from $\mathbb{R}\times\mathbb{R}$ to the ring of Continuous functions
Does there exist an injective ring homomorphism from the ring $$\mathbb{R}\times\mathbb{R}$$ to the ring of continuous functions over $$\mathbb{R}$$?
I know that $$\mathbb{R}\times\mathbb{R}$$ is a field. So, any ring homomrphism should have either identity or the whole field as a kernel. So, does it imply that there is such a homomorphism? I am unable to think of any examples. Any hints. Thanks beforehand.
• $\mathbb R \times \mathbb R$ is not a field. Remember, the direct product of fields is not going to be a field in general, because, for example, $(1,0) \cdot (0,1) = (0,0)$. – астон вілла олоф мэллбэрг Dec 10 '18 at 12:10
• @астонвіллаолофмэллбэрг oh, thanks for that fact. So, then, it is not even an integral domain! – vidyarthi Dec 10 '18 at 12:11
• By the definition of the direct product, two homomorphisms from $\mathbb R$ to the ring of continuous functions over $\mathbb R$ give rise to a homomorphism from $\mathbb R \times \mathbb R$ to the latter space, with the two maps as components/projections. However, the pair of maps being injective does not imply that the combined map will be injective. – астон вілла олоф мэллбэрг Dec 10 '18 at 12:18
• @астонвіллаолофмэллбэрг ok, any good examples? – vidyarthi Dec 10 '18 at 12:23
Let us make a general statement, and then apply it to this case.
Definition : Given a ring $$R$$, an element $$e \in R$$ is said to be idempotent, if $$e^2 = e$$. Note that $$0,1$$ are idempotent. Any other idempotent will be referred to as non-trivial.
Let $$R,S$$ be rings such that $$R$$ has a non-trivial idempotent but $$S$$ does not have any non-trivial idempotent. Then, any homomorphism $$\phi : R \to S$$ is not injective. | {
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Proof : Let $$e \in R$$ be a non-trivial idempotent, which is to say that $$e^2 = e$$, and $$e \neq 0,1$$. Let us look at $$\phi(e)$$. Then, $$(\phi(e))^2 = \phi(e^2)= \phi(e)$$, so $$\phi(e)$$ is idempotent in $$S$$, hence equals zero or one. if $$\phi(e) = 0$$ then $$\phi$$ is not injective as $$e \neq 0$$. If $$\phi(e) =1$$, then $$\phi(1-e) = 0$$, and again injectivity is contradicted.
You can use the general statement in lots of places. Try to find candidates for $$R$$ and $$S$$.
In our case, $$S$$ , as the ring of continuous functions over $$\mathbb R$$,contains no non-trivial idempotent, since if $$f^2 = f$$, then for each $$x$$ we have $$f(x) = 0$$ or $$1$$. However, as $$f$$ is continuous, $$f(\mathbb R)$$ is connected, hence an interval, but has to be a subset of $$\{0,1\}$$, which is discrete. Consequently, $$f$$ is either identically zero or identically one, hence is a trivial idempotent.
However, $$\mathbb R \times \mathbb R$$ contains the non-trivial idempotent $$(0,1)$$. Hence, the general statement gives the result you have desired.
As to non-injective maps, there are many of them. As mentioned earlier, consider $$f,g \in S$$ not necessarily distinct. Then, $$(a,b) \to af + bg$$ is a homomorphism. It is not difficult to see that every homomorphism is of this form.
Also, think about what happens if you treat $$R$$ and $$S$$ as just abelian groups under addition. In that case, can you find an injective group homomorphism between the two? Note that the concept of idempotent cannot be defined without the multiplication, so this might be an interesting side question. | {
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Suppose there is an injective ring homomorphism. The unit element $$(1,1)$$ of $$\mathbb{R} \times \mathbb{R}$$ maps to the unit element of $$\mathcal{C} (\mathbb{R} )$$, i.e. the constant function $$1_{\mathbb{R}}$$. If $$f,g$$ are the images of $$(1,0) , (0,1)$$ respectively, then $$f+g = 1_{\mathbb{R}}$$ and $$fg = 0$$. This shows that $$f(1_{\mathbb{R}}-f) = 0 \implies f = f^2$$. Since $$f$$ is continuous, $$f$$ is identically $$0$$ or identically $$1$$. The first case contradicts injectivity and so does the second because then $$g = 0$$. So there does not exist any injective ring map.
• +1: Very nice argument! And from this we readily find that the only ring homomorphisms $\Bbb R\times\Bbb R\to\mathcal C(\Bbb R)$ are $(x,y)\mapsto x\cdot 1_{\Bbb R}$ and $(x,y)\mapsto y\cdot 1_{\Bbb R}.$ – Cameron Buie Dec 10 '18 at 18:36 | {
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# An algebra question
• June 16th 2010, 11:12 AM
econlover
An algebra question
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n
Remarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.
• June 16th 2010, 11:28 AM
earboth
Quote:
Originally Posted by econlover
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n
Remarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.
1. Use the laws of powers:
$\frac{(2 \cdot 5^n)^2 + 25^n}{5^n} = \frac{4 \cdot 5^{2n} + 5^{2n}}{5^n}= \frac{5 \cdot 5^{2n}}{5^n}=5 \cdot 5^n = 5^{n+1}$
• June 16th 2010, 11:33 AM
econlover
slight question
Thanks.
I understand most of it but how did you move along from the 1st step of your answer to the second one
that is, in specific terms, where did the 4 go?
• June 16th 2010, 11:39 AM
SpringFan25
considering the numerator (top of fraction) only:
$4 . 5^{2n} + 5^{2n} = (4+1).5^{2n} = 5 \times 5^{2n} = 5^{2n+1}$
• June 16th 2010, 11:41 AM
econlover
Thanks for the message. But I guess you misinterpret the sign of this "."
It is not 4.5 (4 decimal 5) but 4 . 5 meaning 4 x 5 = 4 times 5
Hope this clears up the misconception
• June 16th 2010, 11:47 AM
econlover
Still don't understand this step. Where did the 1 come from ? What about the positive sign?
Can someone explain? Thanks!
• June 16th 2010, 12:18 PM
SpringFan25
it is only factorising. i was not interpreting your dot as a decimal.
in general:
4x + x = (4+1)x = 5x
You have
$4 \times 5^{2n} + 1 \times 5^{2n}$ = $5 \times 5^{2n}$
Which says: "4 lots of $5^{2n}$ plus 1 lot of $5^{2n}$" equals "5 lots of $5^{2n}$"
• June 17th 2010, 02:03 PM
bjhopper
Quote:
Originally Posted by econlover
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n | {
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# The limit of $(n!)^{1/n}/n$ as $n\to\infty$ [duplicate]
(Proof necessary)
$$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$$
I don't have an answer yet, but I know it exists, and is less than $1$.
Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back
Any help is appreciated
• Is this your limit: $\lim_{n\rightarrow\infty} \frac{(n!)^{\frac{1}{n}}}{n}$? – Jared Sep 17 '14 at 20:15
• Hint: If the limit exists, let's say its $L$, then $$\lim_{n \to \infty} \frac{n!}{n^n}=L^n$$ – Oria Gruber Sep 17 '14 at 20:18
• Stirling's formula will help you – ThePortakal Sep 17 '14 at 20:18
• See this question for something very similar. It can be done without Strilings formula. – Winther Sep 17 '14 at 20:32
• @Oria: The value of a limit (if it exists) cannot depend on the limiting variable. So "$\displaystyle\lim_{n \to \infty}\frac{n!}{n^n} = L^n$" is definitely incorrect. – JimmyK4542 Sep 17 '14 at 20:51
Put $$a_n = \frac{n!^{1/n}}{n}$$
then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$
where we have used $\log n! = \log 1 + \log 2 + \ldots + \log n$. The sum above is a Riemann sum for the integral $\int_0^1\log x dx$ so
$$\lim_{n\to\infty} \log a_n = \int_0^1\log x dx = [x\log x - x]_0^1 = -1$$
and it follows that $$\lim_{n\to\infty}a_n = \frac{1}{e}$$ | {
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and it follows that $$\lim_{n\to\infty}a_n = \frac{1}{e}$$
• Hey winther I don't understand how you're jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). – Calvin Hammond Sep 17 '14 at 21:28
• @CalvinHammond $$\log n! - n \log n = \log 1 + \log 2 + \cdots + \log n - (\log n + \log n + \cdots + \log n),$$ where the number of terms in the parentheses is $n$. So $$\log n! - n \log n = \log \frac{1}{n} + \log \frac{2}{n} + \cdots + \log \frac{n}{n}$$ using the identity $\log x - \log y = \log \frac{x}{y}$. The rest follows. – heropup Sep 17 '14 at 21:42
• Yes that is. Thanks for explaining it @heropup – Winther Sep 17 '14 at 21:46
• Could you explain the remain sum more please. – Calvin Hammond Sep 17 '14 at 21:56
• Notice that you have to be a little careful in applying this technique to improper integrals: math.stackexchange.com/questions/449103/… – user84413 Sep 18 '14 at 16:51
Here is a simple elementary proof I found, but first of all, some lemmas:
• This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$
• You can try to prove this inequality yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1}\\$
• This inequality is the one I'm going to use though because it gives a much tighter bound on our sequence and that's just more fun, though you could use the second inequality without change in proof. I could give you the proof if needed:$\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1/2}\\$
Ok, so here is the proof: | {
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Ok, so here is the proof:
• We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i} \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i}$
• Now, we use our inequalities to bound our sequence. First, an upper bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \leq \prod_{i=1}^{n-1}e^{1}=e^{n-1}$
• Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
• Now, since $\frac{n!^{1/n}}{n}=(\frac{n^n}{n!})^{-1/n}$, we get: $\displaystyle e^{\frac{1}{n}-1} \leq \frac{(n!)^{1/n}}{n} \leq e^{\frac{1}{n}-1} \cdot \sqrt[2n]{n}$
• Finally, by the squeeze theorem, we get $\displaystyle e^{0-1} \leq \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \leq e^{0-1} \cdot 1$
• Hence, $\displaystyle \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=e^{-1}$
I know there are simpler proofs, but this one is elementary and I feel like it gives you the direct intuition as to why it's true. | {
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Stirling's Approximation $$n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ Which means that $n!$ is asymptotically equivalent to $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ as $n$ approaches $\infty$. If your familiar with asymptotic formulas, then you'd also know that this implies that $$\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} =1$$ Now, using the algebraic laws of limits, we have $$\lim_{n\to\infty} n!=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ $$\lim_{n\to\infty} e^n=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n^n}{n!}\right)$$ $$\lim_{n\to\infty} e=\lim_{n\to\infty} (2\pi n)^{\frac{1}{2n}}\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)$$ So now $$e=\left[\lim_{n\to\infty} e^{\ln(2\pi n)^{\frac{1}{2n}}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{\ln(2\pi n)}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]$$ $$=\left[\lim_{n\to\infty} e^{\frac{\frac{d}{dn}\ln(2\pi n)}{\frac{d}{dn}2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{1}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]$$ $$=e^0\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]=1\cdot \lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}$$ Thus $$e=\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}$$ And now we can easily see that $$\lim_{n\to \infty} \frac{(n!)^{\frac{1}{n}}}{n}= \lim_{n\to \infty} \frac{1}{\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)}=\frac{1}{e}$$ Let me know if you have any questions.
Let $\displaystyle a_n=\frac{n!}{n^n}$.
Then $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}$, | {
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so $\;\;\;\;\displaystyle\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$ $\hspace{.2 in}$ (since $\frac{a_{n+1}}{a_n}\rightarrow L\implies(a_n)^{\frac{1}{n}}\rightarrow L)$.
• This is a great proof, but you would need to prove that the two forms of the limit are equivalent. – Fujoyaki Sep 17 '14 at 22:46
• Thanks; you are right that I am using the result that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=L$. See math.stackexchange.com/questions/69386/… – user84413 Sep 17 '14 at 22:49
Hint. Take the natural logarithm of $\dfrac{(n!)^{1/n}}{n}$ and obtain $$\frac{\log 1+\log 2+\cdots+\log n}{n}-\log n=\frac{\log (1/n)+\log(2/n)+\cdots+\log(n/n)}{n} \\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right) \to \int_0^1 \log x\,dx=-1.$$
Hint (Sterling): $$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$
• Yes that is my limit. I'm using this to help a friend but I don't know much about to how to solve this myself. So I'm asking for a full answer. – Calvin Hammond Sep 17 '14 at 20:28 | {
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# How to minimise the cost of guessing a number in a high/low guess game?
In a high/low guess game, the "true" number is one of $$\{1,\cdots,1000\}$$. You'll be told if your guess is $$<,>$$ or $$=$$ the true number for each guess you make, and the game terminates when you guess out the true number. Suppose the following three scenarios:
1. If your guess is either lower or higher, you pay $$\1$$ in both cases. If your guess is correct, you pay nothing and the game ends.
2. If your guess is higher, you pay $$\1$$; if your guess is lower you pay $$\2$$. If your guess is correct, you pay nothing and the game ends.
3. If your guess is higher, you pay $$\1$$; if your guess is lower you pay $$\1.5$$. If your guess is correct, you pay nothing and the game ends.
In these three scenarios, what is, respectively, the minimum number of $$\$$ you must have to make sure you can find the true number?
Formally, define a space of all guess strategies $$\Sigma$$. We can then identify the cost $$C$$ as a function $$C:\Sigma\times \{1,\cdots,1000\}\to\Bbb N_+,\quad v\mapsto C(S,v)$$ that maps the pair $$(S,v)$$ (where $$v$$ is the "true" number) to the cost it's going to take under this arrangement. Our problem is then to find $$\min_{S} \max_{v} C(S,v).$$
Can this min-max problem be analytically solved? What would be the corresponding optimal strategy then?
For scenario 1, my intuition is to use bisection search, which, at worst, costs $$\10$$ ($$2^{10}=1024$$). But I cannot prove this is indeed optimal. For the second scenario, since making an under-guess means a higher cost, maybe it's more optimal to use a "right-skewed" bisection, i.e. you keep making guesses at the $$2/3$$-quantile point. But this is just (rather wild) intuition, not anything close to a proof. | {
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For this game, the most natural representation of a strategy is a rooted binary tree, with vertices labeled by guesses (i.e. numbers from $$\{1, \ldots, N=1000\}$$), and each vertex having at most two edges (connecting it to the guess to make next, for "lower" and "higher" cases).
Let the penalties you pay for "lower/higher" cases be $$L$$ and $$H$$, respectively. Now consider an arbitrary $$N$$, and consider an optimal tree $$T(N)$$ (that reaches the minimum worst-case total penalty). Its root is labeled with some guess $$R$$, $$1\leqslant R\leqslant N$$, its left subtree can be chosen to be [isomorphic to] $$T(R-1)$$, and its right subtree isomorphic to $$T(N-R)$$ (assuming $$T(0)$$ is empty).
So if $$P(N,L,H)$$ is the optimal penalty, then $$P(0,L,H)=P(1,L,H)=0,\\ P(N,L,H)=\min_{1\leqslant R\leqslant N}\max\begin{cases}L+P(N-R,L,H)\\ H+P(R-1,L,H)\end{cases}\quad(N>1)$$ which allows one to compute $$P(1000,1,1)=\color{red}{9},\quad P(1000,2,1)=14,\quad P(1000,1.5,1)=11.5$$ (note $$9$$ but not $$10$$ as you've suggested; the decision process is not just a bisection but has extra "$$=$$" outcomes). The trees can be computed along the way.
Regarding the asymptotic behavior (with $$N\to\infty$$), if we let $$R(N,L,H)$$ be the "argmin" (say, the smallest root of optimal trees), one can show that the limits $$\alpha=\alpha(L,H)=\lim_{N\to\infty}\frac{P(N,L,H)}{\ln N},\quad\beta=\beta(L,H)=\lim_{N\to\infty}\frac{R(N,L,H)}{N}$$ exist and satisfy $$H+\alpha\ln\beta=L+\alpha\ln(1-\beta)=0.$$ In particular, $$\beta(2,1)=(\sqrt{5}-1)/2$$, and $$\beta(1.5,1)$$ is the root of $$\beta^3=(1-\beta)^2$$. | {
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• Thanks. This makes sense. There seems to be no way to compute such a complicated recursion analytically though. Did you use DP for this? – Vim May 11 at 12:30
• Yes, I did. Probably this still can be analyzed further, but I didn't do that. – metamorphy May 11 at 12:38
• IIRC, your recursion also gives the optimal strategy when you keep track of the partitions chosen at each recursion node. – Vim May 11 at 13:41
For the (2,1) case, I found the worst-case cost increases by 1 at every Fibonacci number, so $$P(F_n,2,1)=1+P(F_n-1,2,1)=1+P(F_{n-1},2,1)$$. These key values $$F_n$$ increase by ratio $$F_n/F_{n-1}\to\phi=(1+\sqrt{5})/2$$.
They grow exponentially; conversely, the maximum cost for $$P(N,2,1)$$ grows logarithmically.
For (1.5,1) the worst-case cost increases by $$0.5$$ at these values: $$P(N,2,1)>P(N-1,1.5,1)\,if\, N=2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,\ldots$$
The recursion in this sequence is $$a_{n+1}=a_{n-1}+a_{n-2}$$. Their ratio, $$a_{n+1}/a_n$$, approaches 1.3246, which is the root of $$P^3=P+1$$, just as $$\phi$$ for the (2,1) case is the root of $$P^2=P+1$$.
• Would you elaborate on how the ratios arise? – Vim May 11 at 12:32
• For $(4/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-3}$, and for $(5/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-4}$ – Empy2 May 11 at 13:00
Here are some worked-out cases for smaller lists and variable penalties.
Suppose you pay $$a$$ for a guess that is too low and $$b$$ for a guess that is too high.
Given a list such as $$\{1,\ldots,1000\}$$, the cost $$C$$ of the optimal strategy is a function only of the number of items $$n$$ in the list. Without loss of generality, assume that $$a\leq b$$. (If you want the case $$a\geq b$$, just reverse the order of the list.)
We can determine the cost of the optimal strategy in a few simple cases: | {
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We can determine the cost of the optimal strategy in a few simple cases:
• $$C(1) = 0$$. There is only one guess, and it is correct.
• $$C(2) = a$$. High guesses are penalized more, so guess the lower element.
• $$C(3) = b$$. Guess the middle element. If you're wrong, you know what the correct element is and it costs at worst $$b$$ total.
For larger values of $$C(n)$$, consider what the optimal first move. Suppose there are $$n$$ elements in the list and you choose index $$1\leq k\leq n$$. In the worst case, you are wrong. If the true answer is lower, you must pay $$b$$ and search through a list of $$k-1$$ elements. If the true answer is higher, you must pay $$a$$ and search through a list of $$n-k$$ elements. So in the worst case, by choosing $$k$$ in the first round, you'll incur a cost of $$\max\left(b + C(k-1),\; a + C(n-k)\right).$$ We can search for an optimal $$k$$ which minimizes this cost in the first move, in which case the recursive solution tells us what the rest of the strategy (and its corresponding cost) is.
Note that the optimal strategy will never be in the latter half of the list. This is because $$C$$ is monotonic. When $$k$$ is in the right half of the list (so that $$n-k \leq k-1$$), we can always shrink the worst-case cost $$\max(b+C(k-1), a+C(n-k))$$ by exchanging $$k-1$$ and $$n-k$$, putting $$k$$ into the symmetric position on the left half of the list.
By this reasoning, we find:
• $$C(4) = \max(2a,b).$$ Choose the second element. If you're too high, you'll pay a cost of $$b$$ but know the true answer. If you're too low, you'll pay a cost of $$a$$, then search through the remaining two elements for a worst-case cost of $$C(2)=a$$. | {
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• $$C(5) = a+b$$. You can choose either the second or third (middle) elements. In the worst case if you choose the middle element, it's too high so you pay $$b+C(2) = a+b$$ to search the two remaining. If you choose the second element, in the worst case it's too low so you pay $$a+C(3) = a+b$$ to search the three remaining.
• $$C(6) = a+b$$. Choose the third element. In the worst case, it's either too high so you pay $$b+C(2) = a+b$$, or it's too low so you pay $$a+C(3)=a+b$$.
• $$C(7) = \min(a+C(4), b+C(3)).$$ As with the case $$n=4$$, the optimal strategy depends on the relative magnitudes of $$a$$ and $$b$$. If $$2b > 3a$$, then choose the third element in the list. Otherwise, choose the fourth (middle). | {
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### Bisection Method Algorithm Matlab | {
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Updated to reflect the latest version of MATLAB, the second edition of this title introduces the theory and applications of the most commonly used techniques for solving numerical problems on a computer. Secant method D. This method requires two initial guesses satisfying. Numerical Methods with Matlab The Bisection Method 4 1. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Design and simulation of three phase induction motor at different load conditions in matlab/simulink. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). m code; Worksheet 04; 10th - 14th September : Tu: Bisection Method (continued) and order of convergence ; Th: Newton's Method and order of convergence; Homework 01 is due on 09/13! Worksheet 05; Code for Newton's Method; 17th - 21st September : Tu: Secant Method with order of. A few steps of the bisection method applied over the starting range [a 1;b 1]. Here is a Matlab function that carries out the bisection algorithm for our cosmx function. It allows the code writer to focus on the logic of the algorithm without being distracted by details of. 2) using the bisection method. 1 Bisection Method In bisection method we reduce begin with an interval so that 0 2[a;b] and divide the interval in two halves,i. It’s very intuitive and easy to implement in any programming language (I was using MATLAB at the time). They announced that the genetic algorithm is better than classical methods. where the value of the function. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. On the other hand, the only difference between the false position method and the bisection method is that the latter uses ck = (ak + bk) / 2. The bisection algorithm should be: Save the interval boundaries; Look if [a,b] has a root. In python (with matplotlib), use the savefig function. Provide the function, 'f' and provide two | {
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In python (with matplotlib), use the savefig function. Provide the function, 'f' and provide two guesses. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. Bisection Method: Develop a MATLAB program to find the root of the following function using the bisection method gm gc tan g-9. Im studying for a math test and on a old test there is a task about bisection. The standard technique is something like Brent's method (see Numerical Recipes in C, Section 9. From these algorithms, the developer has to explore and exploit the algorithm suitable under specified constraints on the function and the domain. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. - Bisection method for bounded searching. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. Here's the code:. It shows with bold stripes the length of the bracketed region. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. 2d Truss Analysis Matlab Program. Matlab has the function ‘fzero’ to find function zeros. The simplest root-finding algorithm is the bisection method: we start with two points a and b which bracket a root, and at every iteration we pick either the subinterval or, where is the midpoint between a and b. LAB 1: the bisection and secant methods In this section you will learn how to implement and analyse the Bisection and Secant methods in MAT-LAB. Like the others methods, we approach this problem by writing the equation in the form f(x) = 0 for some function f(x). Learn more about bisection, bisection method, algorithm. newton raphson method matlab pdf. We have provided MATLAB program for Bisection | {
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method, algorithm. newton raphson method matlab pdf. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. The Bisection Method & Intermediate Value Theorem. m (after that day assignments should be put into the mbox of Yinglun ZHU) (1)The original demonstration of Newton’s method was done by Newton almost 350 years ago. The bisection search. f(c)<0 then let b=c, else let a=c. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. What are the applications of the bisection. Fixed Point Iteration 8 1. The setup of the bisection method is about doing a specific task in Excel. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. 6524; m = 73. The Steepest Descent Algorithm for Unconstrained Optimization and a Bisection Line-search Method Robert M. Problems 197. So let's take a look at how we can implement this. Essentially, the root is being approximated by replacing the. ) (Use your computer code) I have no idea how to write this code. Using this simple rule, the bisection method decreases the interval size iteration by iteration and reaches close to the real root. Learn how genetic algorithms are used to solve optimization problems. In mathematics, the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. In this method, there is no need to find the. Learn more about matlab. A next search interval is chosen by comparing and nding which one has zero. where the value of the function. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. All the Fortran 90 programs listed here are corresponding to the Fortran 77. The equation is of form, f(x) = 0. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. % % | {
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string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. % % OUTPUT: approximate solution p or. Again, as before, Newton’s method does not always converge, but when it does, it does so faster (p = 2) than the bisection method (p = 1) and the secant method (𝑝= (1+√5)⁄2). 2D 3D Algorithms ASCII C# C++ Cellular Automata Clustering Cryptography Design Patterns Electronics game Image Processing Integral Approximation Java JavaFX Javascript LED Logic Gates Matlab Numerical Methods Path Finding Pygame Python R Random Root Finding R Shiny Sound UI Unity. It was observed that the Bisection method converges at the 14th iteration while Newton methods. 5 Secant Method 189. The bisection method depends on the Intermediate Value Theorem. Bisection Algorithm Let , - i. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. The algorithm applies to any continuous function. B The comparative results are shown in table 3. GitHub Gist: instantly share code, notes, and snippets. This scheme is based on the intermediate value theorem for continuous functions. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. In general, Bisection method is used to get an initial rough approximation of solution. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Bisection method in MATLAB A video of programming the code in realtime. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). % Newton's Method algorithm! n = 2;! nfinal = N + 1; % Store final iteration if tol is reached before N iterations! Newton's Method MATLAB Implementation. The algorithm, created by | {
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tol is reached before N iterations! Newton's Method MATLAB Implementation. The algorithm, created by T. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further. 2 Bisection method 2. Comparative Study Of Bisection, Newton-Raphson And Secant Methods Of Root- Finding Problems International organization of Scientific Research 2 | P a g e Given a function f x 0, continuous on a closed interval a,b , such that a f b 0, then, the function f x 0 has at least a root or zero in the interval. Root approximation through bisection is a simple method for determining the root of a function. Freund February, 2004 1 2004 Massachusetts Institute of Technology. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Before you start, review the \Introduction to MATLAB" notes. The chance of convergence with such a small precision depends on the calculatord: in particular, with Octave, the machine precision is roughly ⋅ −. Numerical rate of convergence of root has been found in each calculation. Learn how genetic algorithms are used to solve optimization problems. Earlier we discussed a C program and algorithm/flowchart of bisection method. The Bisection Method The Bisection Method at the same time gives a proof of the Intermediate Value Theorem and provides a practical method to find roots of equations. m needed for Homework 4. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Essentially, the root is being approximated by replacing the. 6: Calculate the ground track of a satellite from its orbital elements. 15625 (you need a few extra steps for ε abs) Applications to Engineering. I wrote his code as part of an article, How to solve equations using python. The bisection method is one of the simplest and most reliable of iterative methods for the | {
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python. The bisection method is one of the simplest and most reliable of iterative methods for the solution of nonlinear equations. Need help with this bisection method code!. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. he gave us this template but is not working. Description. The problem is that it seems like the teachers recommended solution to the task isn't quite right. This Demonstration shows the steps of the bisection root-finding method for a set of functions. f = @(x) (cos(x)); a = input( 'Please enter lower. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. This is a very simple and powerful method, but it is also relatively slow. 2 (Bisection Method). Bisection Method C Program Bisection Method MATLAB Program. If a change of sign is found, then the root is calculated using the Bisection algorithm (also known as the Half-interval Search). t is the root of the given function if f (t) = 0; else follow the next step. The method is also called the interval halving method. This method will divide the interval until the resulting interval is found, which is extremely small. Powered by Create your own unique website with customizable templates. 000013273393044 9 1. 21 dcm_to_ypr. The following Matlab project contains the source code and Matlab examples used for bisection method. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. Remark: 𝑝𝑝. So if you need MATLAB programming homework help, feel free to | {
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and that it is relatively. Remark: 𝑝𝑝. So if you need MATLAB programming homework help, feel free to ask for a quote. Set 1: The Bisection Method Set 2: The Method Of False Position. X +4 X i mark the bracket. Bisection Method of Solving a Nonlinear Equation. % Using the. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. The bisection search. 4 Basis of Bisection. 1shows the several first iterations of the bisection algorithm. Bayen Bisection algorithm to aproximate an equation result. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). Bisection method. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. The Secant Method One drawback of Newton's method is that it is necessary to evaluate f0(x) at various points, which may not be practical for some choices of f. Divide the interval [a, b]. 3 Algorithms and convergence 2. Assumption: The function is continuous and continuously differentiable in the given range where we see the sign change. The algorithm for this is given as follows: Choose a;b so that f(a)f(b) <0 1. Bisection Method The Bisection method is a root finding algorithm. Essentially, the root is being approximated by replacing the. The problem I encounter is that the same computation must be performed on each element of a large array(~1. where the value of the function. By the Nested Interval Property of Real Numbers the sequence of Nested Intervals converge to a unique point, which should therefore be r. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. MATLAB Tutorial – Roots of Equations ES 111 1/13 FINDING ROOTS OF EQUATIONS Root finding is a skill that is particularly well | {
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Equations ES 111 1/13 FINDING ROOTS OF EQUATIONS Root finding is a skill that is particularly well suited for computer programming. In these lectures details about how to use Matlab are detailed (but not verbose) and. m This program will implement Euler’s method to solve the differential equation dy dt = f(t,y) y(a) = y 0 (1) The solution is returned in an array y. Suppose we want to solve the equation. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. Learn more about bisection method, homework. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. 00064404356011 -0. Chapter 6 Finding the Roots of Equations The Bisection Method Copyright © The McGraw-Hill Companies, Inc. 11) uses fzero to calculate the root for this heat capacity example. " by Timmy Siauw and Alexandre M. The software, matlab 2009a was used to find the root of the function for the interval [0,1]. Suppose that f(¢) is a continuous function defined over an interval [a;b] and f(a) and f(b) have opposite signs. This is a quick way to do bisection method in python. function p_min=bisection(func,int,iter,tol_x,tol_f) % It calculates the zero of a regular real function with one variable. The convergence rate of the bisection method could possibly be improved by using a different solution estimate. Bisection method- code stops after one iteration. So the abscissa of point where the chords cuts the x-axis (y=0) is given by,. Because of this, it is often used to roughly sum up a solution that is used as a starting point for a more rapid conversion. Bisection Method Issues/Help. % p_min is the solution and represents the abscissa's value of the zero. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our | {
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$[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. Given these facts. Blog Archive. 6 in the text. It’s take a first approximation by apply two times the Bisection method and complete a correct approximation by use the Newton-Raphson method. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Noanyother restrictionsapplied. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. What are the applications of the bisection. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The x-coordinate of this point is the average of the positive and negative guesses. Newest vertex bisection. However, if there are several solutions present, it finds only one of them, just as Newton's method and the secant method. to a bug in. "In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. There are classical root-finding algorithms: bisection, false position, Newton-Raphson, modified Newton-Raphson, secant and modified secant method, for finding roots of a non-linear equation f(x) = 0 [7,8,9,10,11]. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Later you will learn how to add it to the calling sequence. Introduction. Bisection Method of finding the roots of an equation is both simple and straight forward - I really enjoyed playing with bisection back in college | {
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is both simple and straight forward - I really enjoyed playing with bisection back in college (oooh yeah ES84 days) and I decided to make a post and implement bisection in scilab. algorithm apply axis bisection method boundary condition button Chebyshev Chebyshev nodes clicking coefficient computation constraints converge curve data points defined depicted in Fig derivative diagonal dialog box difference approximation differential equation eigenvalues eigenvectors element end end Example Figure find first flops. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. It will helpful for engineering students to learn Bisection method MATLAB program easily. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. MATLAB coding of all methods. - Bisection method for bounded searching. Examples illustrate important concepts such as selection, crossover, and mutation. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. 6524; m = 73. It is a very simple and robust method, but it is also relatively slow. 1 and ε abs = 0. Here we are required an initial guess value of root. The Algorithm for The Bisection Method for Approximating Roots Fold Unfold. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. This method is also known as Regula Falsi or The Method of Chords. Python Bisection Method. This is a quick way to do bisection method in python. The number of iterations we will use, n, must satisfy the following | {
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do bisection method in python. The number of iterations we will use, n, must satisfy the following formula:. Wednesday, September 28, 11. It requires two initial guesses and is a closed bracket method. Bisection method add iteration table into my code. We are going to find the root of a given function, with bisection method. m with contents. The basic idea is a follows. Matlab will spit out that the root in this interval = '6'. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. Algorithm of Bisection Method [YOUTUBE 9:47] Example of Bisection Method [YOUTUBE 9:53] Advantages & Drawbacks of Bisection Method [YOUTUBE 8:31] MULTIPLE CHOICE TEST : Test Your Knowledge of Bisection Method PRESENTATIONS. For searching a finite sorted array, see binary search algorithm. The problem I encounter is that the same computation must be performed on each element of a large array(~1. m - matlab file that defines Equation (2. Numerical rate of convergence of root has been found in each calculation. This algorithm is a new approach to compute the roots of nonlinear equations f(x)=0, by propose hybrid algorithm between the Bisection algorithm and Newton-Raphson algorithm. The method is based on the following algorithm: Initialization: The bisection method is initialized by specifying the function , the interval [a,b], and the tolerance > 0. Step 3: If f(a). In this tutorial we are going to develop pseudocode for Bisection Method so that it will be easy while implementing using programming language. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Numerical Integration: Rectangle Method. Secant method d. The convergence to the root is slow, but is assured. This series of video | {
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Method. Secant method d. The convergence to the root is slow, but is assured. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Newton's Method in Matlab. Online calculator. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. Bisection method In short, the bisection method will divide one triangle into two children triangles by connecting one vertex to the middle point of its opposite edge. Write a program that calculates the root(s) of a non-linear equation using the bisection method and also the secant method. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. This book serves as a textbook for a first course in numerical methods using MATLAB to solve problems in mechanical, civil, aeronautical, and electrical engineering. Algorithm for Bisection Method: Input function and limits. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. The problem is that it seems like the teachers recommended solution to the task isn't quite right. 6 Newton Method for a System of Nonlinear Equations 191. Find more Mathematics widgets in Wolfram|Alpha. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. "Bisection Method and Algorithm for Solving The Electrical Circuits" Iteration,Bisection Method, Fortran, C, MatLab. This scheme is based on the intermediate value theorem for continuous functions. So, it has a. Also, Newton’s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. in the case of MATLAB®, 16 digits). 000003315132176 10 1. The idea is simple: divide the interval in two, a solution must exist within one subinterval, | {
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10 1. The idea is simple: divide the interval in two, a solution must exist within one subinterval, select the subinterval where the sign of. 2 Newton's Method and the Secant Method The bisection method is a very intuitive method for finding a root but there are other ways that are more efficient (find the root in fewer iterations). Im studying for a math test and on a old test there is a task about bisection. Richard Brent's improvements to Dekker's zeroin algorithm, published in 1971, made it faster, safer in floating point arithmetic, and guaranteed not to fail. Using the Bisection method to find the negative root (only) of the equation ,3C2 — ex — O (a) [5 points) Choose an initial interval [a, b]. 000000828382262 11 1. Also, Newton’s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. Online calculator. 6524; m = 73. Figure 3: Synthetic seismic section displayed in “time from datum” and computed from the log section shown above. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Unless the roots of an equation are easy to find, iterative methods that can evaluate a function hundreds, thousands, or millions of times will be required. 2d Truss Analysis Matlab Program. ^3 - 2; exists. Then faster converging methods are used to find the solution. The basic idea is to switch between inverse quadratic interpolation and bisection based on the step performed in the previous iteration and based on inequalities gauging the difference between guesses:. The algorithm of bisection method is such that it can only find one root between a defined interval. Learn more about bisection method loop. Coding a bisection algorithm using matlab (numerical. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection | {
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matlab (numerical. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Bisection – separate files and embedded functions c. then there is at least one real root in the interval. Access Ebooks on other topics. So let's take a look at how we can implement this. Need help with this bisection method code!. It is a very simple and robust method, but it is also relatively slow. 4 Basis of Bisection. m Algorithm 4. As we point out in the introduction, we will mainly discuss newest vertex bisection and include longest edge bisection as a variant of it. The number of iterations we will use, n, must satisfy the following formula:. Another was to say “root. 21 dcm_to_ypr. Bisection method in MATLAB A video of programming the code in realtime. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Learn the algorithm of the bisection method of solving nonlinear equations of the form f(x)=0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Introduction a. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Solutions to the Exercises from "An Introduction to MATLAB and Numerical Methods for Engineers. Find the midpoint of a and b, say "t". Bisection method b. , Vasiliou, P. Learn more about bisection. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. m, instructions how to run it, an example of a file myfunction. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. Initialization: nd [a 1;b 1] ˆ[a;b], with f(a 1)f(b 1) <0, set i= 1. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. A next search interval is chosen by comparing and nding which one | {
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+2 X-2-4 X-3 X +3 f(x) x Figure 1. A next search interval is chosen by comparing and nding which one has zero. Blog Archive. The first algorithm that I learned for root-finding in my undergraduate numerical analysis class (MACM 316 at Simon Fraser University) was the bisection method. So let's take a look at how we can implement this. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. In mathematics, the bisection method is a root-finding algorithm which repeatedly bisects an interval then selects a subinterval in which a root must lie for further processing. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. MATLAB Programming Assignment Help, Root ?nding using the bisection method, In many applications, including ?nancial mathematics, ?nding zeros of a function f(x) = 0 (4) is paramount. The bisection search. This is achieved by selecting two points A and B on that interval. The convergence to the root is slow, but is assured. Analysis of the Problem. He used it for finding roots of cubic polynomials. m Algorithm 4. Golden section search Fibonacci search Bisection method Secant method Bracketing Chapter 5 Gradient Methods. Input, output. The c value is in this case is an approximation of the root of the function f (x). They are the secant method, bisection method, and newton's method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. The problem is that it seems like the teachers recommended solution to the task isn't quite right. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. In our improved hybrid algorithm, we compute the x-intercept x using the Newton-Raphson method at the midpoint of the previous | {
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we compute the x-intercept x using the Newton-Raphson method at the midpoint of the previous interval. It is a very simple and robust method, but it is also relatively slow. Open methods: Newton-Raphson method, Secant method. MATLAB has the function fzero which performs this bisection algorithm. Numerical Methods using MATLAB, 3e, is an extensive reference offering hundreds of useful and important numerical algorithms that can be implemented into MATLAB for a graphical interpretation to help researchers analyze a particular outcome. If a function is continuous between the two initial guesses, the bisection method is guaranteed to converge. Midpoint Method. 000000828382262 11 1. Fixed Point Method Using Matlab Huda Alsaud King Saud University Huda Alsaud Fixed Point Method Using Matlab. Analysis of the Problem. If the guesses are not according to bisection rule a message will be displayed on the screen. We set [a 0;b 0] = [a;b]. Blog Archive. Lecture 31-33 - Rootfinding Table of Contents 31. A few steps of the bisection method applied over the starting range [a 1;b 1]. m and newton. The setup of the bisection method is about doing a specific task in Excel. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. Implement the Bisection algorithm elegantly and easily (3 answers) Closed 3 years ago. ————————————————. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b, and f(a) and f(b) are of opposite signs, that is,. Root Search with the bisection method. The bisection method in Matlab is quite straight-forward. Joseph DeSimone, Applied Mathematics Graduate Student. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in | {
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Student. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. Essentially, the root is being approximated by replacing the. The instructions of the problem are: Use bisection method to find a root of the function $$\sin x + x \cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. Maple für Akademiker. 23 ground_track. Pseudo-code is a simple way to represent an algorithm in a logical and readable form. 5: Calculation of the state vector from the orbital elements. 6 in the text. Suppose we want to solve the equation. 2) using the bisection method. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. Bisection method add iteration table into my code. But they're not live. If (b i+1 a i+1)=2 > , set i= i+1 and go to step 1 4. GitHub Gist: instantly share code, notes, and snippets. We may minimize a convex f : → by finding a point at which f ′ = 0. (original given interval) Thanks for contributing an answer to Code Review Stack Exchange! Bisection method for finding the root of a function. Repeat steps 3 and 4 100 times. 5 Single Variable Newton- Raphson Method 9. m) and see how we can compute the root to a polynomial using this method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. This tutorial covers in depth algorithm for Bisection Method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guaranteed method for finding real root of non-linear equations. The bisection method is guaranteed to converge to a root of the function f, if the function is continuous between the lower and upper bounds. It will helpful for engineering students to learn Bisection method MATLAB program easily. In simple terms, these | {
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for engineering students to learn Bisection method MATLAB program easily. In simple terms, these methods begin by attempting to evaluate a problem using test (“false”) values for the variables, and then adjust the. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 −4. Bisection Method Example. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. Another class of mesh refinement method, known as regular refinement, which divide one triangle into 4 similar small triangles, is implemented in uniformrefine. Bisection Method // C++ code Posted: January 31, 2012 by muhammadakif in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Also, a good intermediate approximation may be discarded. It is the slowest algorithm provided by the library, with linear convergence. Initialization: nd [a 1;b 1] ˆ[a;b], with f(a 1)f(b 1) <0, set i= 1. Designing Robot Manipulator Algorithms. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. m Algorithm 4. Graphical method useful for getting an idea of what's going on in a problem, but depends on eyeball. You may not use MATLAB's built-in functions for finding roots -- instead, please implement two different algorithms. Algorithms in this toolbox can be used to solve general problems All algorithms are derivative-free methods Direct search: patternsearch Genetic algorithm: ga Simulated annealing/threshold acceptance: simulannealbnd, threshacceptbnd Genetic Algorithm for multiobjective optimization: gamultiobj | {
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simulannealbnd, threshacceptbnd Genetic Algorithm for multiobjective optimization: gamultiobj Kevin Carlberg Optimization in Matlab. For more videos and resources on this topic, please visit http. Let m = (L+H)/2. Roots (Bisection Method) : FP1 Edexcel January 2012 Q2(a)(b) : ExamSolutions Maths Tutorials - youtube Video. I am trying to solve the equation f(x) = x^3 + x - 1 , by using Bisection method within the interval [ 0 , 1] , i have succeeded to generate a code to solve this equation but by using " while " function for looping , i need some one to help me to solve it by using " for " function , could any one help me to do that ? the code is :. "In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. This is calculator which finds function root using bisection method or interval halving method. If the guesses are not according to bisection rule a message will be displayed on the screen. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. In this project we use MATLAB to analyze some of the numerical techniques. Im studying for a math test and on a old test there is a task about bisection. This scheme is based on the intermediate value theorem for continuous functions. It was observed that the Bisection method converges at the 14th iteration while Newton methods. In the secant method, it is not necessary that two starting points to be in opposite sign. 5 Single Variable Newton- Raphson Method 9. Next, we study some known numerical algorithms those can be used to find the | {
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Raphson Method 9. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton–Raphson algorithm and fixed point algorithm. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. t is the root of the given function if f (t) = 0; else follow the next step. Step 2: Let c=(a+b)/2. The program assumes that the provided points produce a change of sign on the function under study. B The comparative results are shown in table 3. The proof that the binary search method works is provided by Bolzano's Bisection Theorem. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The root is then approximately equal to any value in the final (very small) interval. At least one root exists between the two points if the function is real, continuous, and changes sign. 3 Newton's and secant methods 2. 2) using the bisection method. But they're not live. Introduction a. Plot error. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Download MatLab Programming App from Play store. Bisection Method // C# code Posted: January 31, 2012 by Shahzaib Ali Khan in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Useful Computational Methods: The Bisection Method - Finding roots by binary search - Unlike the guess-and-check method, we start with two initial values - one value a below √Q and another value b above √Q, where Q is a positive real number. ContentsDirk DekkerZeroin in AlgolThe test functionBisectionSecant methodZeroin algorithmZeroin in MATLABReferencesDirk DekkerI. As a starting point, let's fix to be the | {
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algorithmZeroin in MATLABReferencesDirk DekkerI. As a starting point, let's fix to be the function cosmx that you just wrote. 1997 CREWES software release CREWES Research Report — Volume 9 (1997) 18-3 Figure 2: The same cross section as above showing the result of the synthesis of an ensemble of new logs. 3 Limits of Accuracy 1. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. 1shows the several first iterations of the bisection algorithm. Designing Robot Manipulator Algorithms. It is a very simple and robust method, but it is also relatively slow. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. The c value is in this case is an approximation of the root of the function f (x). It requires two initial guesses and is a closed bracket method. Always Converges: like Bisection, it. In this project we use MATLAB to analyze some of the numerical techniques. It shows with bold stripes the length of the bracketed region. Newton's method is an iterative method. Error = x- (a+b)/2. The Bisection Method is used to find the zero of a function. The x-coordinate of this point is the average of the positive and negative guesses. Download MatLab Programming App from Play store. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. com 9/27/01. To code the bisection algorithm. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. The method is also called the interval halving method. Basic Bisection Algorithm: 1. We also check whether f(a) = 0 or f(b) = 0, and if so return the value of a or b and exit. Image: The Bisection | {
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whether f(a) = 0 or f(b) = 0, and if so return the value of a or b and exit. Image: The Bisection Method explained. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Here’s what we do: As with the bisection algorithm, start by choosing an interval [a,b] in which we. The bisection method can be easily adapted for optimizing 1-dimensional functions with a slight but intuitive. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 −4. f = @(x) (cos(x)); a = input( 'Please enter lower. Use The Following Pseudocode For The Bisection Method To Write MATLAB Code To Approximate The Root Of F(x) = Pt - X - 2, Interval (0,2), Tolerance 10-3, Maximum Number Of Iterations 50. m to determine the root of Equation (2. The convergence to the root is slow, but is assured. Repeat steps 3 and 4 100 times. Analysis of the Problem. The Bisection Method will cut the interval into 2 halves and check which. f(c)<0 then let b=c, else let a=c. Data scientists use a bisection search algorithm as a numerical approach to find a quick approximation of a solution. Steven Chapra's Applied Numerical Methods with MATLAB, third edition, is written for engineering and science students who need to learn numerical problem solving. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. The brief algorithm of the bisection method is as follows: Step 1: Choose a and b so that f(a). m - matlab file to determine the root of Equation (2. This method will divide the interval until the resulting | {
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file to determine the root of Equation (2. This method will divide the interval until the resulting interval is found, which is extremely small. (b) [5 points) Determine the number of steps N you should take to find the answer with tolerance 0. You have seen how Matlab functions can return several results (the root and the number of iterations, for example). Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. For others, an algorithm of Alefeld, Potra, and Shi is used. So in order to use live solutions, we're going to look at the Bisection Method and then the Golden Section Search Method. Before you start, review the \Introduction to MATLAB" notes. the Matlab code bisection. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. The file EULER. Implement the bisection method to find a zero of the function over [0,1]; Implement the Newton's method to find a zero of the function over [0,1]; Implement the secant method to find a zero of the function over [0,1]. ^3 - 2; exists. This process involves finding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Consult the MATLAB TA's if you have any questions. Powered by Create your own unique website with customizable templates. I found where it was in the directory and added the folder to the path so when I entered it again I now get: C:\Users\Lulu\Documents\MATLAB\Numerical Optimisation\bisection. Viewed 601 times 0 $\begingroup$ To code the bisection algorithm. This method is suitable for finding the initial values of the Newton and Halley's methods. 5: Calculation of the state vector from the orbital elements. Design and simulation of three phase induction motor at different load conditions in matlab/simulink. ) (Use your computer code) I have no idea how to write this code. Root Search with the | {
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) (Use your computer code) I have no idea how to write this code. Root Search with the bisection method. Context Bisection Method Example Theoretical Result The Root-Finding Problem A Zero of function f(x) We now consider one of the most basic problems of numerical approximation, namely the root-finding problem. 7 Symbolic Solution for Equations 193. The bisection method is an iterative algorithm used to find roots of continuous functions. Open methods: Newton-Raphson method, Secant method. Analysis of the Problem. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Readers can code the algorithms in the programs of their choice. The bisection method is an enclosure type method for finding roots of a polynomial f(x), i. - Matlab: function end -vs- SciLab: function endfunction - Matlab: [first, second, third] -vs- Scilab: [third, second, first] Lectures for Spring 2009 Intro to Matlab+ Bisection + Newton Method. MATLAB M-files for implementation of the discussed theory and algorithms (available via the book's website) Introduction to Optimization, Fourth Edition is an ideal textbook for courses on optimization theory and methods. The bisection method in math is the key finding method that continually intersect the interval and then selects a sub interval where a root must lie in order to perform the more original process. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). If the guesses are not according to bisection rule a message will be displayed on the screen. As a starting point, let's fix to be the function cosmx that you just wrote. - Use abstracted Autonomous Guided Vehicles (AGVs) system with P controller and supervisory controller as a case study to validate the conflict-driven fault detection method in | {
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and supervisory controller as a case study to validate the conflict-driven fault detection method in Matlab. Bisection Method MATLAB Program Note: Bisection method guarantees the convergence of a function f(x) if it is continuous on the interval [a,b] (denoted by x1 and x2 in the above algorithm. The program assumes that the provided points produce a change of sign on the function under study. This is a repository where i put all of the implementation that i have done in numerical analysis. A simple improvement to the bisection method is the false position method, or regula falsi. We start from the 2D case sketched in [3] and the approximation scheme presented in [3, 6], and then we extend the reconstruction scheme of separatrices in the. 000013273393044 9 1. Bisection Method Example. The equation is of form, f(x) = 0. We have developed such an algorithm and it is given in the M-file regfals. f(a)*f(b) < 0. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. He used it for finding roots of cubic polynomials. In simple terms, these methods begin by attempting to evaluate a problem using test (“false”) values for the variables, and then adjust the. Algorithm To find a solution. Bisection Method C Program Bisection Method MATLAB Program. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Algorithm Input: computable f(x) and [a;b], accuracy level. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. Includes methods used in MATLAB, Mathcad, Mathematica, and various software libraries. Shown here, it is a function, and it crosses the X-axis at just before 2. In order to avoid the shortcoming of the hybrid algorithm[1], we suggest an improved hybrid | {
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2. In order to avoid the shortcoming of the hybrid algorithm[1], we suggest an improved hybrid algorithm. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. 6 was used to nd the root of the function, f(x) = cosx xexp(x) on a close interval [0;1] using the Bisection method and Newton's method the result was compared. There is the graphical interface too. Bisection Method Bisection Method for finding roots of. ————————————————. This example shows how to generate HDL code from MATLAB® design implementing an bisection algorithm to calculate the square root of a number in fixed point notation. Bisection Method. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. raphson method. This is the same as the slope, k 2 , from the second order midpoint method. It is a very simple and robust method, but it is also rather slow. This is a very simple and powerful method, but it is also relatively slow. Experiment 1. Bisection Method http//numericalmethods. What are the applications of the bisection. This is the first of a three part series. Numerical analysis I 1. Bisection method- code stops after one iteration. Calculates the root of the given equation f(x)=0 using Bisection method. Set r i= (a i+ b i)=2; 2. We won't dwell into explaining what each of them does but will jump straight into explaining the secant method. The basic idea is a follows. Methods by Young and Mohlenkamp, 2018 Bisection Method-- 4 Iterations by Hand (example) Bisection Method-- 4 Iterations by Hand (example) MATLAB Tutorial Part 6 Bisection Method Root finding | {
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Method-- 4 Iterations by Hand (example) MATLAB Tutorial Part 6 Bisection Method Root finding matlab4engineers. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. Method of Steepest Descent Analysis of Optimization Algorithms Analysis of Gradient Methods. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. The algorithm always selects a subinterval which contains a root. The bisection method is an application of the Intermediate Value Theorem (IVT). It is called the bisection method, and it is used for finding roots of a function f (that is, points c where f(c)=. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Method http//numericalmethods. - Bisection method for bounded searching. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. m; the Matlab code fixedpoint. Matlab will spit out that the root in this interval = '6'. Also use Euler's method for the same problem, and compare your results. Interpolation and approximation (12 lecture hours) 3. We almost have all the tools we need to build a basic and powerful root-finding algorithm, Newton's method*. In mathematics, the bisection method is a root-finding method that applies to any. The algorithm is iterative. bisect(list, element, begin, end). The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function. 23 ground_track. bisection method using log10(x)-cos(x) Program to read a Non-Linear equation in one variable, then | {
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bisection method using log10(x)-cos(x) Program to read a Non-Linear equation in one variable, then evaluate it using Bisection Method and display its kD accurate root Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton–Raphson algorithm and fixed point algorithm. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. GitHub Gist: instantly share code, notes, and snippets. m, instructions how to run it, an example of a file myfunction. Also, this method closely resembles with Bisection method. I found it was useful to try writing out each method to practice working with MatLab. Copy to clipboard. changes sign from. Newton's method requires both the function value and its derivative, unlike the bisection method that requires only the function value. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen. How can we qualify more generally which method. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. The algorithm of bisection method is such that it can only find one root between a defined interval. It is also known as Binary Search or Half Interval or Bolzano Method. Here we consider a set of methods that find the solution of a single-variable equation , by searching iteratively through a neighborhood of the domain, in which is known to be located. jf(r i)j<. This is more a problem | {
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a neighborhood of the domain, in which is known to be located. jf(r i)j<. This is more a problem of the algorithm than a MATLAB problem. In Matlab help functions written: The algorithm, created by T. It is a very simple and robust method, but it is also relatively slow. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. Find more Mathematics widgets in Wolfram|Alpha. Bisection algorithm The algorithm itself is fairly straightforward and "fast" in some sense: the number of iterations is roughly Log2 of the ratio of the initial interval length and the desired accuracy. | {
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It is also known as an order of the polynomial. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.For a univariate polynomial, the degree of the polynomial is simply the highest exponent occurring in the polynomial. • If the variable does not have an exponent, the degree is 1. Second Degree Polynomial Function. Trot transitions on a circle: Do transitions between collected and medium trot on a circle, being careful not to let the tempo change. The computer is able to calculate online the degree of a polynomial. Revise the sentences if necessary. The degree of a polynomial with a single variable (in our case, ), simply find the largest exponent of that variable within the expression. A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n … Expression of Interest Form To undertake a Higher Degree by Research in the Faculty of Science, Engineering and Built Environment. Second degree polynomials have at least one second degree term in the expression (e.g. With thousands of questions available, you can generate as many Degree of expression Worksheets as you want. abstract-algebra polynomials rational-functions. To identify the kind of algebraic expression and determine the degree, variables and constant in. Degree words are traditionally classified as adverbs, but actually behave differently syntactically, always modifying adverbs or adjectives and expressing a degree… • If a term does not contain any variable, the degree is 0. See more. The negative and weak positive staining rates in normal appearing mucosa, adenoma, and carcinoma were 42.5%, 71.4%, and 82.3%, respectively, indicating a decreased degree of KLF4 expression over the course of progressive transformation of normal cells into malignant derivatives. Therefore, the degree of this expression is . | {
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of normal cells into malignant derivatives. Therefore, the degree of this expression is . Degrees of Formality (Cont.) The degree of is 3. The correct answer is: [B]: → " 2xy⁴ + 4x²y³ – 6x³y² – 7x⁴" . The First Amendment of the U.S. Constitution protects the rights of individuals to freedom of religion, speech, press, petition, and assembly. 2. a letter to a Member of Parliament:Thank you for your help in this matter. $\begingroup$ To me, and this is standard terminology in theoretical CS, the degree of a boolean function would mean the degree of its polynomial (Fourier) representation. The highest value of the exponent in the expression is known as Degree of Polynomial. The degree of reaction (rth), or reaction ratio, is a parameter used for multistage turbomachinery defining the ratio of the static head (see Clearance gap pressure) to the fall head (of turbines) or the pump head (of centrifugal pumps).It is an indication of how the static pressure is distributed between impeller and stage. Download free printable Degree of expression Worksheets to practice. Solve this set of printable high school worksheets that deals with writing the degree of binomials. Read "Prevalence and degree of expression of the carbapenemase gene (cfiA) among clinical isolates of Bacteroides fragilis in Nottingham, UK, Journal of Antimicrobial Chemotherapy" on DeepDyve, the largest online rental service for scholarly research with thousands of … The degree ofreaction is then expressed as[For axial machines thenThe degree of reaction can also be written in terms of the geometry of theturbomachine as obtained bywhere is the vane angle of rotor outlet and is the vane angle of statoroutlet. Solve an algebraic expression with fractions. Degree of Binomials. It is sum of exponents of the variables in term. 5 Read the sentences below and decide whether or not they use the degree of formality appropriate for the given situation. Degree definition, any of a series of steps or stages, as in a | {
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appropriate for the given situation. Degree definition, any of a series of steps or stages, as in a process or course of action; a point in any scale. Example: x 3 y + x 2 + y x 3 y has degree 4 (3 for x and 1 for y) x 2 has degree 2 y has degree 1 So highest degree is 4, thus polynomial has degree 4 Degree of reaction (R) is an important factor in designing the blades of a turbine, compressors, pumps and other turbo-machinery. Substituting in this expression all numbers we already know, we obtain. And finally, the expression of degree by means of comparison, which is a special case of degree specification. Gene variability and degree of expression of vaccine candidate factor H binding protein in clinical isolates of Neisseria meningitidis. Degree of Polynomial - definition Degree of Polynomial is highest degree of its terms when Polynomial is expressed in its Standard Form. Click hereto get an answer to your question ️ Find the degree of the expression [ x + (x^3 - 1)^1/2]^5 + [ x - (x^3 - 1)^1/2]^5 Calculating the degree of a polynomial. For instance, the degree of 8x is 1. $$\frac{x^2+1}{6x-2}$$ If the above question has a degree, please tell me the difference between the degree of a polynomial and the degree of an expression? On comparing people or things, as bearers of a certain quality or characteristic, we do it by means of degree specification, thus in terms of positive, comparative, and superlative comparison. Correct answer to the question Which algebraic expression is a polynomial with a degree of 5? The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.To find the degree all that you have to do is find the largest exponent in the polynomial.Note: Ignore coefficients-- coefficients have nothing to do with the degree of a polynomial. Exploring these different trots can help you to increase your horse’s collection, cadence and expression. The collection has to have the same degree … For instance, the | {
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collection, cadence and expression. The collection has to have the same degree … For instance, the degree … Cox proportional hazards model was used to assess the relationship between the degree of expression of DNMT1 and overall survival after adjusting for relevant covariates. 1. a note to a co-worker:The meeting is at 10 sharp.Don’t be late. • If a term has more than one variable, the degree is equal to the sum of the exponents of all its variables. The calculator may be used to determine the degree of a polynomial. The quadratic function f(x) = ax 2 + bx + c is an example of a second degree polynomial. Freedom of expression refers to the ability of an individual or group of individuals to express their beliefs, thoughts, ideas, and emotions about different issues free from government censorship. Thank you for your interest in studying a Higher Degree by Research (HDR) course at Deakin University. But it seems unlikely that the term is used in this sense here; also I do not see a nice way … Remember that a polynomial is any algebraic expression that consists of terms in the form $$a{x^n}$$. There are no higher terms (like x 3 or abc 5). For the reaction : 2HI(g) ⇌ H2 (g)+ I2 (g), the degree of dissociated (α) of HI(g) is related to equilibrium constant Kp by the expression: asked Jan 6, 2020 in Chemistry by Mousam ( 52.8k points) While finding the degree of the polynomial, the polynomial powers of the variables should be either in ascending or descending order. Find the degree of each term and then compare them. If you're interested in studying the Associate Degree of Advanced Manufacturing, please send an expression of interest to us at adaminfo@uts.edu.au. Degree of Polynomials. Here's how you would do it: (x + 3)/6 = 2/3 First, cross multiply to get rid of the fraction. The degree of a polynomial is the largest exponent. Free Class 7 Degree of expression Worksheets. Note: This answer choice is a polynomial with a degree of "5" .Note that there are 2 (TWO) | {
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Note: This answer choice is a polynomial with a degree of "5" .Note that there are 2 (TWO) variables in this polynomial expression The expression of DNMT1 was examined using immunohistochemistry, and the degree of expression of DNMT1 was expressed as a percentage of cells positive for DNMT1 and its intensity. Another way to write the last example is $- 8{x^0}$ Written in this way makes it clear that the exponent on the $$x$$ is a zero (this also explains the degree…) and so we can see that it really is a polynomial in one variable. In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. Once assessed, you will also need to apply for 609996 Associate Degree of Advanced Manufacturing via UAC. The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. The highest power is the degree of … To obtain the degree of a polynomial defined by the following expression x^3+x^2+1, enter : degree(x^3+x^2+1) after calculation, the result 3 is returned. The degree of is 6. Polynomial is a mathematical expression consisting of variables, constants that can be combined using mathematical operations addition, subtraction, multiplication and whole number exponentiation of … In turbomachinery, Degree of reaction or reaction ratio (R) is defined as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage.. If you want to solve an algebraic expression that uses fractions, then you have to cross multiply the fractions, combine like terms, and then isolate the variable. 2x 2, a 2, xyz 2). Degree of Dissociation. 4x3 – StartFraction 2 Over x EndFraction 2y3 + 5y2 – 5y 3y3 – StartRoot 4 y EndRoot Question: what is the degree of the following expression? The Fixed Class of Degree Words "[An] example of words that | {
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the degree of the following expression? The Fixed Class of Degree Words "[An] example of words that don't fit neatly into one category or another is degree words. The degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. In Freemasonry, the third and highest degree is that of master mason, attained after a stiff examination, and several writers speculate that this may be the source of the late nineteenth-century expression for an inquisition. There are 2 offer rounds remaining for UAC, so please check the submission deadlines with UAC. Kelly A(1), Jacobsson S, Hussain S, Olcén P, Mölling P. Author information: (1)Department of Clinical Medicine, School of Health and Medical Sciences, Örebro University, Sweden. Which algebraic expression is a polynomial with a degree of 3? To give or be subjected to intensive questioning and/or rough treatment. 3x5 + 8x4y2 – 9x3y3 – 6y5 2xy4 + 4x2y3 – 6x3y2 – 7x4 8y6 + y5 – 5xy3 + 7x2y2 – x3y – 6x4 –6xy5 + 5x2y3 – x3y2 + 2x2y3 – 3xy5 - e-eduanswers.com Worksheets that deals with writing the degree of the polynomial powers of the variables term..., the polynomial powers of the variables in term [ B ]: → 2xy⁴ + 4x²y³ 6x³y²... In studying a higher degree by Research ( HDR ) course at Deakin University Advanced via... – 6x³y² – 7x⁴ '' the given situation is 1 polynomials have at least one second degree term in expression! Degree by Research ( HDR ) course at Deakin University have at least one second degree polynomial t late... Its Standard Form, compressors, pumps and other turbo-machinery Manufacturing via UAC Over x 2y3. Letter to a Member of Parliament: thank you for your interest studying! Letter to a Member of Parliament: thank you for your interest in studying a higher degree Research! A co-worker: the meeting is at 10 sharp.Don ’ t be.. May be used to determine the degree is 1 in ascending or descending order variables | {
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you for your help in this expression all numbers we know. Or descending order, so please check the submission deadlines with UAC instance, the of. Many degree of Formality ( Cont. determine the degree of a polynomial is any algebraic is. + 5y2 – 5y 3y3 – degree of expression 4 y EndRoot Degrees of Formality appropriate the. A polynomial term shows being raised to anything larger than seven of its when. Worksheets to practice each term and then compare them Standard Form and then compare them seventh! 3Y3 – StartRoot 4 y EndRoot Degrees of Formality appropriate for the given situation decide whether or they. Uac, so please check the submission deadlines degree of expression UAC is able to calculate online the degree reaction... The expression ( e.g is raised to the seventh power, and no other in this expression is to... 5Y 3y3 – StartRoot 4 y EndRoot Degrees of Formality appropriate for the given.... Generate as many degree of its terms when polynomial is highest degree of expression Worksheets to practice ax. Is: [ B ]: → 2xy⁴ + 4x²y³ – 6x³y² – 7x⁴ '' horse... • If a term does not contain any variable, the polynomial, the degree 8x! 5 ) the correct answer to the question Which algebraic expression is as... Or be subjected to intensive questioning and/or rough treatment of all its variables 7x⁴.... Have at least one second degree polynomials have at least one second degree in! Expression and determine the degree of 8x is 1 Read the sentences below and decide whether or not use. The correct answer is: [ B ]: → 2xy⁴ 4x²y³. You will also need to apply for 609996 Associate degree of polynomial is the degree is 1 of... + bx + c is an important factor in designing the blades of a polynomial is any expression! One second degree polynomial: thank you for your interest in studying a higher degree by (... 2 + bx + c is an important factor in designing the blades of a second term! A { x^n } \ ) interest in studying a higher degree by Research ( HDR ) course Deakin! { x^n } \ ) | {
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# Mathematical field, show that 0·a = 0, (-1)·a = -a, …
Based on the axioms for a mathematical field, the wiki article states that 0·a = 0 and (-1)·a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.
https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition
Also, it would seem logical that if a ≠ b, and if c ≠ 0, then c·a ≠ c·b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) . | {
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• These are in fact standard easy arguments (your guess in the last sentence is correct). I'm sure someone will step up and provide them. Here's a hint for the first one: consider $(0 + 0) \times a$. I'm curious: what prompts your curiousity? A course you're taking? – Ethan Bolker Apr 22 '17 at 0:12
• @EthanBolker - a question came up at another forum. An "addition" table was given for a finite field with 4 numbers (0,1,2,3), where the "addition" turns out to be exclusive or, and the problem was asking to produce the multiplication table, based on the axioms rather than knowledge of GF(4). That prompted me to wonder how the "consequences" as noted in the wiki article are derived, and what operations are considered acceptable as derivations for field math. – rcgldr Apr 22 '17 at 0:42
• @EthanBolker - I'm aware that algebra works with finite finite fields (I've worked with RS ECC), but never thought about the derivations based on the axioms. For that other forum question, after taking into account 0·a = a·0 = 0 and 1·a = a·1 = a, only 4 products in the multiplication table (2·2, 2·3, 3·2, 3·3) needed to be determined, so this could be done by trial and error, where an axiom would fail if the wrong set of 4 products was chosen. Noting that every number has a multiplicative inverse, meant that two of those products = 1 => 2·3 = 3·2 = 1, which simplified the problem. – rcgldr Apr 22 '17 at 1:06
For $0 \times a$: $$0 \times a = (0+0)\times a = 0\times a + 0 \times a .$$ Now whatever $0 \times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \times a = 0$.
For $(-1) \times a$: $$0 \times a = (1 + (-1)) \times a = 1 \times a + (-1) \times a = a + (-1) \times a$$ but $a$ has a unique additive inverse $-a$.
If $c \ne 0$ then it has a multiplicative inverse $d$. Then $$ca = cb \implies ca - cb = c(a-b) = 0 \implies dc(a-b) = 0 \implies a-b = 0 \implies a = b.$$ | {
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• Shouldn't the third part somewhere include a - b ≠ 0 or perhaps a - b = e ≠ 0 ? – rcgldr Apr 22 '17 at 0:26
• @rcgldr No. I proved your claim by reasoning with the contrapositive form. If $a \ne b$ then the last entry in my chain of correct implications is false, so the first entry must be false. – Ethan Bolker Apr 22 '17 at 0:30
• OK, thanks for the explanation. I was also considering let a-b = e ≠ 0, then ca - cb = ce ≠ 0, since c ≠ 0 and e ≠ 0. – rcgldr Apr 22 '17 at 0:35
I presume that by "-a" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a. Again, that follows from the distributive law. (1+ (-1))a= 0a= 0. But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also. "a+ (-1)a= 0" is precisely what is meant by "additive inverse of a".
The statement "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is most easily proved by proving the "contrapositive". For any statement "if p then q", the contrapositive is the statement "if not q then not p" and it is easy to show in general that a statement is true if and only if its contrapositive is true.
The contrapositive of "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is "if ca= cb then a= b". To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0. By the distributive law, c(a- b)= 0. Since c is not 0 we must have a- b= 0 so a= b.
• Should the contrapositive be stated as $$\text{If ca=cb the either a=b or c=0}?$$ – Juniven Apr 22 '17 at 0:37
$(-1)a=-a$ means $a+(-1)a=0$ as additive inverses are unique.
$a+(-1)a=1*a+(-1)a$ (existence of multiplicative identity.
$=(1+(-1))a$ (distributive property)
$=0*a$. Remains to show $0*a=0$ for all $a$
$0*a= (0+0)*a$ (Associativity and definition of additive property.
$=0*a+0*a$.
$0a=0a+0a$
$0a+(-0a)=0a+0a+(-0a)$ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)
$0=0a+0=0a$(additive inverse and implied associtivity.) | {
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$0=0a+0=0a$(additive inverse and implied associtivity.)
So $a+(-1)a=0$. So $(-1)a=-a$ (and $a= -(-1 (a))$.) | {
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This document presents the close relationship between Spearman correlation and Pearson correlation. Namely, that Spearman correlation is just a Pearson correlation on $$rank$$ed $$x$$ and $$y$$. It is an appendix to the post “Common statistical tests as linear models”.
TL;DR: Below, I argue that this approximation is good enough when the sample size is 12 or greater and virtually perfect when the sample size is 20 or greater.
# 1 Simple example
Here, I start by creating some mildly correlated data. Note that the distribution of the data does not matter since everything is ranked, so the results are identical for non-normal data.
data = MASS::mvrnorm(50, mu=c(2, 2), Sigma=cbind(c(1, 0.4), c(0.4, 1)))
x = data[,1]
y = data[,2]
plot(rank(y) ~ rank(x))
Now let’s test it
# Three ways of running the same model
spearman = cor.test(x, y, method='spearman')
pearson = cor.test(rank(x), rank(y), method='pearson') # On ranks
linear = summary(lm(rank(y) ~ rank(x))) # Linear is identical to pearson; just showing.
# Present the results
print('coefficient rho is exactly identical:')
rbind(spearman=spearman$estimate, ranked_pearson=pearson$estimate,
ranked_linear=linear$coefficients[2]) print('p is approximately identical:') rbind(spearman=spearman$p.value,
ranked_pearson=pearson$p.value, ranked_linear=linear$coefficients[8])
## [1] "coefficient rho is exactly identical:"
## rho
## spearman 0.1848259
## ranked_pearson 0.1848259
## ranked_linear 0.1848259
## [1] "p is approximately identical:"
## [,1]
## spearman 0.1982118
## ranked_pearson 0.1988046
## ranked_linear 0.1988046
The correlation coefficients are exact. And the p-values are quite close! But maybe we were just lucky? Let’s put it to the test:
# 2 Simulate for different N and r
The code below does the following: | {
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# 2 Simulate for different N and r
The code below does the following:
1. Generate some correlated data for all combinations of the correlation coefficient ($$r = 0, 0.5, 0.95$$) and sample sizes 1 $$N = 6, 8, 10, ... 20, 30, 50, 80$$.
2. Compute coefficients and p-values for Spearman correlation test and Pearson correlation on the ranked values, many times for each combination.
3. Calculate the number of errors.
library(tidyverse)
# Settings for data simulation
PERMUTATIONS = 1:200
Ns = c(seq(from=6, to=20, by=2), 30, 50, 80) # Sample sizes to model
rs = c(0, 0.5, 0.95) # Correlation coefficients to model
# Begin
D = expand.grid(set=PERMUTATIONS, r=rs, N=Ns) %>% # Combinations of N and r
mutate(
# Use the parameters to generate correlated data in each row
data = map2(N, r, function(N, r) MASS::mvrnorm(N, mu=c(0, 4), Sigma=cbind(c(1, r), c(r, 1)))),
# Tests
pearson_raw = map(data, ~cor.test(rank(.x[, 1]), rank(.x[, 2]), method='pearson')),
spearman_raw = map(data, ~cor.test(.x[, 1], .x[, 2], method = 'spearman')),
# Tidy it up
pearson = map(pearson_raw, broom::tidy),
spearman = map(spearman_raw, broom::tidy),
) %>%
# Get estimates "out" of the tidied results.
unnest(pearson, spearman, .sep='_') %>%
select(-data, -pearson_raw, -spearman_raw)
# If you want to do a permutation-based Spearman to better handle ties and
# overcome issues on how to compute p (computationally heavy).
# Corresponds very closely to R's spearman p-values
#mutate(p.perm = map(data, ~ coin::pvalue(coin::spearman_test(.x[,1] ~ .x[,2], distribution='approximate')))) %>%
#unnest(p.perm)
head(D)
### 2.0.1 Inspect correlation coefficients
Just as we saw in the simple example above, correlation coefficients (estimate and estimate1) are identical per definition. We can see that by calculating the difference and see that it is always zero:
summary(D$estimate - D$estimate1)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## | {
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summary(D$estimate - D$estimate1)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
##
Yup. The minimum difference is zero. The maximum is zero. All of them are identicalNo need to look further into that.
### 2.0.2 Inspet p-values
Before getting started on p-values, note that there are several ways of calculating them for Spearman correlations. This is the relevant section from the documentation of cor.test:
For Spearman’s test, p-values are computed using algorithm AS 89 for n < 1290 and exact = TRUE, otherwise via the asymptotic t approximation. Note that these are ‘exact’ for n < 10, and use an Edgeworth series approximation for larger sample sizes (the cutoff has been changed from the original paper).
You can get beyond all of that by doing permutation-based tests at the cost of computational power (see outcommented code above). However, there is almost virtually perfect correspondence, so I just rely on R’s more computationally efficient p-values.
Below, I plot the difference between Spearman-derived p-values and ranked-Pearson-derived p-values. I have chosen not to plot as a function or r in the name of simplicity because it makes no difference whatsoever, but feel free to try it out yourself.
library(patchwork)
# A straight-up comparison of the p-values
p_relative = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value, color=N)) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
geom_hline(yintercept=0.05, lty=2) +
labs(title='Absolute relation', x = 'Spearman p-value', y = 'Pearson p-value') +
#coord_cartesian(xlim=c(0, 0.10), ylim=c(0, 0.11)) +
theme_gray(13) +
guides(color=FALSE)
# Looking at the difference (error) between p-values
p_error_all = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
labs(title='Error', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005)) +
theme_gray(13) +
guides(color=FALSE) | {
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} |
# Same, but zoomed in around p=0.05
p_error_zoom = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
labs(title='Error zoomed', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005), xlim=c(0, 0.10)) +
theme_gray(13)
# Show it. Patchwork is your friend!
p_relative + p_error_all + p_error_zoom
Figure: The jagget lines in the leftmost two panels are due to R computing “exact” p-values for Spearman when N < 10. For samples larger than N>=10, an approximation is used which gives a smoother relationship.
It is clear that larger N gives a closer correspondence. For small p, the Pearson-derived p-values are lower than Spearman, i.e. more liberal. Importantly, when N=12, this effect is at most 0.5% in the “significance-region” which I deem acceptable enough for the present purposes. When N=20, the error is at most 0.2% for the whole curve and there is no difference when N=50 or N=100.
# 3 Conclusion
The correlation coefficients are identical.
The p-value is exact enough when N > 10 and virtualy perfect when N > 20. If you are doing correlation studies on sample sizes lower than this, you have larger inferential problems than those posed by the slight differences between Spearman and Pearson. | {
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# Negative roots when simplifying surds
#### NeedingWD40
##### New member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
#### Dr.Peterson
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.
So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.
Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
#### Jomo
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.
#### NeedingWD40 | {
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#### NeedingWD40
##### New member
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.
So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.
Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
Ah! That's why the formula for solving a quadratic equation explicitly says + or -⎷(b2-4ac), otherwise only the positive root would be found. Thank you very much, I'm clearer now
#### JeffM
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
You are absolutely correct that
$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$
But the technical definition of the square root function of a non-negative real number is:
$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$
That is why if we get an equation that looks like
$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$
That plus or minus sign is there for a reason.
#### NeedingWD40 | {
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That plus or minus sign is there for a reason.
#### NeedingWD40
##### New member
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.
Thank you, I'm clearer now.
#### NeedingWD40
##### New member
You are absolutely correct that
$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$
But the technical definition of the square root function of a non-negative real number is:
$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$
That is why if we get an equation that looks like
$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$
That plus or minus sign is there for a reason.
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
#### tkhunny
##### Moderator
Staff member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
"problem includes ... no plus or minus" - Not a fan.
What is $$\displaystyle \sqrt{4}$$? It is 2
I do not ever recall seeing such a simplification problem as: What is $$\displaystyle \pm\sqrt{4}$$? It is +/- 2.
Let the context be your guide.
Don't make up things that don't exist. $$\displaystyle \sqrt{4} = 2$$ and that is NOT the same as "+/-2".
Don't miss things that do exist. $$\displaystyle Solve\;x^{2} = 4$$. $$\displaystyle x = +2\;or\;x = -2$$ and that is the same as "x = +/-2".
Also, keep your eyes on that "+/-" symbol. It means several different things in different contexts.
#### Otis | {
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#### Otis
##### Senior Member
I'm not sure whether you've yet learned anything about solving algebraic equations or the concept of a function (you posted on the Pre-Algebra board), but here's my comments, anyways.
When you are given a surd, it always represents the positive root (only is exception is √0).
When you are solving an equation, sometimes you need to introduce surds (like the algebraic step of taking the square root of each side of an equation). In this case, you need to consider both the positive and negative root. Otherwise, you might miss solution(s).
As Jeff wrote, we express a positive root as √n, and when we need to express the negative root we write -√n. Cheers :cool:
#### JeffM
##### Elite Member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
Please read the other responses to this post because they are good and will give you different perspectives.
$$\displaystyle \sqrt{a} \ge 0$$ by definition if we are dealing with real numbers.
So, we do not assume that the square root is the primary root (meaning a non-negative number), we know that it is a non-negative number by definition.
However, if the square root of a is positive, we know that the square of the square root's additive inverse is also a. In notation,
$$\displaystyle \sqrt{a} * \sqrt{a} = a = (-\ \sqrt{a}) * (-\ \sqrt{a}).$$
So if we have an equation like
$$\displaystyle x^2 = c \ge 0$$,
we have no way to know from the equation itself whether
$$\displaystyle x = \sqrt{c} \text { or } x = - \ \sqrt{c}.$$
It may be that either solution works or that only one of the two works, but that must be decided based on information not contained in the equation itself. | {
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Uncertainty about which answer applies does not mean that there is any uncertainty about what the surd means. To clarify where the uncertainty lies we say
$$\displaystyle x = \pm \sqrt{c}.$$
The sign of the surd is certain. It is the sign of x that is uncertain.
Last edited:
#### pka
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$
#### JeffM | {
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#### JeffM
##### Elite Member
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$
I like this convention that the surd can only be applied to non-negative real numbers. It would avoid a lot of confusion.
Unfortunately, conventions are necessarily extra-individual. And the surd is frequently used in a broader sense.
Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
$$\displaystyle \sqrt[3]{-\ 8}$$
or is that prohibited as well. If so, is there a convenient notation to deal with it?
#### pka
##### Elite Member
Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
$$\displaystyle \sqrt[3]{-\ 8}$$
or is that prohibited as well. If so, is there a convenient notation to deal with it?
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$
#### JeffM
##### Elite Member
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$
Ahh so your school's rule is | {
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$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$
$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$
#### pka
##### Elite Member
Ahh so your school's rule is
$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$
$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.
There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$
#### lookagain
##### Senior Member
NeedingWD40 said:
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
The above is incorrect. You need grouping symbols, such as below. I used asterisks for the products.
e.g.
⎷300 - ⎷48 =
⎷(100*3) - ⎷(16*3) =
10⎷3 - 4⎷3 =
6⎷3
Last edited:
#### topsquark
##### Full Member
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.
There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$
$$\displaystyle = 2 \cdot \left ( e^{ 3i \pi/2 + 2 i \pi n } \right ) ^{1/4}$$
$$\displaystyle = 2 \cdot e^{3i \pi /8 + i \pi n / 2}$$ | {
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$$\displaystyle = 2 \cdot e^{3i \pi /8 + i \pi n / 2}$$
$$\displaystyle = 2 \cdot e^{3i \pi / 8} \cdot i^n$$
$$\displaystyle = 2 (i ^n) \left ( cos ( 3 \pi /8 ) + i~sin (3 \pi / 8) \right )$$
where n is an integer.
Perhaps it's a bit messy but I don't have to think so hard about any negative signs.
-Dan
#### pka
##### Elite Member
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$
-Dan
I did a very poor job explaining our objections. No one would ever need to use $$\displaystyle \sqrt[4]{-16 i }.$$
On the other hand there could be a need to solve $$\displaystyle z^4+16\bf{i}=0$$
There are four roots of that equation. If $$\displaystyle \rho=2\exp\left(\frac{-\pi\bf{i}}{8}\right)$$ then $$\displaystyle \rho^4+16\bf{i}=0$$, in other words one solution.
Let $$\displaystyle \zeta=\exp\left(\frac{2\pi\bf{i}}{4}\right) \$$ now the collection $$\displaystyle \rho\cdot\zeta^k,~k=1,2,3$$ are the other three roots(solutions).
#### NeedingWD40
##### New member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!
So, to attempt to summarise:
⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
#### JeffM
##### Elite Member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!
So, to attempt to summarise:
⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
That is very good. I'd make one slight emendation to your third point. I'd phrase it as
The context of the equation should clarify if both solutions are valid or else which one of the two is valid. | {
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I would not assume that the negative solution is the only one that may not apply. In some cases, what is classified as positive and negative is arbitrary. It is of course true that we usually classify in a way that ensures a valid answer is non-negative, but that is not guaranteed.
Very minor point. You seem to have it. Good job. | {
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# Trick to find multiples mentally
We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$? | {
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-
You'll want to see this KCd blurb. – Guess who it is. Dec 8 '11 at 15:49
I've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. – Thomas Andrews Dec 8 '11 at 16:04
How can we find divisors of 23? Very easily: they are just $\pm 1$ and $\pm 23$. Presumably, you want to recognize multiples of 23? – Arturo Magidin Dec 8 '11 at 16:11
For $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. – André Nicolas Dec 8 '11 at 16:16
@ArturoMagidin Yes, I meant multiples. It's corrected. – Oltarus Dec 9 '11 at 8:46
One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \$ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\$ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\$ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\$ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad$ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1$ | {
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$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad$ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5$
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad$ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2$
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad$ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0$
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).
-
One approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$
-
I generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. – Qiaochu Yuan Dec 8 '11 at 18:28
@QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. – Ross Millikan Dec 8 '11 at 18:40
In general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$ | {
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More generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\equiv 1\pmod n$. Then $X\equiv 0\pmod n$ if and only if $Xm\equiv 0\pmod n$. But if $X=Bu+v$, then $Xm\equiv u+mv\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.
In the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.
Or, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.
In general, you can always find a $k$ such that $10^k=1\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.
Usually, you want to find a relatively small pair $(m,k)$ so that $m10^k\equiv \pm 1\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.
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The following is a simple method to check divisibility by $7$ or $13$:
Cut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.
For example
$12345631241$ leads to
$$241-639+345-12=-65 \,.$$
Thus our number is a multiple of $13$, but not a multiple of $7$.
This works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.
Method 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.
Let say that this multiple is $a_1..a_k1$. | {
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Let say that this multiple is $a_1..a_k1$.
Then you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.
The trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...
A simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .
Also, for small numbers the following
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I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,
$$10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$
The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$
To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1) + \cdots$$ Since all this numbers $(10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.
Using this approach we can derive divisibility of any integer.
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This can be explained using Horner's method as below. z=a0 + a1 ×10 + a2 ×102 + a3 × 10^3+⋯+ an × 10^n
i.e. Z = ((((...(an × 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0
( Shift-add representation of number)
Z % 7 = ((((...(an × 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0 (Mod 7)
Now evaluate from left to right using modulo algebra. | {
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Now evaluate from left to right using modulo algebra.
To know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .
Regard, | {
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1. ## Help with reducing quadratic equation to final terms
$x^4 - 6x^2 + 8 = 0
$
$x^2=t$
$x^4=t^2$
a= 1, b = 6, c = 8
$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$
$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$
$\Rightarrow x = \frac {3 \pm \sqrt {2}}{}$
How do I reduce this further to get the final answer or is my answer correct? I always get confused. Thanks in advance.
2. Originally Posted by lilrhino
$x^4 - 6x^2 + 8 = 0
$
$x^2=t$
$x^4=t^2$
a= 1, b = 6, c = 8
$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$
$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$
$\Rightarrow x = \frac {3} \pm \frac { \sqrt {2}}{2}$
How do I reduce this further to get the final answer. I always get confused. Thanks in advance.
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here
you replaced $x^2$ with $t$, so you should have
$t^2 - 6t + 8 = 0$ .........now factor
$\Rightarrow (t - 4)(t - 2) = 0$
$\Rightarrow t = 4 \mbox { or } t = 2$
But $t = x^2$
$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$
$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
3. Hello, lilrhino!
You're making it unnecessarily complicated . . . and wrong.
$x^4 - 6x^2 + 8 \:= \:0$
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$
And we have two equations to solve:
. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$
. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
4. Originally Posted by Jhevon
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here
you replaced $x^2$ with $t$, so you should have
$t^2 - 6t + 8 = 0$ .........now factor
$\Rightarrow (t - 4)(t - 2) = 0$
$\Rightarrow t = 4 \mbox { or } t = 2$
But $t = x^2$
$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$ | {
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$\Rightarrow t = 4 \mbox { or } t = 2$
But $t = x^2$
$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$
$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
5. Originally Posted by lilrhino
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
6. Originally Posted by Soroban
Hello, lilrhino!
You're making it unnecessarily complicated . . . and wrong.
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$
And we have two equations to solve:
. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$
. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
I do that often unfortunately. Math is not my strongest subject, so I sometimes make things more complicated that necessary. Thanks for your response. | {
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7. Originally Posted by Jhevon
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
I'm not easily shocked, I may spend more time thinking about it though . It's easier the way Soroban demonstrated it actually, so I'm not shocked. Thanks... | {
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# Tiling a square with rectangles
Consider the set of all the rectangles with dimensions $$2^a\times 2^b\,a,b\in \mathbb{Z}^{\ge 0}$$. We want to tile an $$n\times n$$ square by rectangles from this set (you can use a rectangle several times). What is the minimum number of rectangles we need?
If $$f(n)$$ is the sum of digits of $$n$$ in base $$2$$, I think we need at most $$f(n)^2$$ rectangles. I have an example for this number: write $$n=2^{a_1}+2^{a_2}+...2^{a_{f(n)}}$$ and split each side to segments with length $$2^{a_1},2^{a_2},...,2^{a_{f(n)}}$$ and consider $$f(n)^2$$ rectangles obtained this way.
On the other hand, you need at least $$f(n)$$ rectangles to tile a raw (or column) so I think you need $$f(n)^2$$ rectangles, but I can't prove it.
Any ideas?
• By $f(n)$ do you mean the sum of the bits in the binary representation of $n$? Apr 3, 2019 at 20:18
• @JohnWaylandBales yes f(n) is the least number such that $n=2^{a_1}+2^{a_2}+...2^{f(n)}$
– ali
Apr 3, 2019 at 20:23
• You mean $f(n)$ is the least number such that $n = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_{f(n)}}$ right? Because $f(n)$ counts the number of terms, but it is not the highest exponent. Apr 4, 2019 at 3:51
• @antkam yes i edited the question
– ali
Apr 4, 2019 at 4:36
MAJOR UPDATE: | {
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MAJOR UPDATE:
What I am about to show is not a proof for the minimum number of rectangles. However, in some cases I found the number of rectangles can be less than $$f(n)^2$$. The smallest $$N×N$$ grid that I have found that can have less than $$f(n)^2$$ rectangles is $$15×15$$, which is displayed below: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&6&6&6&6&6&6&6&6\\ \hline 1&1&1&1&2&2&3&7&8&9&9&10&10&10&10\\ \hline 11&11&11&11&11&11&11&11&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline \end{array}$$ | {
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The method used in the above $$15×15$$ square can be generalized not just to other squares but to rectangles as well. In order to make full use of this method, I will expand the op's method to rectangles. Let the length of a rectangle be equal to $$m$$ units and the width be $$n$$ units. Then the number of base-2 rectangles used to cover a $$m × n$$ rectangle by the op's method is $$f(m)f(n)$$. My method isn't fundamentally different from the op's method. It splits the $$m×n$$ rectangle into five sub-rectangles, then the op's method is applied to each of the five rectangles. In some cases the number of base-2 rectangles that covers the five sub-rectangles is less than the number of base-2 rectangles that cover original $$m$$×$$n$$ rectangle using the op's method. The five rectangles are arranged so that their are two pairs of rectangles that occupy the corners and one rectangle that is in the middle (not touching the perimeter). Each pair of rectangles are the same size and orientation but in opposite corners. (Top left and bottom right or Top right and bottom left.) The length and width of the five rectangles are constructed from two other unit lengths $$a$$ and $$b$$. $$a$$ is the smallest number such that $$m+a$$ is a power of two. $$b$$ is the smallest number such that $$n+b$$ is a power of two. This means that $$f(m+a)$$ and $$f(n+b)$$ are each one. The length and width of the two rectangles in the first pair are $$f\left(\frac{m+a}{2}\right)$$ and $$f\left(\frac{n-b}{2}\right)$$ respectively. The length and width of the two rectangles in the second pair are $$f\left(\frac{m-a}{2}\right)$$ and $$f\left(\frac{n+b}{2}\right)$$ respectively. The formula for the total number of base-2 rectangle used is $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)$$. | {
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Note that if a square with a length of $$n$$ units is of the form $$2^xy$$ where $$x,y\in\Bbb{N}|x\ge 1,y\ge 1$$ and $$y$$ is odd. Then the number of base-2 rectangles used for both the op's method and my method are the the same as the number of base-2 rectangles used for a square of length $$y$$ because each of the dimensions of the sub-rectangles can be multiplied by $$2^x$$. For example if we want to determine how many base-2 rectangles is rectangles are required to cover a $$30×30$$ square using my method. We just use the $$15×15$$ example near the top of this post and multiply the length and width of each base-2 rectangle by $$2$$. So this means the $$30×30$$ square requires the same number of base-2 rectangles as the $$15×15$$ square. So the problem can be simplified to just rectangles where $$m$$ and $$n$$ are odd.
A simple inequality can be made which would indicate which method uses less base-2 rectangles. Let $$N_l$$ be the number of ones in the number for length of the rectangle in binary and $$N_w$$ be the number of ones in the width in binary $$\bigl($$or more simply $$N_l=f(m)$$ and $$N_w=f(n)\bigr)$$. Also Let $$Z_l$$ be the number of zeros in the number for length of the rectangle in binary, $$Z_w$$ be the number of zeros in the width in binary. My method uses less rectangles than the op when $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)\lt f(m)f(n)$$
$$f\left(\frac{m+a}{2}\right)=1$$ $$f\left(\frac{n+b}{2}\right)=1$$ $$f\left(\frac{m-a}{2}\right)=N_l-1$$ $$f\left(\frac{n-b}{2}\right)=N_w-1$$ $$f(m)=N_l$$ $$f(n)=N_w$$ $$f(a)=Z_l+1$$ $$f(b)=Z_w+1$$
With the above substitutions the inequality can be changed to:
$$2(N_l-1)+2(N_w-1)+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$2N_l+2N_w-4+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$(Z_l+1)(Z_w+1)\lt N_lN_w-2N_l-2N_w+4$$ $$(Z_l+1)(Z_w+1)\lt (N_l-2)(N_w-2)$$ | {
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In the specific case of the square (where the length equals the width) my method uses less base-2 rectangles than the op when the number ones in the binary representation of the length is at least four more than than the number of zeros. To get the maximum utility out of my method the inequality shouldn't only be applied to the entire length and width of the main square it should also be applied to components of the square. For example consider the square $$1927×1927$$. The binary representation of 1927 is 11110000111. There are three more ones than zeros in this number so my method would normally break even with the op, covering the square with 49 base-2 rectangles. There is a way to cover the square using less base-2 rectangles by spliting the square into four rectangles $$1920×1920$$, $$1920×7$$, $$7×1920$$, and $$7×7$$. Splitting this way doesn't change the net result of the op's method. The first three sub rectangles satisfies the inequality. If I use my method on the first three sub rectangles I use 13, 11, and 11 base-2 rectangles respectively. Using the op's method on the last sub rectangle then counting up all of the base-2 rectangles I can cover the $$1927×1927$$ square using 44 base-2 rectangles.(13+11+11+9) So if a combination of sub-strings in the binary value of the length and width satisfies the inequality like it did three times with the sub rectangles then my method will use less base-2 rectangles than the op's method. Finding the minimum number of base-2 rectangles for some squares will inevtably involve searching for the best way to split the square. | {
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For large enough squares the worst digit combination where my method does no better than the op is a block of three ones and the rest are alternating zeros and ones. For example the square $$\require{enclose}\enclose{horizontalstrike}{343×343}$$, its binary representation is 101010111. This square requires 36 base-2 rectangles and is tied for most number of required base-2 rectangles amoung the nine digit squares. This means that a upper bound can be made for the minimum number of rectangles required. Let $$\enclose{horizontalstrike}{d_l}$$ be the number of digits in the binary representation of the length of the rectangle. ($$\enclose{horizontalstrike}{d_l=N_l+Z_l}$$) Let $$\enclose{horizontalstrike}{d_w}$$ be the number of digits in the binary representation of the width of the rectangle. ($$\enclose{horizontalstrike}{d_w=N_w+Z_w}$$) Then the upper bound is: $$\enclose{horizontalstrike}{\left(\left\lceil\frac{d_l}{2}\right\rceil+1\right)\left(\left\lceil\frac{d_w}{2}\right\rceil+1\right)}$$
I conjecture that the combination of my method and the op's method is the optimal way of minimizing the number of base-2 rectangles. The only way that someone might use be able to use less rectangles is to find a another way of spliting the square into sub-rectangles such that using the op's method on those sub-rectangles uses less base-2 rectangles than using my method and the op's method on the whole square. | {
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Rob Pratt's(RP's) post shows that there is a third method for covering the $$n×n$$ square with less base-2 rectangles than my method or the op's method for some $$n×n$$ squares. In order to describe how many rectangles RP's method uses I will continue to use the the term $$b$$ from my method (where $$b$$ is the smallest number such that $$b+n$$ is a power of 2). I will also need a new sets of terms $$c_k$$ and $$s_k$$ where $$k\in\Bbb{N}|1\le k\le f(b)$$. $$c_1$$ is the value of left most ones digit of b in binary form. $$c_2$$ is the value of the second ones digit from the left of b in binary form. $$c_3$$ is the value of the third ones digit from the left of b in binary form. Etc. $$s_v=\sum_{j=1}^vc_v$$. For example if $$n=23$$ then $$b=9$$, $$c_1=8$$, $$c_2=1$$, $$s_1=8$$, $$s_2=9$$. RP's method uses $$2f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}\right)+f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}+s_k\right)+f\left(\frac{n-b}{2}\right)f\left(\frac{n+b}{2}-s_k\right)+f(b)f(b-s_k)$$
base-2 rectangles. Each $$f(•)f(•)$$ product contains the length and width of each of the sub-rectangles that covers the square inside the f function. RP's method has $$k$$ ways of covering the $$n×n$$ square one for each $$s$$ element. Obviously the particular $$s_k$$ element that uses the least number of base-2 rectangles according to the above formula is the one that is used for the minimum. A sufficient condition for when RP's method uses less base-2 rectangles than both my method and the op's method when the binary representation of $$n$$ has at least three more ones than zeros, the second digit to the left is a zero, and the spliting method that was mentioned for the $$1927×1927$$ square doesn't apply. | {
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• absolutely brilliant!! IMHO well worth the bounty. :) Apr 11, 2019 at 2:02
• What does your method obtain for $n\in\{23,30,31\}$? Mar 5, 2020 at 22:36
• @Rob_Pratt 16,13, and 17 base-2 rectangles respectively Mar 5, 2020 at 23:30
• @RobPratt I realized that the way I explained it in my edited post it doesn't show how n=30 is 13 base-2 rectangles with my method. I will edit accordingly. Mar 5, 2020 at 23:37
For fixed $$n$$, you can solve this problem via integer linear programming as follows. Let $$R$$ be the set of rectangles. For $$i\in\{1,\dots,n\}, j\in\{1,\dots,n\}$$, let $$R_{i,j}\subset R$$ be the subset of rectangles that contain cell $$(i,j)$$. Let binary decision variable $$x_r$$ indicate whether rectangle $$r\in R$$ is used. The problem is to minimize $$\sum_r x_r$$ subject to: \begin{align} \sum_{r \in R_{i,j}} x_r &= 1 &&\text{for i\in\{1,\dots,n\}, j\in\{1,\dots,n\}} \\ x_r &\in \{0,1\} &&\text{for r \in R} \end{align}
Here are several optimal values that differ from $$f(n)^2$$: $$\begin{matrix} n &15 &23 &30 &31 &46 &47 &55 &59 &60 &61 &62 &63\\ \hline f(n)^2 &16 &16 &16 &25 &16 &25 &25 &25 &16 &25 &25 &36\\ \text{optimal} &13 &15 &13 &17 &15 &19 &20 &20 &13 &20 &17 &21\\ \end{matrix}$$ | {
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For example, here is an optimal solution for $$n=23$$ with $$15$$ rectangles (shown here with hex values $$0$$ through $$E$$ for compactness): $$\begin{matrix} 0&0&0&0&0&0&0&0&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&1&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2\\ \end{matrix}$$
• I should have mentioned this earlier but good job finding this. As of when this comment being posted you are the only one who has helped me with this problem. If no one else posts am answer by the end of the bounty grace period you will receive the bounty. Also I have made a formula for your method in my most recent edit that I just made, you might want to take a look. Mar 8, 2020 at 23:13
• Thanks. I added a few more values $< f(n)^2$ just now. Mar 9, 2020 at 23:35
NOTE:This doesn't work, the induction hypothesis is too strong (and false).
Lets first consider a more general question, where we tile a rectangle $$R$$ by smaller rectangles, where all vertices are points in an (ambient) integer lattice. | {
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We have a row of rectangles $$T_i$$ touching the bottom edge of $$R$$, and each of these has a top edge $$e_i$$. For each $$T_i$$ we define the number $$\lambda(T_i)$$ to be the minimal number of our tiling rectangles that intersect any column starting in $$T_i$$.
Our first claim is that for the total number of rectangles in $$R$$, denoted $$r(R)$$, we have $$\sum_i \lambda(T_i) \leq r(R)$$
Lets prove this by induction on the height of the rectangle $$R$$ (drawing a picture may help see whats happening). First, if the height is $$1$$, then we are done trivially.
So now for the inductive step, let $$R_0$$ have height $$n$$, and consider the edges $$e_i$$ that have minimal height, and define $$a$$ to be this height. This is to say, they border the $$a$$th row, if the first row is the bottom row of $$R_0$$. Say that we have $$k$$ minimal edges $$e_i$$ bordering this row.
We now consider the new rectangle $$R_0'$$ we obtain by chopping off the first $$a$$ rows of $$R$$. On one hand, this has strictly smaller height, so we have, by induction and our definition of $$k$$: $$\sum_i \lambda(T_i') \leq r(R_0)-k$$
But each rectangle on the bottom row of $$R_0'$$ is either one of the rectangles of $$R_0$$, chopped, but not removed, or a rectangle of $$R_0$$ lying above one of our minimal edges $$e_i$$.
Now note if our original $$T_i$$ is chopped but not removed, $$\lambda(T_i)=\lambda(T_i')$$, and if our original $$T_j$$ is removed (so top edge has minimal height), then $$\lambda(T_j')=\lambda(T_j)-1$$, where $$T_j'$$ is any of the rectangles lying directly over $$T_j$$. Thus, adding $$k$$ to both sides of our previous equality, we have:
$$\sum_i \lambda(T_i) \leq \sum_i \lambda(T_i')+k\leq r(R_0)$$
So we are done by induction. | {
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$$\sum_i \lambda(T_i) \leq \sum_i \lambda(T_i')+k\leq r(R_0)$$
So we are done by induction.
So for your case, note that each column must have at least $$f(n)$$ rectangles in it, and note the bottom row has at least $$f(n)$$ rectangles. This follows since $$f(n)$$ is the minimal number of powers of two needed to express $$n$$. Thus, $$f(n)^2\leq r(R)$$ in your case.
• I think your original claim is false.I don't know how to send a picture in comment but you can easily draw $3\times 4$ counter examples(two horizontal dominos and two vertical dominos in first two rows and two $1\times1$square and a domino on the third row).the problem is on your induction step the rectangles above two removed rectangle may not be distinict.
– ali
Apr 5, 2019 at 9:41
• your last statement have counter example too.if each row intersect k rectangle and each column intersect k rectangle doesn't mean we need $k^2$ rectangle.
– ali
Apr 5, 2019 at 9:43
• True, I'll leave this up in case someone can make this approach work. Apr 5, 2019 at 10:03 | {
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# Finding a tricky composition of two piecewise functions
I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:
$$f(x) = \begin{cases} 2x+1, & \text{if x \le 0} \\ x^2, & \text{if x > 0} \end{cases}$$
$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$
My main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine.
In this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.
The first branch point is associated with $f$, and happens at $x=0$. We note that: $$g\circ f(x) = \begin{cases} g(2x+1), & \text{if x≤ 0} \\ g(x^2), & \text{if x > 0} \end{cases}$$
Let's just consider each case separately. First, $g(2x+1)$ when $x≤0$. We remark that $x≤0\implies 2x≤0\implies 2x+1≤1$ so we see that $x≤0\implies g(2x+1)=-2x-1$.
Now consider $g(x^2)$ when $x>0$. We see that we hit a branch point at $x=\sqrt 2$. Specifically, $x<\sqrt 2\implies x^2<2$ so $x<\sqrt 2\implies g(x^2)=-x^2$. Of course $x≥\sqrt 2 \implies x^2≥2$ so $x≥\sqrt 2\implies g(x^2)= 5$. Combining all this we see that $$g\circ f(x) = \begin{cases} -2x-1, & \text{if x ≤ 0} \\ -x^2, & \text{if 0< x < \sqrt 2}\\ 5 & \text{if x≥\sqrt 2} \end{cases}$$ | {
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As a sanity check, we try the special case $x=1$. We note that $g\circ f(1)=g(f(1))=g(1)=-1$ as desired. I advise picking a few other special values just to check.
• Thanks for your answer. You say that "[w]e see that we hit a branch point at $x = \sqrt 2$," but how do we see this? I don't know how one would conclude this from any of the information given. – Linear Pedant Jul 22 '16 at 22:19
• @LinearPedant The branch point happens when the argument of $g$ reaches $2$. Because the argument is $x^2$, the branch point is at $x=\sqrt2$. – f'' Jul 22 '16 at 23:41
• @LinearPedant Did the reply from f'' help? It is exactly right. We know that $g(x)$ has a branch point at $x=2$, thus $g(x^2)$ has a branch point at $x^2=2$, a.k.a. $x=\sqrt 2$ (since we already know $x>0$ so we can ignore the negative square root). – lulu Jul 23 '16 at 1:45
• @lulu @f" Yes, thank you. This reply greatly helped me understand the problem better. – Linear Pedant Jul 23 '16 at 1:52
$$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases}$$
Therefore
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$
So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.
$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases}$$
Let's look at one piece of $f$ at a time. On the first piece, $2x + 1 \ge 2$ means $x \ge 1/2$. But this is impossible because $x \le 0$ on the first piece. Also on the first piece, $2x + 1 < 2$ means $x < 1/2$. Well, on the first piece we have $x \le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \le 0$.
On the second piece, $x^2 \ge 2$ means $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$. On the second piece we always have $x > 0$, therefore we have $x^2 \ge 2$ when $x \ge \sqrt{2}$. Also on the second piece, $x^2 < 2$ means $-\sqrt{2} < x < \sqrt{2}$. And since on the second piece we always have $x > 0$, then we must have $0 < x < \sqrt{2}$ in order to have $x^2 < 2$. | {
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Putting it all together so far, we have the following:
$$f(x) \ge 2 \text{ if and only if } x \ge \sqrt{2}$$
$$f(x) < 2 \text{ if and only if } x \le 0 \text{ or } 0 < x < \sqrt{2}$$
Notice that this last one can be simplified but we need to keep them separate. Why is this? We'll see as we continue. Recall:
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$
Therefore:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0 \text{ or } 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
Separating the conditions gives us:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0\\ -f(x), & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
And we need to do this because $f(x)$ itself is different for $x \le 0$ and $0 < x < \sqrt{2}$. Finally, we end up with:
$$h(x) := g(f(x)) = \begin{cases} -(2x+1), & x \le 0 \\ -x^2, & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
• Ack, someone beat me to it. I guess that's what happens when I let myself get distracted. – tilper Jul 22 '16 at 15:59
You're correct about the value of $g(f(x))$ when $x\leq 0$; since $f(x)$ will be at most $2\cdot0+1=1$, $g$ is only going to evaluate $f(x)$ according to the definition for $x<2$. Testing for cases here is a good approach, and you've just resolved the $x\leq0$ case. When $x>0$, consider the values of $f(x)$: when will they be less than $2$, and when will they be greater? This will determine where $g(f(x))$ takes on its values.
$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$ In finding $g(f(x))$ with $g$ defined as above, one should remember that $$g(f(x)) = \begin{cases} -x, & \text{if f(x) < 2} \\ 5, & \text{if f(x) \ge 2} \end{cases}$$ with $f(x)$, not $x$, being the thing about which one asks when it is $<2$ and when it is $\ge 2$. | {
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(The calculus text by Salas and Hille had a problem like this that used to get assigned in every academic term by vast numbers of professors who taught the course to thousands of students. And every time an hour would be spent answering the students questions about it, which in every case said something like "I didn't even know where to start", and that's an hour that could have been spent on something that would help the students understand calculus. It was an even worse waste of time and effort than what usually goes on in the kind of math courses where the students don't want to learn math.) | {
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# Can every proof by contradiction also be shown without contradiction?
Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?
As an aside, how is proving by contradiction viewed in general by ‘advanced’ mathematicians. Is it a bit of an ‘easy way out’ when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn’t fond of proof by contradiction.
## Answer
To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$.
There are two key logical principles that express different parts of what we call “proof by contradiction”:
1. The principle of explosion: for any statement $A$, we can take “$\bot$ implies $A$” as an axiom. This is also called ex falso quodlibet.
2. The law of the excluded middle: for any statement $A$, we can take “$A$ or $\lnot A$” as an axiom.
In proof theory, there are three well known systems:
• Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to “direct proof”, because it does not let us leverage a negation for any purpose.
• Intuitionistic logic includes minimal logic and the principle of explosion
• Classical logic includes intuitionistic logic and the law of the excluded middle | {
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• Classical logic includes intuitionistic logic and the law of the excluded middle
It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.
The scheme “If $A$ implies a contradiction, then $\lnot A$ must hold” is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says “if $A \to \bot$ then $A \to \bot$”. But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.
Attribution
Source : Link , Question Author : sonicboom , Answer Author : Carl Mummert | {
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