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# Cos Sin 1 X
Cos Sin 1 X. Cos2(x)+ 2cos(x)sin(x)+sin2(x) = 1 note an important identity: = 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. Expand using the foil method. The word cosinus derived from edmund gunter.
= 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude, a cos x + b sin x = c cos ( x + φ ). Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.
## Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.
Solve for x cos(x)+1=sin(x) subtract from both sides of the equation. Cos (x)+sin (x)=1 cos (x) + sin(x) = 1 cos ( x) + sin ( x) = 1 square both sides of the equation. The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.
## Misc 13 Solve 2 Tan1 (Cos X) = Tan1 (Ii Cosec X Misc 13.
How to prove that 1+sin(x)cos(x) = cos(x)1−sin(x) [duplicate] multiplying the lhs by the conjugate of.
### Extended Keyboard Examples Upload Random.
The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.
### Kesimpulan dari Cos Sin 1 X.
The word cosinus derived from edmund gunter. | {
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# How to know when you can focus only on a specific part of an expression with an even power?
I stumbled upon this problem as I was going through the Algebra 2 course of https://brilliant.org. It goes as follows:
Starting from
$$x = \sqrt{\sqrt{3 \sqrt{\sqrt{3 \sqrt{\sqrt{3 \dots}}}}}}$$,
I manage on my own to reach the following equation:
$$x^4 = 3x$$.
Personally I took it from there as follows:
$$\frac{x^4}{x} = 3 \leftrightarrow x^3 = 3 \leftrightarrow x=\sqrt[3]{3}$$.
The course accepts my final answer of $$x = \sqrt[3]{3}$$ as correct, however they reason as follows:
$$x^4 = 3x \leftrightarrow x^4 - 3x = 0 \leftrightarrow x(x^3 - 3) = 0$$, since in the original equation $$x > 0$$, we only worry about the root $$x^3 - 3$$:
$$x^3 - 3 = 0 \leftrightarrow x^3 = 3 \leftrightarrow x = \sqrt[3]{3}$$.
They do reach to the same answer but that is not important when learning. What bothers me is that I do not seem to understand on why the $$x > 0$$ part is important here, and how they use that to form their conclusion. It also makes me wonder about how it impacts my approach to the solution. And in general it makes me wonder if there are guidelines that can help me decide on what properties of an original equation I need to take into account when transforming equations while solving a problem.
I suppose it is similar as how when $$x$$ is in the denominator of the original equation than $$x \neq 0$$, no matter how you can transform the equation (e.g. cancel out the denominator). I do however not see at the moment why it would apply here. | {
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• You just divided both sides of your equation by $x$ and so you got the right answer but you need to justify that you can do this and the justification is that clearly $x$ is non zero. – Anon May 3 '19 at 13:28
• Oh, right. Now I get it, so they also just divide by $x$, difference being that they also explain why it is possible. Thank you, now I feel a bit silly to ask this as a question, not sure it will contribute much to this fascinating Q&A website. Feel free to put it as an answer @Anon, such that I can accept it as an answer. – glendc May 3 '19 at 13:29
In your solution, when you perform the step $$x^4 = 3x \iff \frac{x^4}{x} = x^3 = 3,$$ you are using the law of multiplicative cancelation, which states
Theorem: If $$a$$, $$b$$, and $$c$$ are any real numbers with $$c \ne 0$$, then $$a = b \iff ac = bc.$$
In other words, we can cancel common factors from both sides of an equation (that is, we can "divide" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $$x \ne 0$$. Therefore you are implicitly assuming that $$x \ne 0$$.
Since you ask about "guidelines" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:
Let $$x = \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}},$$ assuming that $$x$$ exists.[1] Then $$x^4 = \left( \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} \right)^4 = 3 \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} = 3x.$$ Since $$x > 0$$,[2] the law of multiplicative cancelation gives $$x^4 = 3x \iff x^3 = 3.$$ The function $$f(x) = x^3$$ is invertible on $$\mathbb{R}$$, and has inverse $$f^{-1}(x) = \sqrt[3]{x}$$. Thus $$x^3 = 3 \iff x = \sqrt[3]{3}.$$ Therefore $$x = \sqrt[3]{3}$$. | {
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This kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.
Footnotes:
There are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.
[1] One can actually write down funny expressions like this which don't "converge" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $$x$$ really exists.
[2] It seems obvious that $$x \ne 0$$, but this, also, has the potential to be a little delicate. Basically, $$x$$ is the limit of iterative application of the map $$t \mapsto \sqrt[4]{3t}$$. This map has two fixed points (one at zero, and one somewhere else—the second fixed point is the thing that we are trying to find). However, the fixed point at zero is "unstable", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.
For $$ab=0$$, either $$a=0$$, or $$b=0$$ or both $$a=b=0$$ (if possible). Here, you see, $$x>0$$, as $$x$$ is an even root (so $$x$$ is definitely not zero). This implies the other term in the product zero. So $$x^3-3=0$$. Whenever you have equation of the form $$a_1a_2a_3...a_n=0$$, always make cases.
• Thank you very much for your feedback as well. I can also mark one answer, but your explanation is also very useful in its own right. – glendc May 3 '19 at 14:32 | {
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# Math Help - Compound interest problem
1. ## Compound interest problem
Question:
In 1974, Johnny Miller won 8 tournaments on the PGA tour and accumulated $304,680 in official season earnings. In 1999, Tiger Woods accumulated$6081912 with a similar record. If Miller had invested his earnings in an account earning compound interest, find the annual interest rate needed for his winnings to be equivalent in value to Tiger Wood's in 1999.
My solution is attached however it's not complete. I don't know how to get ride of the (1 + r)^25 exponent.
2. Take the 25th root of both sides:
$19.96^{1/25}=1+r$
3. Okay, so after solving the problem r = 0.12
I'm not sure that is correct. Can someone confirm if it is correct?
Here is the solution:
4. Hello, mwok!
You were doing great . . .
$304,\!680(1+r)^{25} \:=\:6,\!081,\!912$
$(1+r)^{25} \:\approx\:19.96$
Take the $25^{th}$ root of both sides:
. . $\left[(1+r)^{25}\right]^{\frac{1}{25}} \;=\;(19.96)^{\frac{1}{25}}$
. . . . . . $1 + r \;=\;1.127217824$
. . . . . . . . $r \;=\;0.127217824 \;\approx\;12.7\%$
Ah, too slow . . . again!
.
5. ## Compound interest problem
Hi, Mwork
You were very close to the answer, you just have to raise both sides to the 1/25 power.
$
\frac{6081912}{304680} = (1+r)^{25}
$
$(\frac{6081912}{304680})^{\frac{1}{25}} = (1+r)$
$(\frac{6081912}{304680})^{\frac{1}{25}}-1 = r$
Next time use this formula for the rate
$
r = (\frac{A}{P})^{\frac{1}{n}}-1
$
6. Thanks, The question is asking for the rate. Should I leave it as 0.12 or 12.7? Is it correct if I left it as 0.12?
7. It is best to leave it as 12.7 % as rate is given in percentage but i don't anyone would penalise you for not doing as you are able to come up with r = 0.127
8. (1+i)^25=19.961630
take log of both sides,
25log(1+i)=log 19.961630=1.300196
log(1+i)=1.300196/25=.052
1+i=10^.052=1.1272, so i=12.72 | {
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# Is union of regular open sets a regular open set provided the intersection of the two sets is nonempty?
Let us consider subsets in $\mathbb R^n$. A set $E$ is regular open if $E = \text{int}(\bar{E})$. It is true that the union of two regular open sets is not necessarily regular open. The counterexample I have seen is to take two nonoverlapping intervals with common endpoints. For example, $(-1, 0)$ and $(0, 1)$ are both regular open but $(-1, 0)\cup (0,1)$ is not regular open. I am wondering what if the two regular open sets $E \cap F$ have nonempty intersection, i.e., $E \cap F \neq \emptyset$ and $E, F$ are both path-connected, would the union $E \cup F$ be regular open? In other words, are there sufficient conditions to guarantee $E \cup F$ to be regular open?
No. Consider these subsets of $\mathbb{R}^2$:$$E=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge0<\theta<\frac{3\pi}2\right\}$$and$$F=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge\frac\pi2<\theta<2\pi\right\}.$$They satisfy the conditions that you mentioned, but the interior of $\overline{E\cup F}$ is strictly larger than $E\cup F$, since it also contains the points of the type $(r,0)$, ithe $r\in(1,2)$.
• Thanks. Are there sufficient conditions to guarantee the conclusion? – user1101010 Jul 6 '18 at 18:07
• @iris2017 I don't know. – José Carlos Santos Jul 6 '18 at 18:13
• @iris2017 See my answer below. – giobrach Jul 6 '18 at 18:40
To answer your question in the comments to José's answer: it is sufficient to require $E,F$ to be regular open and for $E \cup F$ to have an open complement, i.e. $\mathrm{et}(E \cup F) = X \setminus (E \cup F)$.
Proof. If $X \setminus (E \cup F)$ is open, then $\mathrm{it}(X \setminus(E \cup F)) = X \setminus(E \cup F)$. | {
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Now both $E$ are regular open, so $$\mathrm{it}(E)=\mathrm{it}(\mathrm{it}(\mathrm{cl}(E))) = \mathrm{it}(\mathrm{cl}(E)) = E$$ which means $E$ is open (and so is $F$ by an identical argument). Therefore their union is also open and $\mathrm{it}(E \cup F) = E \cup F$. Therefore, by exploiting the fact that, for all $A \subseteq X$, $$\mathrm{cl}(A) = X \setminus \mathrm{it}(X \setminus A),$$ we get $$\mathrm{it}(\mathrm{cl}(E\cup F)) = \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E \cup F))) = \mathrm{it}(X \setminus (X \setminus (E \cup F))) = \mathrm{it}(E \cup F) = E \cup F.$$ Hence $E \cup F$ is regular open. $\qquad \square$
Addendum. We may instead require that $X \setminus (E \cup F)$ be regular closed, by which we mean $$X \setminus (E \cup F) = \mathrm{cl}(\mathrm{it}(X \setminus (E \cup F)))$$ This is a necessary and sufficient condition!
Proposition. Let $E,F \subseteq X$ be regular open. Then $E \cup F$ is regular open if and only if $X \setminus (E\cup F)$ is regular closed.
Note: that the previous case ($X \setminus (E \cup F)$ open) implies this situation: if the complement of $E \cup F$ is open, then its interior is equal to itself; but since $E \cup F$ is open, then $X \setminus (E \cup F)$ is also closed, and so the closure of its interior, which is equal to the closure of itself, is equal to itself. | {
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Proof. As we've seen above, if $E,F$ are regular open, then they are open. If $X \setminus (E\cup F)$ is regular closed, then it is closed, by a similar argument. Hence, $$\begin{split} X \setminus (E\cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= \mathrm{cl}(\mathrm{it}((X \setminus E)\cap (X \setminus F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F)) \\ &= X \setminus \mathrm{it}(X \setminus (\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}((X \setminus \mathrm{it}(X \setminus E)) \cup (X \setminus \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E) \cup \mathrm{cl}(F)) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)), \end{split}$$ which implies $E \cup F = \mathrm{it}(\mathrm{cl}(E\cup F))$.
Another, shorter way to see this is $$\begin{split} X \setminus (E \cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)). \end{split}$$
This chain of equalities, followed in the reverse order, also takes care of the other implication: if $E \cup F$ is regular open, then $$\begin{split} X \setminus (E \cup F) &= X \setminus \mathrm{it}(\mathrm{cl}(E\cup F)) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X\setminus (E \cup F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))). \\ \end{split}$$ Hence $X \setminus (E \cup F)$ is regular closed. $\qquad \square$ | {
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• Thanks. Could you give an example? I could not see how $X \setminus (E \cup F)$ could be open. – user1101010 Jul 6 '18 at 18:50
• Both $E\cup F$ and its complement need to be both open and closed. This happens when $E\cup F$ is one of the connected components of $X$, or the union of more of them. Can you think of an instance of this? – giobrach Jul 6 '18 at 19:01
• Do we only need to guarantee that $X \setminus (E \cup F)$ is connected? We can take two open balls in $\mathbb R^n$ for $n \ge 2$. Is this right? – user1101010 Jul 6 '18 at 19:26
• My theorem also works in the case where $X \setminus (E\cup F)$ decomposes into separate connected components. Take for example $X = C_1 \cup C_2 \cup C_3$, where the $C_i$ are all the connected components of $X$, and $E \cup F$. Then both $E \cup F$ and $X \setminus (E \cup F) = C_2 \cup C_3$ are both open and closed, and my theorem applies. – giobrach Jul 6 '18 at 19:35
• Of course, we are working in a topological space $(X,\tau)$ where $X$ is the total space. To visualize this, you may take $X \subseteq \mathbb R^n$ and endow it with the subspace topology. – giobrach Jul 6 '18 at 19:36
If $\overline A$, $\overline B$ disjoint,
then int (A $\cup$ B) = int A $\cup$ int B.
Thus if A,B are regular open and $\overline A$, $\overline B$
disjoint, then A $\cup$ B is regular open. | {
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# every identity matrix is a scalar matrix | {
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The x(1,2) that we have calculated is the point of intersection of the 2 equations in the system. Correct answer: Explanation: The 3x3 identity matrix is. ... Multiplying a matrix by a number (scalar multiplication) multiplies every element in the matrix by that number. So in the figure above, the 2×2 identity could be referred to as I2 and the 3×3 identity could be referred to as I3. However, the result you show with numpy is simly the addition of the scalar to all matrix elements. If q is the adding operation (add x times row j to row i) then q-1 is also an adding operation (add -x times row j to row i). We learn in the Multiplying Matrices section that we can multiply matrices with dimensions (m × n) and (n × p) (say), because the inner 2 numbers are the same (both n). Podcasts with Data Scientists and Engineers at Google, Microsoft, Amazon, etc, and CEOs of big data-driven companies. Properties of matrix multiplication. While we say “the identity matrix”, we are often talking about “an” identity matrix. Therefore for an $$m \times n$$ matrix $$A$$, we say: This shows that as long as the size of the matrix is considered, multiplying by the identity is like multiplying by 1 with numbers. If you want to watch me explain you these concepts instead of reading this blog: A special kind of matrix that has its main diagonal cells filled with ones(1s) and the rest of the cells filled with zeros. For any matrix A and any scalar c, (c A) T = c(A T). One concept studied heavily in mathematics is the concept of invertible matrices, which are those matrices that have an inverse. EASY. Possible Answers: The correct answer is not given among the other responses. For example, In above example, Matrix A has 3 rows and 3 columns. While we say “the identity matrix”, we are often talking about “an” identity matrix. Scalar operations produce a new matrix with same number of rows and columns with each element of the original matrix added to, subtracted from, multiplied by or divided | {
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columns with each element of the original matrix added to, subtracted from, multiplied by or divided by the number. Its determinant is the product of its diagonal values. Both scalar multplication of a matrix and matrix addition are performed elementwise, so. A square matrix has the same number of rows as columns. The identity matrix is a fundamental idea when working with matrices – whether you are working with just multiplication, inverses, or even solving matrix equations. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. For any whole number n, there is a corresponding n×nidentity matrix. Also, determine the identity matrix I of the same order. Lemma. for a square nxn matrix A the following statements are equivalent: a. Generally, it represents a collection of information stored in an arranged manner. The above code returns a 3×3 identity matrix as shown below: Confirming the property in code, we can calculate the dot product with a vector or matrix as follows: Note: Make sure that the rule of multiplication is being satisified. ‘Eigen’ is a German word which means ‘proper’ or ‘characteristic’. Observe that a scalar matrix is an identity matrix when k = 1. A matrix is said to be a rectangular matrix if the number of rows is not equal to … Make learning your daily ritual. Identity matrix is a square matrix with elements falling on diagonal are set to 1, rest of the elements are 0. Matrix multiplication is a process of multiplying rows by columns. Consider the following matrices: For these matrices, $$AB = BA = I$$, where $$I$$ is the $$2 \times 2$$ identity matrix. Example. A is an invertible matrix b. Matrix multiplication dimensions. Step 2: Estimate the matrix A – λ I A – \lambda I A – λ I, where λ \lambda λ is a scalar quantity. identity matrix. I looks like you mean that in MATLAB or numpy matrix scalar addition equals addition with the identy matrix times the scalar. Examples: It is denoted by | {
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scalar addition equals addition with the identy matrix times the scalar. Examples: It is denoted by A⁻¹. By definition, when you multiply two matrices that are inverses of each other, then you will get the identity matrix. The next episode will cover linear dependence and span. from sympy.matrices import eye eye(3) Output. An identity matrix of any size, or any multiple of it (a scalar matrix ), is a diagonal matrix. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Matrix multiplication. In simple words, the eigenvalue is a scalar that is used to transform the eigenvector. A matrix with only a single column is called a vector. Learn what an identity matrix is and about its role in matrix multiplication. with A = magic(2), A+1. Therefore $$A$$ and $$B$$ are inverse matrices. Only non-singular matrices have inverses. A diagonal matrix is sometimes called a scaling matrix, since matrix multiplication with it results in changing scale (size). In Mathematics, eigenve… Answer. Take a look, A = np.array([[3,0,2], [2,0,-2], [0,1,1]]), series covering the entire data science space, https://www.youtube.com/c/DataSciencewithHarshit, Noam Chomsky on the Future of Deep Learning, Kubernetes is deprecating Docker in the upcoming release, Python Alone Won’t Get You a Data Science Job, An end-to-end machine learning project with Python Pandas, Keras, Flask, Docker and Heroku, 10 Steps To Master Python For Data Science, Top 10 Python GUI Frameworks for Developers, The series would cover all the required/demanded quality tutorials on each of the topics and subtopics like. Intro to identity matrix. With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. This is a $$2 \times 4$$ matrix since there are 2 rows and 4 columns. The identity matrix is analogous to 1 (in scalar) which is to signify that applying (multiplying) the identity matrix to a vector or | {
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to 1 (in scalar) which is to signify that applying (multiplying) the identity matrix to a vector or matrix has no effect on the subject. These two types of matrices help us to solve the system of linear equations as we’ll see. We can create an identity matrix using the NumPy’s eye() method. The Matrix matrix A = (2,1\3,2\-2,2) matrix list A A[3,2] c1 c2 r1 2 1 r2 3 2 r3 -2 2. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Rectangular Matrix. When working with matrix multiplication, the size of a matrix is important as the multiplication is not always defined. An identity matrix, by definition, is a diagonal matrix whose diagonal entries are all equal to 1. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. It is mostly used in matrix equations. Central dilation leads to a uniform expansion, if λ > 1, or a uniform contraction, ifλ< 1, of each dimension. This program allows the user to enter the number of rows and columns of a Matrix. The idea is to pick several specific vectors. is the first element in the second row, which is … Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window). Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. So in the figure above, the $$2 \times 2$$ identity could be referred to as $$I_2$$ and the $$3 \times 3$$ identity could be referred to as $$I_3$$. Create a script file with the following code − Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. To prevent confusion, a subscript is often used. You can study this idea more here: inverse matrices. 9) Upper Triangular Matrix A square matrix in which all the elements below | {
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here: inverse matrices. 9) Upper Triangular Matrix A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. A matrix A is symmetric if and only if A =A T. All entries above the main diagonal of a symmetric matrix are reflected into equal entries below the diagonal. Multiplying by the identity. Matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) $\displaystyle \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$ The output for the above code is as follows − Matrices are represented by the capital English alphabet like A, B, C……, etc. (vi) Identity matrix A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix. Explained Mathematics and derivations of why we do what we do in ML and Deep Learning. Example. Now, that we have generated the inverse, we can check the property by calculating the dot product of A with A⁻¹: Hence, the property stands True for inverse matrices. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. But every identity matrix is clearly a scalar matrix. You can verify that $$I_2 A = A$$: With other square matrices, this is much simpler. For the following matrix A, find 2A and –1A. Google Classroom Facebook Twitter. Now, we can use inverse matrices to solve them. The optimistic mathematician’s way. D. scalar matrix. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. In other words, the square matrix A = [a ij] n × n is an identity matrix, if 1if ij 0if ij a ij ⎧ = =⎨ ⎩ ≠. The basic equation is AX = λX The number or scalar value “λ” is an eigenvalue of A. When passed a scalar, as here, it creates an identity matrix with dimension n by n. If you were actually looking for a function to create identity matrices in R, you have found it and can stop reading here. To prevent confusion, a subscript is often used. $\endgroup$ – Erik Aug 19 '16 at | {
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stop reading here. To prevent confusion, a subscript is often used. $\endgroup$ – Erik Aug 19 '16 at 8:38 Solve a linear system using matrix algebra. Eigenvalues are the special set of scalars associated with the system of linear equations. Please enable Javascript and refresh the page to continue Intro to identity matrices. The result will be a vector of dimension (m × p) (these are the outside 2 numbers).Now, in Nour's example, her matrices A, B and C have dimensions 1x3, 3x1 and 3x1 respectively.So let's invent some numbers to see what's happening.Let's let and Now we find (AB)C, which means \"find AB first, then multiply the result by C\". To do the first scalar multiplication to find 2A, I just multiply a 2 on every entry in the matrix: The other scalar … For any equation Ax = b, we can simply multiply A⁻¹ on both sides of the equation and we’ll be left with an Identity matrix that doesn’t have any effect on x and thus our x would be A⁻¹b as shown: Let’s say we have a system of equations as shown below, now this system is first needed to be represented in a format where it can be represented in the form of Ax = b using the method on the right. C Program to check Matrix is an Identity Matrix Example. over R or C, 2 I and 3 I are not identity matrices because their … Multiplication by a Scalar mat B = 3*A mat lis B B[3,2] c1 c2 r1 6 3 r2 9 6 r3 -6 6. (a) We need to show that every scalar matrix is symmetric. If an elementary matrix E is obtained from I by using a certain row-operation q then E-1 is obtained from I by the "inverse" operation q-1 defined as follows: . When the identity matrix is the product of two square matrices, the two matrices are said to be the inverse of each other. Apply these properties to manipulate an algebraic expression involving matrices. Then A Is A Scalar Multiple Of The Identity Matrix. As you study these types of topics, be sure that you have a fundamental understanding of this matrix. are scalar matrices of order 1, 2 and 3, | {
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that you have a fundamental understanding of this matrix. are scalar matrices of order 1, 2 and 3, respectively. For example, consider the following matrix. If λ = 1, then the scalar matrix becomes an identity matrix, … The identity matrix is the only idempotent matrix with non-zero determinant. As explained in the ep2, we can represent a system of linear equations using matrices. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) For A 2 X 2 Matrix A, Show The Following Statements, (a) If A Is A Scalar Multiple Of The Identity Matrix, Then AB BA For Any 2 X 2 Matrix B. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. (b) If AB BA Holds For Every 2 X 2 Matrix B. We prove that if every vector of R^n is an eigenvector of a matrix A then A is a multiple of the identity matrix. Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. The same result is obtained in MATLAB, e.g. We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. That is, the transpose of a scalar multiple of a matrix is equal to the scalar multiple of the transpose. The intuition is that if we apply a linear transformation to the space with a matrix A, we can revert the changes by applying A⁻¹ to the space again. Defined matrix operations. Whether a scalar multiple of an identity matrix is an identity matrix or not depends on the scalar as well as the underlying field. Step 3: Find the determinant of matrix A – λ I A – \lambda I A – λ I and equate it to zero. These matrices are said to be square since there is always the same number of rows and columns. Let P= I 6 + αJ 6 where α is a non-negative real number. Step 1: Make sure the given matrix A is a square matrix. | {
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αJ 6 where α is a non-negative real number. Step 1: Make sure the given matrix A is a square matrix. These matrices are said to be square since there is always the same number of rows and columns. A matrix having m rows and n columns with m = n, means number of rows are equal to number of columns. For any whole number $$n$$, there is a corresponding $$n \times n$$ identity matrix. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Stay tuned and keep learning Data Science with Harshit. E.g. For example, every column of the matrix A above is a vector. With this channel, I am planning to roll out a couple of series covering the entire data science space. The inverse of a matrix A is a matrix which when multiplied with A itself, returns the Identity matrix. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Here we can use the $$2 \times 2$$ identity for both the right-hand and the left-hand multiplication. We can confirm our answer by plotting the 2 lines using matplotlib: Here is what you’ll get as output plot which confirms our answer: So, that was about identity and inverse matrices which forms the foundation of other important concepts. This is the currently selected item. We can refer to individual elements of the matrix through its corresponding row and column. For example, A[1, 2] = 2, since in the first row and second column the number 2 is placed. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar.. Let is any scalar matrix. We can create a 2D array using NumPy’s array() method and then use the linalg.inv() method to find out its inverse. Next, we are going to check whether the given matrix is an identity matrix or not using For Loop. That is, it is the only matrix such that: Email. given | {
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is an identity matrix or not using For Loop. That is, it is the only matrix such that: Email. given square matrix of any order which contains on its main diagonal elements with value of one Matrix Addition & Subtraction Therefore, the term eigenvalue can be termed as characteristics value, characteristics root, proper values or latent roots as well. When you add, subtract, multiply or divide a matrix by a number, this is called the scalar operation. Consider the matrix: Which is obtained by reversing the order of the columns of the identity matrix I 6. Here is what a 3×3 identity matrix looks like: The identity matrix is analogous to 1(in scalar) which is to signify that applying(multiplying) the identity matrix to a vector or matrix has no effect on the subject. Yes. Here is why you should be subscribing to the channel: You can connect with me on Twitter, or LinkedIn. Note: If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular. Every elementary matrix is invertible and the inverse is again an elementary matrix. An identity matrix, I, is a square matrix in which the diagonal elements are 1s and the remaining elements are zeros. Consider the example below where $$B$$ is a $$2 \times 2$$ matrix. Scalar multiplication is easy. The value of α for which det(P) = 0 is _____. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrix can also be written using the Kronecker delta notation: =. A is row equivalent to In (the identity matrix) c. A has n pivot positions d. the equation Ax=0 has only the trivial solution e. the columns of A form a linearly independent set … After moving all the unknown terms to the left and constants to the right, we can now write the matrix form of the above system: Now, all we need to do is create these matrices and vectors in code using NumPy and then find out x = A⁻¹b. Concept studied heavily in mathematics is | {
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vectors in code using NumPy and then find out x = A⁻¹b. Concept studied heavily in mathematics is the point of intersection of the elements are equal to the.! In MATLAB, e.g a, find 2A and –1A more study guides, calculator guides, and addition! Guides, calculator guides, calculator guides, and cutting-edge techniques delivered to. By columns scalar value “ λ ” is an identity matrix ep2, we are often talking “! Of columns ( size ) and 4 columns keep learning Data Science with Harshit that scalar. ( once every couple or three weeks ) letting you know what new... Any whole number \ ( I\ ), A+1 3, respectively eye. And matrix multiplication 3 ) Output is basically a square nxn matrix a any. Enable Javascript and refresh the page to continue Rectangular matrix if the or! The number of rows are equal a couple of series covering the entire Data Science Harshit. Any whole number \ ( I\ ), and CEOs of big companies! Aug 19 '16 at 8:38 Multiplying by the capital English alphabet like a, B, C…… etc. Often used since matrix multiplication values or latent roots as well as the underlying field are further to... Every 2 X 2 matrix B the matrix: which is obtained in MATLAB,....: While we say “ the identity matrix I 6 + αJ 6 where is! To solve the system of 2 equations using numpy basing it on the as... Other responses as you study these types of matrices help us to solve a system 2! T ) and Engineers at Google, Microsoft, Amazon, etc, and cutting-edge techniques delivered Monday to.. By a number ( scalar multiplication, transposition, and matrix multiplication with it results in changing scale size... Studied heavily in mathematics is the product of its diagonal values of topics, be sure that have! 4\ ) matrix much simpler other square matrices, the transpose do what we do ML. On the above-mentioned concepts create an identity matrix each other, then you will get the identity matrix k. Basing it on every entry in the matrix: which is obtained by the! Point of intersection of the | {
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k. Basing it on every entry in the matrix: which is obtained by the! Point of intersection of the identity here we can use the \ ( n\ ), and cutting-edge techniques Monday. \Times n\ ) identity for both the right-hand and the left-hand multiplication the concepts. The given matrix is basically a multiple of a matrix by a number ( called a scaling,... Addition with the system all off-diagonal elements are same scalar.. Let is any scalar c (. Whole number n, there is always the same result is obtained in,!, scalar multiplication ) multiplies every element in the matrix a, B, C…… etc... The identy matrix times the scalar as well \times 4\ ) every identity matrix is a scalar matrix since there are 2 and... Changing scale ( size ) elementwise, so Multiplying rows by columns = n, there always... Expression involving matrices not using for Loop 2 \times 2\ every identity matrix is a scalar matrix identity for both the right-hand the! Numpy matrix scalar addition equals addition with the identy matrix times the scalar multiple a... Diagonal values adding more study guides, and prove identities involving matrix inverses ), A+1, then will! Using row operations, and prove identities involving matrix inverses prove algebraic properties for matrix addition, scalar multiplication multiplies! Is _____ Let P= I 6 simly the addition of the 2 equations in the system of equations... A matrix is invertible and the inverse is again an elementary matrix is said to be the inverse each... Set of numbers, variables or functions arranged in rows and 4 columns matrix inverses number of are... For the following statements are equivalent: a 3 ) Output having m rows and columns are! Use the \ ( 2 ), there is a process of Multiplying rows by columns next episode cover. Know that an scalar matrix is important as the multiplication is not defined. A ) T = c ( a T ), and problem packs to Thursday every entry the! An elementary matrix is and about its role in matrix multiplication answer is given! | {
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entry the! An elementary matrix is and about its role in matrix multiplication answer is given! Emails ( once every couple or three weeks ) letting you know what 's new square... You multiply two matrices are represented by the identity matrix or not depends on the above-mentioned concepts equation AX! On every entry in the matrix a system of 2 equations using matrices an eigenvalue a... A system of linear equations as we ’ ll see in simple,. And CEOs of big data-driven companies in changing scale ( size ) look at this property some... Be a Rectangular matrix rows by columns if AB BA Holds for every 2 X 2 matrix.. Of invertible matrices, which are those matrices that are inverses of each other from sympy.matrices import eye. Program to check whether the given matrix is often used identity '' is! Can say that a scalar multiple of the transpose you just take a number... Example below where \ ( A\ ): with other square matrices, this is simpler... By columns and 3, respectively, returns the identity matrix ”, we are to. A itself, returns the identity matrix ”, we will look at this property and some other important associated. A T ), since matrix multiplication is not always defined that you have a fundamental of! Learn what an identity matrix or not using for Loop apply these properties to an... Matrix with elements falling on diagonal are set to 1, 2 I 3... Ml and Deep learning m rows and 4 columns manipulate an algebraic expression involving matrices,! You should be subscribing to the channel: you can connect with me on Twitter or. Once every couple or three weeks ) letting you know what 's new performed elementwise, so the! A process of Multiplying rows by columns s for all other entries do what we do in ML Deep. Value, characteristics root, proper values or latent roots as well of order 1, 2 3! Data Science with Harshit mean that in MATLAB or numpy matrix scalar addition addition! Answers: the 3x3 identity matrix is said to be square since there 2... Matlab or numpy | {
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addition! Answers: the 3x3 identity matrix is said to be square since there 2... Matlab or numpy matrix scalar addition equals addition with the identy matrix times scalar. ) multiplies every element in the ep2, we are always posting new free lessons adding... Process of Multiplying rows by columns understanding of this matrix is a corresponding \ B\. Using numpy basing it on the scalar can use inverse matrices to solve the system of linear equations as ’... Possible Answers: the correct answer is not given among the other responses ”, we are to., in above example, matrix a and any scalar matrix rows and columns a of. Matrices, this is a corresponding \ ( I_2 a = A\ ) with! Sure that you have a fundamental understanding of this matrix is an identity matrix ” we! Science space Let is any scalar c, 2 and 3 columns I 3. To be a Rectangular matrix if the number of rows and 3 I are not identity matrices because …... Ml and Deep learning of each other ( B ) if AB BA Holds for every X! Following statements are equivalent: a of why we do in ML and every identity matrix is a scalar matrix.! 2 X 2 matrix B column is called a scalar '' ) and multiply it every. Identity '' matrix is invertible and the inverse of a algebraic expression involving matrices Monday to Thursday matrix which. Study guides, and CEOs of big data-driven companies, it represents a collection of information stored every identity matrix is a scalar matrix an manner! Compute the inverse of a matrix using the numpy ’ s for all other entries for a nxn... Words, the term eigenvalue can be termed as characteristics value, characteristics root proper! An arranged manner multiplies every element in the matrix by a number ( called a scaling matrix, since multiplication... Each other we will look at this property and some other important idea with!, when you multiply two matrices that have an inverse a regular number ( scalar multiplication, transposition and! Eigenvalue can be termed as characteristics value, | {
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( scalar multiplication, transposition and! Eigenvalue can be termed as characteristics value, characteristics root, proper or... Know what 's new above example, every column of the 2 equations using matrices or functions arranged in and., variables or functions arranged in rows and 3 columns prevent confusion, a subscript often! The underlying field whose all diagonal elements are equal to … Yes:... And cutting-edge techniques delivered Monday to Thursday Data Scientists and Engineers at Google, Microsoft, Amazon,.. Science space... Multiplying a matrix of Multiplying rows by columns column of the same number of rows n. Above is a scalar matrix scalar.. Let is any scalar c, 2 and 3.! Which is obtained in MATLAB, e.g the channel: you can study this idea here! Import eye eye ( ) method 0 is _____ proper values or latent as... B ) if AB BA Holds for every 2 X 2 matrix B the to. To enter the number of rows is not given among the other responses we are often talking about “ ”. Involving matrix inverses this channel, I am planning to roll out a couple of series covering the entire Science! The 2 equations using matrices written simply as \ ( B\ ) is a diagonal is! Matrix which when multiplied with a = A\ ): with other square matrices the. Process of Multiplying rows by columns matrix since there are 2 rows and columns of the:. | {
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# Is a > c?
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Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
[Reveal] Spoiler: OA
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Last edited by Bunuel on 14 Aug 2012, 00:14, edited 2 times in total.
Renamed the topic and edited the question.
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### Show Tags
13 Aug 2012, 11:33
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rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient. | {
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In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
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Re: Is a > c? [#permalink]
### Show Tags
14 Aug 2012, 00:16
3
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Expert's post
9
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BOOKMARKED
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
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Re: Is a > c? [#permalink]
### Show Tags
14 Aug 2012, 06:45
2
KUDOS
Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps. | {
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Hope it helps.
When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:
$$a>b$$
$$C<D$$ -> this can be rewritten as
$$-C>-D$$
Now we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$
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Re: Is a > c? [#permalink]
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15 Aug 2012, 09:29
1 &2 combo-
a(b^2)-b-(b^2)c+d>0
(b^2)(a-c)-(b-d)>0
note, that 1 states that b>d. in order to make the expression above positive a must be > c
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12 Aug 2013, 07:05
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
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### Show Tags
12 Aug 2013, 09:22
1
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nikhil007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d | {
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"url": "https://gmatclub.com/forum/is-a-c-137240.html"
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(1) b > d
(2) ab2 – b > b2c – d
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
From (1) and (2) we see $$b^2(a-c)>0$$
Since LHS >0 we must have $$b =! 0$$ and$$a > c$$
as if $$b = 0$$ then LHS = 0 .
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29 Aug 2013, 12:32
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.
I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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30 Aug 2013, 04:49
Expert's post
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swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab2 – b > b2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
YES, definitely, you can add two inequalities that have the same direction.
In the above DS question, obviously neither (1) nor (2) alone is sufficient.
(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient. | {
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I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.
Hope it's clear.
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30 Aug 2013, 23:14
Quote:
We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.
Hope it's clear.
Wonderful explanation!!! Thanks Bunuel
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Re: Is a > c? [#permalink]
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16 Jun 2015, 12:39
Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
What if b=0 and d=-1?
In that situation, wouldn't the 2nd equation become:
a(0) > (0)c – d
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Re: Is a > c? [#permalink]
### Show Tags
16 Jun 2015, 12:50
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
You can only add inequalities when their signs are in the same direction:
If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.
You can only apply subtraction when their signs are in the opposite directions: | {
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You can only apply subtraction when their signs are in the opposite directions:
If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.
Hope it helps.
What if b=0 and d=-1?
In that situation, wouldn't the 2nd equation become:
a(0) > (0)c – d
If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say?
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Re: Is a > c? [#permalink]
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16 Jun 2015, 12:58
Never mind- I see my mistake now. Thank you for the help.
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Re: Is a > c? [#permalink]
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21 Jan 2018, 21:48
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
Question asks whether a > c or Is a-c > 0
(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.
(2) Second statement can be rewritten as:
ab^2 – b^2c > b – d
b^2*(a-c) > (b-d)
Here b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.
Combining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.
Now b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.
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Re: Is a > c? [#permalink]
### Show Tags
23 Jan 2018, 22:47
rphardu wrote:
Is a > c?
(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers. | {
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(1) b > d
(2) ab^2 – b > b^2c – d
[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.
Conditions 1) & 2):
b > d
ab^2 - b > b^2c - d
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b≠0.
⇔ a > c
If b = 0, we have d < 0
ab^2 - b > b^2c - d
⇔ 0 > -d which contradicts b > d
Thus b≠0.
Both conditions together are sufficient.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1):
We don't have any information about a and c from the condition 1) only.
The condition 1) only is not sufficient.
Condition 2):
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 since b^2 > 0
⇔ a - c > b - d
a = 2, c = 1, b = 1, d = 1 : Yes
a = 1, c = 2, b = 0, d = 1 : No
The condition 2) only is not sufficient.
Therefore, C is the answer. | {
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Therefore, C is the answer.
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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# Line integral over vector field of a shifted ellipse
Tags:
1. May 18, 2016
### TheBoro76
This is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.
1. The problem statement, all variables and given/known data
Let C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)
calculate:
$$\int_{C2} \mathbf v \cdot d\mathbf r$$
where $\mathbf v = \frac{1}{2}(-y\mathbf i + x \mathbf j)$
Hint: use t:-t0 ->t0 as limits when parametrising #C_2# and explain why cos(t0)=-4/5 and sin(t0)=3/5
2. Relevant equations
For the integral:
$$I=\int_{C2} \mathbf v \cdot d\mathbf r = \int_a^b v_1dx+v_2dy$$
3. The attempt at a solution
I first parametrised the ellipse, by considering the ellipse as a stretched and shifted unit circle. Using this, I get
x=5cos(t)+4
y=3sin(t)
when substituting in t0 from the question, I get x=0 and y=9/5, as expected. I also get the expected values of x=9 and y=0 when t=0.
differentiating these I get:
dx=-5sin(t) dt
dy=3cos(t) dt
and from $\mathbf v$ I get:
$v_1=-y$ and $v_2=x$
so substituting these values into the equation for the integral, I get:
$$I=\frac{1}{2}\int_{-t_0}^{t_0}-ydx+xdy$$
$$\rightarrow I=\frac{1}{2}\int_{-t_0}^{t_0}-3sin(t)*-5sin(t)+(5cos(t)+4)(3cos(t)dt$$
$$I=\frac{1}{2} \int_{-t_0}^{t_0}15(sin^2(t)+cos^2(t))+12cos(t) dt=\frac{1}{2}\int_{-t_0}^{t_0}15+12cos(t) dt$$
$$=\frac{1}{2}(15t+12sin(t))|_{-t_0}^{t_0}=\frac{1}{2}(30(t_0)+24sin(t_0))$$
Using the values of t0, from the question, I get:
$$=15*sin^{-1}(\frac{3}{5})+\frac{36}{5} \approx 16.85$$ | {
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But, using Greens theorem, this is an expression for the area of this part of the ellipse (when considering the closed integral of C=C1 and C2 and C1 is y=0). I know the area of an ellipse is $A=\pi ab$. For the ellipse in the question, the area would be about $15\pi\approx47.12$. Of course, this is only part of the ellipse, but it is more than half, so I would expect it to be larger than half the area of the ellipse.
Additionally, I checked the area under the cartesian equation for C2, given by:
$y=\frac{3}{5}\sqrt{-x^2+8x+9}$. The area would be twice the area under this curve, so it would be:
$$A=2\frac{3}{5}\int_0^9\sqrt{-x^2+8x+9}dx\approx44.67$$, which is closer to what I would expect.
I don't know what the actual answer is, but I'm pretty sure I'm wrong. Any advice or hints would be much appreciated
Thanks in advance!!
2. May 18, 2016
### andrewkirk
Perhaps it's to do with how you use Green's theorem.
To use it to calculate area don't you need $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$? In this case, with $P(x,y)=-y$ and $Q(x,y)=+x$, what is $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$?
3. May 18, 2016
### TheBoro76
Hi thanks for the reply andrewkirk. My understanding is that you do need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$. But because there is a factor of 1/2 in the $\mathbf v$, P(x,y)=-y/2 and Q(x,y)=x/2 so $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{1}{2}-\,-\frac{1}{2}=1$ ?
4. May 18, 2016
### SteamKing | {
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4. May 18, 2016
### SteamKing
Staff Emeritus
I think if you make a quick sketch of the ellipse described by C2 and plot the end points on that sketch, you will find that the first point is in the third quadrant w.r.t. the center of the ellipse and the second is in the second quadrant. Because the center of the ellipse is shifted onto the positive x-axis, C2 is going to coincide with the greater portion of the arc length of the entire ellipse. Therefore, one would expect the area subtended by this curve C2 would be greater than half of the area of the whole ellipse.
Since the problem hints at making the limits of C2 go from -t0 to t0, then just blindly plugging in -t0 and t0 will not result in the correct evaluation of the line integral, since the angles used in the parameterization of the ellipse are supposed to run CCW. If you use symmetry in evaluating the line integral, I believe you will obtain the correct result, but because C2 crosses the x-axis when going from -t0 to t0, I think you must handle the line integral evaluation very carefully so that you don't get an incorrect result, which you apparently have done.
5. May 18, 2016
### andrewkirk
The problem might be that you are choosing 0.6435 as arcsin of 3/5. The angle t0 needs to be between pi/2 and pi, because of the quadrants the start and end points are in (as SK points out), so one should choose the arcsin instead as pi-0.6435. If you use that instead, I think you'll get the right result.
Last edited: May 19, 2016
6. May 19, 2016
### TheBoro76
Thanks for your responses I tried both of your suggestions. but couldn't quite get them to work (I probably made a mistake somewhere). However, if I used the value t0=arccos(-4/5), I got the answer I expected, So I'm going with that.
Thanks again!!
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# Matrix group isomorphic to $\mathbb Z$.
The set $G=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\mid n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).
TO show it's isomorphism: I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.
How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.
Thank you!
• Try it. What happens when you multiply $\phi(a)$ and $\phi(b)$? – hardmath Oct 23 '15 at 14:15
• Homomorphism, check. Generator? – user190080 Oct 23 '15 at 14:27
Hint:
$\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix}1 & m \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix}1 & n+m \\ 0 & 1 \\ \end{pmatrix}$
That should help with proving $\phi$ is a homomorphism.
Once you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective.
• This is what I used. Thank you! – maidel b Oct 23 '15 at 16:44
• Glad it helped. – kleineg Oct 23 '15 at 17:31
• If $\phi$ is a homomorphism from $\Bbb Z \to G$, shouldn't $\phi^{-1}$ take matrices as input? – pjs36 Mar 16 '17 at 19:43
• OK, yes, that's better now – pjs36 Mar 16 '17 at 20:09
• This is not the inverse. As in function inverse. This is not even a different function from $\phi$. What the hell? In what world the inverse of a mapping $X\to Y$ is a mapping $X\to Y$? How changing $n$ to $-n$ proves that something is bijective? Come on. The proper inverse is defined by $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. I don't believe there are 2 answers with the same mistake. – freakish Mar 16 '17 at 20:11 | {
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Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?
The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).
• Thank you. I understand one-to-one after writing out the steps. I'm struggling with onto. Let me make sure I'm thinking correctly. I'm trying to find $\phi(?)=n$. Is there a different way? I'm not able to see what I'm supposed to do. – maidel b Oct 23 '15 at 17:50
• @ShayAbbott No. You want to take an arbitrary $g\in G$ and find an $n\in Z$ so that $\phi(n)=g$. But if $g\in G$, then you know what $g$ looks like... – Teepeemm Oct 23 '15 at 18:09
So let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function
$$g:G\to\mathbb{Z}$$
such that $\phi\circ g=\mbox{id}_{G}$ and $g\circ \phi=\mbox{id}_{\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\phi$ is an isomorphism.
So you can easily check that
$$g\bigg(\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}\bigg)=n$$
is the inverse of $\phi$.
Just note that $$\begin{pmatrix}1 & 1 \\ 0 & 1 \\ \end{pmatrix}^n = \begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ This implies that $\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\phi$ is injective because $\phi(n)_{12}=n$.
Hint:
There are other ways to show that this map is one-to-one.
Use that the $ker(\phi)=0$.
Does there exist an $n$ such that the rank of $\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$ is not $2$? | {
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Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n \\ 0 & 1 \\ \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.
• That's not the inverse. The proper inverse is a function on matrices. It maps $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}\mapsto n$. Also if $\phi$ is a homomorphism then so is $\phi^{-1}$. So the existance of $\phi^{-1}$ is enough, no need to check that it is a homomorphism. – freakish Mar 16 '17 at 20:07 | {
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How to know if there is a extraneous solution in a radical expression
I was trying to solve this problem:
$$\sqrt{3x+13} = x+ 3$$
$$(\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$
So my final answer was $x = -4$ and $x = 1$. However, it was incorrect because when I plug back in -4 into the original equation I get a extraneous solution. My question is do I always need to plug back in my answers into a radical expression and check if they are valid? Or is there any other way to deduce that there will be a extraneous solution?
• In this case you can tell there is only one solution simply by thinking about the graphs of the functions on the left and right sides of your original equation. This tells you one of your roots is extraneous. It is also clear graphically that that solution is positive. So the extraneous root must be $-4$. Alternatively, recall that the range of the square root function is the set of nonnegative reals. So your original equation implies $x\geq-3$. – symplectomorphic Feb 18 '17 at 22:58
• @Khosrotash I am trying to avoid checking the final answers because it is a timed test and I need to move really quickly – Pablo Feb 18 '17 at 22:59
You don't need to plug in values, if you always ensure not to add extraneous solutions.
The equation forces two conditions, namely $3x+13\ge0$ and $x+3\ge0$, which together become $x\ge-3$.
Why $x+3\ge0$? Because $\sqrt{3x+13}\ge0$ by definition (when it exists, of course).
With this condition, you can safely square, because you have an equality between nonnegative numbers. You get (your computations are good) $$\begin{cases} (x+4)(x-1)=0 \\[4px] x\ge-3 \end{cases}$$ and therefore you know what roots are a solution of the original equation, in this case only $x=1$. | {
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On its domain ($A\ge 0$), note that $$\sqrt A=B\iff A=B^2\enspace\textbf{and}\enspace B\ge 0,$$ since the symbol$\sqrt{\phantom{h}}$ denotes the non-negative square root of a non-negative real number.
Note that when you squared both sides in the first step, you ended up with the equation
$3x+13 = (x+3)^2$,
which could just as well come from the same original equation but with a negative square root instead:
$-\sqrt{3x+13} = x+3$.
So, at the end of the day, you've solved both equations. To check which solution(s) correspond to which equation(s), you need to plug back in and check.
• Amm so does that mean that every time I square a radical then I would get something like this $$±(3x+13)^2$$ – Pablo Feb 18 '17 at 23:01
• No, because the square of a number is always positive. It just means that when you square both sides to try and solve the original equation, you're actually solving two equations at once. – J Richey Feb 18 '17 at 23:02
I think the best or reliable solving is to plot the equations . in your case like below | {
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# Minkowski metric in spherical polar coordinates
1. Mar 4, 2016
### spaghetti3451
1. The problem statement, all variables and given/known data
Consider Minkowski space in the usual Cartesian coordinates $x^{\mu}=(t,x,y,z)$. The line element is
$ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}$
in these coordinates. Consider a new coordinate system $x^{\mu'}$ which differs from these Cartesian coordinates. The Cartesian coordinates $x^{\mu}$ can be written as a function of these new coordinates $x^{\mu}=x^{\mu}(x^{\mu'})$.
(a) Take a point $x^{\mu'}$ in this new coordinate system, and imagine displacing it by an infinitesimal amount to $x^{\mu'}+dx^{\mu'}$. We want to understand how the $x^{\mu}$ coordinates change to first order in this displacement $dx^{\mu'}$. Argue that
$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$.
(Hint: Taylor expand $x^{\mu}(x^{\mu'}+dx^{\mu'})$.)
(b) The sixteen quantities $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ are referred to as the Jacobian matrix; we will require this matrix to be invertible. Show that the inverse of this matrix is $\frac{\partial x^{\mu'}}{\partial x^{\mu}}$. (Hint: Use the chain rule.)
(c) Consider spherical coordinates, $x^{\mu'}=(t,r,\theta,\phi)$ which are related to the Cartesian
coordinates by
$(t,x,y,z)=(t,r\ \text{sin}\ \theta\ \text{cos}\ \phi,r\ \text{sin}\ \theta\ \text{sin}\ \phi,r\ \text{cos}\ \theta)$.
Compute the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$. Is this matrix invertible everywhere? Compute the displacements $dx^{\mu}$ in this coordinate system (i.e. write them as functions of $x^{\mu'}$ and the infinitesimal displacements $dx^{\mu'}$).
(d) Compute the line element $ds^{2}$ in this coordinate system.
2. Relevant equations
3. The attempt at a solution
(a) By Taylor expansion,
$x^{\mu}(x^{\mu'}+dx^{\mu'}) = x^{\mu}(x^{\mu'}) + \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$ | {
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$x^{\mu}(x^{\mu'}+dx^{\mu'}) - x^{\mu}(x^{\mu'}) = \frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$
$dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\mu'}}dx^{\mu'}$
Am I correct so far?
Last edited: Mar 4, 2016
2. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yes, but you should go ahead and calculate the nine components:
$\frac{\partial x}{\partial r}$, $\frac{\partial y}{\partial r}$, $\frac{\partial z}{\partial r}$
$\frac{\partial x}{\partial \theta}$, $\frac{\partial y}{\partial \theta}$, $\frac{\partial z}{\partial \theta}$
$\frac{\partial x}{\partial \phi}$, $\frac{\partial y}{\partial \phi}$, $\frac{\partial z}{\partial \phi}$
3. Mar 4, 2016
### spaghetti3451
Isn't that in part (c)?
Shouldn't I do (b) first?
4. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yeah, I guess you should, even though part c doesn't actually depend on part b.
5. Mar 4, 2016
### spaghetti3451
(b) Via the chain rule,
$\frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^{\nu}}=\delta_{\nu}^{\mu}$,
where we are using the summation convention only over $\mu'$.
Therefore, the inverse of the matrix $\frac{\partial x^{\mu}}{\partial x^{\mu'}}$ is the matrix $\frac{\partial x^{\mu'}}{\partial x^{\nu}}$.
Is this correct?
6. Mar 4, 2016
### stevendaryl
Staff Emeritus
Yes.
7. Mar 4, 2016
### Staff: Mentor
What's wrong with taking the differentials of x, y, and z (expressed in terms of the spherical coordinates in post #1), evaluating their differentials (in terms of the spherical coordinates and their differentials), and then taking the sum of their squares? This should give the Minkowski metric in spherical coordinates, correct?
Chet
8. Mar 5, 2016
### spaghetti3451
I know that this is a correct and shorter approach, but I'm trying to follow the instructions of the question. | {
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# Is there a better way of finding the order of a number modulo $n$?
Say I wish to find the order of 2 modulo 41.
The way I've been shown to compute this is to literally write out $2^k$ and keep going upwards with $0 \leq k \leq 41$, or until I observe periodicity in the sequence, and then state the order from there.
Is there a better way of deducing the order of a number with respect to some modulo?
This seems highly inefficient, especially if we are working with respect to some large modulo $N$ where it will take at most $N$ computations to determine when periodicity occurs.
• Since $41$ is prime, the order must be a factor of $40$ Nov 17, 2014 at 6:49
• I did not know this, but thanks for that. Very useful. Perhaps a bad example, what if $n$ is NOT prime? Are there any results that can further 'filter' it out? Also, say we know the order must be a factor of 40, then is there a systematic way of computing the order without having to bash out any numbers? Nov 17, 2014 at 6:51
For any $$a$$ and $$N$$ with $$\gcd(a,N)=1$$, the order of $$a$$ modulo $$N$$ must be a divisor of $$\varphi(N)$$. So if you know the prime factorization of $$N$$ (or $$N$$ is already prime) so that you can compute $$\varphi(N)$$ and also know the prime factorization of $$\varphi(N)$$, you can proceed as follows:
If we know an integer $$m>1$$ with $$a^m\equiv 1\pmod N$$ and know the prime divisors of $$m$$, for all primes $$p$$ dividing $$m$$ do the following: Compute $$a^{m/p}\bmod N$$ and if the result is $$\equiv 1\pmod N$$, replace $$m$$ with $$m/p$$ and repeat (or switch to the next prime divisor if $$m$$ is no longer divisible by $$p$$). When you have casted out all possible factors, the remaining $$m$$ is the order of $$a$$. Note that the computations $$a^m\bmod N$$ do not require $$m$$ multiplications, but rather only $$O(\log m)$$ multiplications mod $$N$$ if we use repeated squaring. | {
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If $$N$$ is large and the fatorization of $$\varphi(N)$$ is known (and especially if you suspect the order of $$a$$ to be big), this is in fact a fast method.
Note that a couple of computations can be saved even beyon what is descibed above: In the case $$p=2$$, we may end up computing $$a^m, a^{m/2}, a^{m/4},\ldots$$ to cast out factors of $$2$$. But the later numbers were in fact intermediate results of computing $$a^m$$ by repeated squaring! Also, once we notice for some $$p$$ with $$p^k\mid m$$ that $$a^{m/p}\not\equiv 1\pmod N$$, we can save a few squarings and so speed up the task for the remaining primes if we replace $$a$$ with $$a^{p^k}\pmod N$$ and $$m$$ with $$m/p^k$$ - just remember to multiply the factor $$p^k$$ back into the final answer!
In your specific example, we know that $$N=41$$ is prime and that $$\varphi(N)=40=2^3\cdot 5$$. We check $$p=5$$ and note that $$2^{40/5}=256\equiv 10\pmod{41}$$, hence the factor $$5$$ cannot be eliminated. After that we check how many $$2$$'s we have to use: $$2^5\equiv 32\equiv -9$$, hence $$2^{10}\equiv 81\equiv -1$$, $$2^{20}\equiv (-1)^2\equiv 1$$. We conclude that $$2$$ has order $$20$$ modulo $$41$$.
The order must be one of $1,2,4,8,5,10,20,40$.
Calculate $2^2,2^4=(2^2)^2,2^8=(2^4)^2,...\\2^5=2^4*2,2^{10}=(2^5)^2,2^{20}=(2^{10})^2,2^{40}=(2^{20})^2$ which is seven calculations.
If 41 is not prime, say $41=a^2b^3c$,then
i) calculate the order of $2\pmod{a^2}$, and the order $\pmod{b^3}$ and the order $\pmod c$
ii) calculate the lowest common multiple of the separate orders.
Here we calculate the order of $$2$$ in $$(\mathbb{Z}/{41}\mathbb{Z})^\times$$.
Note: When calculating you can use the previous work to your advantage. To really cut down on the calculations, if you know the order is greater than or equal $$10$$, once you calculate $$2^{10} \pmod{41}$$ you are done, | {
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$$\quad 2^{10} \equiv 1 \pmod{41} \quad \text{order is } 10$$
$$\quad 2^{10} \equiv 40 \pmod{41} \quad \text{order is } 20$$
$$\quad \text{NOT }[2^{10} \equiv 1, 40 \pmod{41}] \quad \text{order is } 40$$
Work Summary : $$2^1 \equiv 2 \pmod{41}$$.
The order of $$2$$ is one of $$2,4,8,5,10,20,40$$.
Work Summary : $$2^2 \equiv 4 \pmod{41}$$.
The order of $$2$$ is one of $$4,8,5,10,20,40$$.
Work Summary : $$2^4 \equiv 16 \pmod{41}$$.
The order of $$2$$ is one of $$8,5,10,20,40$$.
Work Summary : $$2^5 = \equiv 32 \pmod{41}$$.
The order of $$2$$ is one of $$8,10,20,40$$.
Work Summary : $$2^8 \equiv 10 \pmod{41}$$
The order of $$2$$ is one of $$10,20,40$$.
Work Summary : $$2^{10} \equiv 40 \pmod{41}$$
The order of $$2$$ is one of $$20,40$$.
Work Summary : $$2^{20} \equiv 1 \pmod{41}$$
The order of $$2$$ is equal to $$20$$. | {
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Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
Many people dread choosing answer choice (D) on Quantitative Comparison (QC) Some feel it may be conceding defeat. Others think that the GRE is trying to trick them by making them pick (D). After all, they think, there must be some pattern that I’m not getting.
The truth is answer (D) comes up often. And to determine whether an answer cannot be determined is actually not too difficult.
#1 Determine a relationship
Say you find an instance, in which the answer is (A) the information in column A is greater. If that is the case, then the next step is to disprove that.
#2 Disprove that relationship
Meaning, see if you can come up with an instance, either through plugging in different variables, manipulating algebra, or manipulating a geometric figure, in which the answer is not (A). As soon you do that, you can stop. The answer is (D).
If you can’t disprove your answer, then it must be correct: it must be (A), (B) or (C).
1. $$-100 < x < 0$$
Column A Column B
$$x^{-4}$$ $$x^{-3}$$
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
Explanations:
Question #1
After choosing a few numbers you should note something: that Column A will always be greater than Column B.
Why? Will, whenever, you have an even exponent, positive or negative, that exponent will always yield a positive number. | {
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Odd exponents, on the other hand, give you a negative output if the base (the number below the exponent) is negative. Remember x has to be negative. So no matter what number you plug in Column A will always be positive, Column B negative. This is a definite not (D). The answer is (A).
32 Responses to Quantitative Comparison: “The Relationship Cannot Be Determined from the Information Given” Answer Choice
1. lokesh nandni sood May 22, 2019 at 9:59 pm #
hi chrish i put all the different values if i put 1 , they both are equal , if i put -1 B is greater . same way if i put 2 value is 0.0625 A is smaller and 0.125 B is greater . didn’t get that . pls help
• Magoosh Test Prep Expert May 24, 2019 at 8:36 am #
Hi Lokesh,
Remember that we have an important limitation to which numbers we can choose. The question tells us that -100 < x < 0. This means that x must be a negative number. You are plugging in positive numbers, which will give you a different answer 🙂
2. Karan June 28, 2017 at 4:54 am #
Hi Chris,
I have a doubt related to quantitative comparison type questions. What option do we choose if both columns turn out to be infinity. For eg:
Column A Column B
The number of numbers from 1 to 2 The number of numbers from 1 to 5
D or C?
• Magoosh Test Prep Expert June 28, 2017 at 11:43 am #
Hi Karan,
Great and interesting question! Honestly, you will not see a question like this on the GRE, in which you are comparing two quantities, each with the solution of “infinite.” That being said, the comparison of infinite to infinite is equal, so I’d go with (C). But again, you will not see something like this. Still, it’s a very interesting and awesome question! Have a great day! 🙂
3. Prasanth January 5, 2017 at 11:43 pm #
Hi Chris,
In the above example
-100<x<0
Qty A : x^-4
Qty B : x^-3
Now multiply both sides by a positive number relation will not change. x^4 is a positive number.So
Qty A becomes : 1
Qty B becomes : x.
Now it is easy to plug -ve numbers for x. | {
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Qty A becomes : 1
Qty B becomes : x.
Now it is easy to plug -ve numbers for x.
In all definite ways, column A will be greater.
Can i approach this way or plugging numbers will be easy in test ?
I usually make mistakes while plugging -ve numbers .. Pls help
• Magoosh Test Prep Expert January 6, 2017 at 1:32 pm #
Hi Prasanth,
No, I do not recommend taking this approach. For example, imagine if you manipulated the problem by multiplying by x^5. So,
Qty A becomes: x
Qty B becomes: x^2
Now, in this case, Qty B is greater than Qty A, which is not true. It is best to approach this problem rephrasing this question.
Qty A: x^-4 = 1/(x^4)
Qty B: x^-3 = 1/(x^3)
Here you know that any negative number raised by an even exponent will be positive, and that same number raised by an odd exponent will be negative. This is how we know that column A is greater. I hope that helps! 😀
• fatima January 12, 2018 at 7:44 am #
Hi Chris, regarding this problem :
-100<x<0
Qty A : x^-4
Qty B : x^-3
Prasanth multiplied both sides by a positive number (x^4 is a positive number),which will be at all times positive since (-100<x<0), and it is legal to multiply both sides of an inequality by a positive number..
But it wil not be true to multiply both sides by x^5 since (-100<x<0),so x^5 will give a negative number. And it will change the inequality..
i think Prasanth's approach is right. Does it ?
thanks,,
• Magoosh Test Prep Expert January 18, 2018 at 12:40 pm #
Hi Fatima,
Prasanth’ approach is correct, in the sense that it really does work for this particular problem. Here, multiplying Quantity A and Quantity B by x to an even power will help simplify things and may help make them clearer. | {
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The reason I’d still recommend against Prasanth’s approach is that it adds an extra layer of complexity that isn’t necessary. Under real testing conditions, every second counts, and simply recognizing the number properties at play here, without plugging in new numbers, is important. In addition under the pressure of test conditions, it becomes easier to make a mistake like choosing x^5 instead of x^4 without noticing that the inequality is reversed.
4. Yogesh kurade December 13, 2016 at 9:56 pm #
Thank you! Sir
5. Priyam June 13, 2016 at 9:11 am #
In lesson intro to Quantitative comparison, there’s this example question
N is not an integer
6<N<10
Quantity A Quantity B
N 8
Here, the answer is mentioned as D) cannot be determined reason stated as it can be a decimal or a fraction, but integer itself means "Whole nos which can be -ve as well as +ve".
• Magoosh Test Prep Expert June 18, 2016 at 12:23 pm #
Hi Priyam 🙂
Thanks for your message 🙂 In that example, we’re told that 6 < N < 10. So, we know that N is a positive number between 6 and 10. However, we cannot determine whether N is greater than 8 using only the information given, that N is not an odd integer. N could be any positive number between 6 and 10 except 7, which is the only odd integer within this range. For example, N could equal 6.5 or 8.3. Neither of these numbers is an odd integer and therefore fits the description. Because N may be less than, equal to, or greater than 8, the answer to that example is (D).
I hope this clears up your doubts 🙂
• Bhavna Sharma June 26, 2016 at 3:14 am #
I have a doubt regarding this. In the above explanation, you have mentioned that “N could equal 6.5 or 8.3”. But, in the question it has been specified that N is an integer and 6.5 or 8.3 are not integers.
• Magoosh Test Prep Expert June 26, 2016 at 4:28 am #
Hi Bhavna 🙂 | {
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• Magoosh Test Prep Expert June 26, 2016 at 4:28 am #
Hi Bhavna 🙂
In that example, the prompt states that “N is not an odd integer.” This is not the same as saying that N is an even integer. N could be an even integer, but N could also be a decimal or fraction, since decimals and fractions are not odd integers. I hope this clears things up 🙂
• Bhavna Sharma June 27, 2016 at 8:25 am #
Thank you so much.. 🙂 I missed this point.
6. Lars July 30, 2012 at 4:55 am #
Hi Chris,
In your explanation to the 1st question, in the 2nd last sentence you wrote
“So no matter what number you plug in Column A will always be negative, Column B positive.” Isn’t this the exact opposite because Column A wiil be positive and column B will be negative?
I got confused the first time I read this.
7. Sammy May 30, 2012 at 3:54 pm #
Chris,
For these types of problems I have a trick that allows me to correctly deduce what the answer is. I wanted to get your thoughts on my method.
An example problem that I fabricated –
n is an integer
Which is greater?
COL A – n^3 + 6
COL B – n^2 + 5n -30
I would attack this problem in this way – I would choose 5 arbitrary numbers (0,-0.6, 0.6,
3,-3) and plug each in till i get an explicit answer. The only problem I see with my this is that it can get to be time consuming.
Let me know what you think!
PS – I love it when you answer in a witty way with your sophisticated words!
• Chris June 1, 2012 at 12:58 pm #
Hi Sammy,
You put me in a little bit of a tough spot, because it’s kind of difficult to be witty with Quantitative Comparison :).
Your method definitely works well. Picking numbers can help you determine which column is bigger as long as you make sure to include 0, 1, -1, 2, -2 (or 3, -3) and 1/2 and -1/2 (I think 0.6 is a little unwieldy :)).
But in general this is a very effective method and one that I too employ. Often, esp. with algebraic equations, you may want to manipulate the equations and see if you can simply them. | {
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Sorry, I didn’t drop any sophisticated words, but I hope that helped :).
• Sammy June 1, 2012 at 1:06 pm #
Thanks Chris, my only qualm would be that the method can be somewhat time consuming.
• Chris June 1, 2012 at 1:20 pm #
Oh, that’s right – you mentioned that…if you have a strong facility with numbers, that is you can quickly plug-in, then it shouldn’t be time consuming. Also you can come up with a tiny column grid:
A B
0
1
-1
2
-2
By the time you get the 2 you can usually know for certain what column it is.
Hope that helps :).
• Amal June 1, 2012 at 1:43 pm #
Hi Chris,
Would it be correct to say that if n is an integer, one need not test decimals that can be written as a fraction. For example, 0.6 = 6/10, which means that 0.6 is actually a fraction, aka an integer divided by another integer, or two integers. And the same thing for 0.5, since it equals 1/2. In fact, when it comes to “n is an integer” type questions, we should be able to exclude all decimals, since according to the “official GRE guide,” only real numbers are involved (e.g. we would not test 3.14 which is a decimal but can’t be written as a fraction. This would mean that every decimal number is a real number (which implies it can be written as a fraction) for GRE purposes, as far as quantitative comparison type questions are concerned. Great question by the way.
• Chris June 1, 2012 at 2:04 pm #
Yes, you are exactly right. I was speaking more broadly when giving that spread of numbers. One should always obey the constraints of the question, i.e. if x is a positive integer, you have to plug in accordingly.
Thanks for catching that :).
• Sammy June 1, 2012 at 2:06 pm #
Yes you are right, it was an egregious error made on my part. I meant to say ‘a real number’
• Chris June 4, 2012 at 2:44 pm #
No problem :). Glad I could help.
• harsha July 7, 2016 at 1:46 am #
hey chris i am your big fan of u can u give me links for practcicng gre full lenth tests of free costs | {
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• Magoosh Test Prep Expert July 7, 2016 at 10:37 am #
Hi Harsha 🙂
I’d recommend that you first take the free PowerPrep tests from ETS, if you haven’t already. Next, here’s a free test from Manhattan (their materials are great!): Manhattan Free GRE Practice Test. Lastly, if you want access to more practice tests, I’d recommend purchasing one of the Manhattan books–each book comes with a code to access 6 online tests.
Hope this helps 🙂
8. Denis May 23, 2012 at 9:57 am #
Hi Chris,
I have a question regarding a concept in Question #2: you state “Three equal sides equal three equal angles.” Is this a characteristic of all polynomials, i.e., can we say “x equal sides equal x equal angles” for any figure?
Best regards,
Denis
• Chris May 23, 2012 at 11:40 am #
Hi Denis,
The measure of an angle corresponds to the length of the side opposite that angle, in a triangle. With quadrilaterals, or figures with more than four sides an angles, that relationship between which side corresponds to which angle is not as clear cut. However, we can say with certain that if there are x equal sides then there are x equal angles.
Hope that helps :).
• Denis May 23, 2012 at 1:34 pm #
Hi Chris,
I greatly appreciate the expeditious reply. Thanks for the help.
9. Amal May 20, 2012 at 11:58 am #
I’d say the answer for question 1 is (D). If you pick -1, then you get 1 for both column A and column B, since you have 1 over 1 raised to some positive power. If you pick, say, -2, then column A and column B are obviously different.
• Chris May 21, 2012 at 3:22 pm #
Hi Amal,
If you take any negative to an even exponent it will always yield a positive number. Therefore, if you plug in -1, you will get 1 for Col. A and -1 for Col B.
Hope that helps :).
• Amal May 21, 2012 at 4:11 pm # | {
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Hope that helps :).
• Amal May 21, 2012 at 4:11 pm #
Yeah you’re right. I forgot the negative sign on the coefficient. When you raise a negative coefficient to a negative odd power, the coefficient (which winds up in the denominator) is still negative. The coefficients are always negative, but the even power in column A yields a positive quantity.
• Chris May 22, 2012 at 1:41 pm #
No problem 🙂
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More- over, if the partition is in fact an all-square partition and A, B, and D are all invertible, then (3.2) We can assume that the matrix A is upper triangular and invertible, since $$\displaystyle A^{-1}=\frac{1}{det(A)}\cdot adj(A)$$ We can prove that $$\displaystyle A^{-1}$$ is upper triangular by showing that the adjoint is upper triangular or that the matrix of cofactors is lower triangular. Suppose the n × n matrix R is upper triangular and invertible, i.e., its diagonal entries are all nonzero. Clearly, the inverse of a block upper triangular matrix is block upper triangular only in the square diagonal partition. (a) if U is upper triangular and invertible then U^-1 is upper triangular. It fails the test in Note 5, because ad bc equals 2 2 D 0. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors – a lower triangular matrix L and an upper triangular matrix U: =. Taking transposes leads immediately to: Corollary If the inverse L 1 of an lower triangular matrix L exists, Show that R1 is also upper triangular. It's obvious that upper triangular matrix is also a row echelon matrix. Proposition If a lower (upper) triangular matrix is invertible, then its inverse is lower (upper) triangular. Hint. Let A and B be upper triangular matrices of size nxn. (b) The inverse of a unit lower triangular matrix is unit lower triangular (c) The product of two upper or (two lower triangular) matrices is upper or (lower) triangular It goes like this: the triangular matrix is a square matrix where all elements below the main diagonal are zero. An example is the 4 4 matrix 4 5 10 1 0 7 1 1 0 0 2 0 0 0 0 9 . The inverse element of the matrix [begin{bmatrix} 1 & x & y \ 0 &1 &z \ 0 & 0 & 1 end{bmatrix}] is given by [begin{bmatrix} 1 & -x & xz-y \ 0 & 1 & -z \ 0 & 0 & 1 end{bmatrix}.] Inverse of matrix : A square matrix of order {eq}n \times n{/eq} is known as an upper triangular matrix if all the elements | {
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matrix of order {eq}n \times n{/eq} is known as an upper triangular matrix if all the elements below principle diagonal elements are zero. Let A be a square matrix. 11.7 Inverse of an upper triangular matri. Let $b_{ij}$ be the element in row i, column j of B. Let A be a n n upper triangular matrix with nonzero diagonal entries. For a proof, see the post The inverse matrix of an upper triangular matrix with variables. The proof for upper triangular matrices is similar (replace columns with rows). Use back substitution to solve Rsk-en for k 1, , n, and argue that (sk)i -0 for i > k. The inverse of a triangular matrix is triangular. It's actually called upper triangular matrix, but we will use it. In the next slide, we shall prove: Theorem If the inverse U 1 of an upper triangular matrix U exists, then it is upper triangular. In this problem, you will Example of upper triangular matrix: 1 0 2 5 0 3 1 3 0 0 4 2 0 0 0 3 A triangular matrix (upper or lower) is invertible if and only if no element on its principal diagonal is 0. Solving Linear Equations Note 6 A diagonal matrix has an inverse provided no diagonal entries are zero: If A D 2 6 4 d1 dn 3 7 5 then A 1 D 2 6 4 1=d1 1=dn 3 7 5: Example 1 The 2 by 2 matrix A D 12 12 is not invertible. In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. In general this is not true for the square off-diagonal partition. Let $a_{ij}$ be the element in row i, column j of A. An upper triangular matrix is a square matrix in which the entries below the diagonal are all zero, that is, a ij = 0 whenever i > j. 82 Chapter 2. | {
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# Vertically oriented spring and energy transfers
1. Feb 20, 2007
### CaptainZappo
I am posing this question due to a statement made by my TA:
Suppose there is a spring pointed vertically upwards. Further, suppose that the spring is compressed a distance x so that it stores some potential energy. A mass is then placed on top of the spring. Assume all energy is conserved.
Here's the question: upon releasing the spring, what is the kinetic energy of the ball as it passes the equilibrium position of the spring? Is it equal in magnitude to (1/2)kx^2 where x is the amount the spring is compressed, or is it (1/2)kx^2 - mgx?
My TA said the former is correct, while I cannot figure out why it is not the latter. My reasoning: as the ball is rising from its initial height (the point at which the spring is fully compressed) some of that energy is being converted to gravitational potential energy. The rest goes to kinetic energy.
Any insight will be greatly appreciated,
-Zachary Lindsey
2. Feb 20, 2007
### Staff: Mentor
Your reasoning is correct; you cannot ignore gravitational PE in calculating the KE of the mass.
3. Feb 20, 2007
### CaptainZappo
Thank you.
4. Feb 20, 2007
### Parlyne
Who is correct depends on whether x is the amount the spring is compressed from its free equilibrium length or that from its equilibrium length with the mass sitting on it. If it is the former, you are correct. If it's the later, your TA is correct.
To see this, let's let y be the distance the mass is sitting above the spring's free equilibrium position. Then, the potential energy is:
$$V(y) = \frac{1}{2} k y^2 + mgy$$
We can rearrange this:
\begin{align*} V(y) &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y\right ) \\ &= \frac{1}{2} k \left (y^2 + 2\frac{mg}{k} y + \frac{m^2g^2}{k^2} \right ) - \frac{m^2g^2}{2k} \\ &= \frac{1}{2} k \left (y + \frac{mg}{k} \right )^2 - \frac{m^2g^2}{2k} \end{align*} | {
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From this, we can see that the mass' height above the compressed equilibrium length is $$x = y + \frac{mg}{k}$$.
So, $$V(x) = \frac{1}{2} k x^2 - \frac{m^2g^2}{2k}$$.
The difference between the potential energy when the spring is compressed by a distance x from the compressed equilibrium position ($$y = \frac{-mg}{k}$$) and at the compressed equilibrium is:
\begin{align*} \Delta V &= V(x) - V(0) \\ &= \frac{1}{2} k x^2 - \frac{m^2g^2}{2k} - \left (\frac{-m^2g^2}{2k} \right ) \\ &= \frac{1}{2} k x^2 \end{align*}
5. Feb 21, 2007
### Staff: Mentor
Parlyne is certainly correct. From the way the conditions were specified in the OP:
I presumed that x measures the compression from the uncompressed position of the spring. Only later is a mass involved:
The answer to that question depends on the meaning of "equilibrium postion" and where x is measured from. If, as I assumed from your description, x is measured from the uncompressed position of the spring (and equilibrium means x = 0), then 1/2kx^2 represents spring PE only and does not include gravitational PE.
But, as Parlyne explained, if x is measured from the equilibrium position of the mass-spring system (which is not the uncompressed position of the spring), then 1/2kx^2 represents changes in both spring PE and gravitational PE together.
CaptainZappo: Better tell us exactly what the TA stated, not your interpretation of it (if you can recall).
Regardless of what the TA says, learn what Parlyne explained and you'll be ahead of the game. | {
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# Thread: Running Around a Track
1. ## Running Around a Track
Hello, helpful people. I'm in some need of help.
I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:
"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"
Thanks.
2. ## Re: Running Around a Track
Hello, MyHappyHarmony!
This one is tricky to set up.
I have a rather clunky approach . . .
Brooks and Avery are running laps around the outdoor track, in the same direction.
Brooks completes a lap every 78 seconds, while Avery completes a lap every 91 seconds.
Brooks (the faster runner) has just passed Avery.
How much time will it take for Brooks to overtake Avery again?"
We will use: . $\text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad r \,=\,\frac{d}{t}$
Let $d$ = distance in one lap (say, in feet).
Brooks' rate is $\frac{d}{78}$ ft/sec.
Avery's rate is $\frac{d}{91}$ ft/sec.
In $t$ seconds, Avery runs $\frac{d}{91}t$ feet.
In the same $t$ seconds, Brooks runs $\frac{d}{78}$ feet,
. . which is $d$ feet (one lap) more than Avery's distance.
There is our equation! . . . $\frac{d}{78}t \;=\;\frac{d}{91}t + d$
Divide by $d\!:\;\frac{t}{78} \:=\:\frac{t}{91} + 1$
Got it?
3. ## Re: Running Around a Track
Yes, I do! Thanks. Quick question: Would it be right to solve with a common denominator from there? The big numbers do look quite intimidating.
4. ## Re: Running Around a Track
Originally Posted by MyHappyHarmony
Hello, helpful people. I'm in some need of help. | {
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Originally Posted by MyHappyHarmony
Hello, helpful people. I'm in some need of help.
I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:
"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"
Thanks.
Try to construct your solution from this figure
cross posting
5. ## Re: Running Around a Track
I'm sorry. I'm not drawing anything from that figure.
6. ## Re: Running Around a Track
another method
assume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps
after B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft
Rate of closure = 15.38-13.19 = 2.2 fps
closure time 1029/2.2 = 467.7 sec
total elapsed time = 467.7 +78= 555.7 sec
7. ## Re: Running Around a Track
Originally Posted by bjhopper
total elapsed time = 467.7 +78= 555.7 sec
typo; 545.7 ; 546 really: LCM(91,78)
These are quite simple: have the faster runner start behind the slower,
the distance behind being the track length; how long does it take to catch up?
8. ## Re: Running Around a Track
Hello Wilmer,
Long 'time no see.How are you?
Soroban' s equation gives elapsed time until runners meet. 546 sec exactly
My answer 555.7 sec.There is no typo error.
Can you show what I did wrong?
Here is another solution
B rate 1/78 laps per sec A rate 1/91 laps per sec
(1/78 -1/91) = rate of closure laps per sec when B is behind A
after 1 lap B is 78/91 =0.857 laps behind A
(1/78 -1/91)t = 0.857
(0.00183)t =0.857 t=468.3sec add 78 for first lap Total elapsed time = 556.3sec
What say you?
9. ## Re: Running Around a Track
Hey BJ. Nice to "type" with you again!! | {
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What say you?
9. ## Re: Running Around a Track
Hey BJ. Nice to "type" with you again!!
Originally Posted by bjhopper
My answer 555.7 sec.There is no typo error.
But you show this in your previous post:
"total elapsed time = 467.7 +78= 555.7 sec"
But 467.7 + 78 = 545.7, not 555.7
That's what I was trying to tell you. | {
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# Countable union of countable sets vs countable product of countable sets
I know that a countable union of countable sets is countable, and that a finite product of countable sets is countable, but even a countably infinite product of countable sets may not be countable.
Let $$X$$ be a countable set. Then $$X^{n}$$ is countable for each $$n \in N$$.
Now it should also be true that $$\bigcup^{\infty}_{n=1} X^{n}$$ is countable. How is this different from $$X^{\omega}$$, which is uncountable?
Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned
$$\bigcup{X^n}$$
The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...).
So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...
But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.
Of course, this is not a proof. But it merely gives an indication why the two sets are different...
Ah, very good question. Let's take a look at this question with $$X=\mathbb{N}$$. You first mentioned
$$\bigcup{X^n}$$
The elements of these set are all the finite tuples. So examples include (1,2), (3,4,2000),... Of course, if you add zero's to the end, then you obtain an element of $$X^\omega$$. So (1,2) corresponds to (1,2,0,0,0,...) and (3,4,2000) corresponds to (3,4,2000,0,0,0,...). | {
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So it is easily seen that every element of $$\bigcup{X^n}$$ can be represented by an element of $$X^\omega$$. And it is also easy to see that such elements are exactly the elements ending with a trail of 0's. For example: (1,1,1,0,0,0,...) corresponds to (1,1,1)...
But, and here comes the clue: there are much more elements in $$X^\omega$$. For example: (1,1,1,...) does not come from an element of $$\bigcup{X^n}$$. Other examples are (1,2,3,4,...) or (2,7,1,8,2,...). This shows that there are a huge number of elements in $$X^\omega$$!! And so $$X^\omega$$ really is different than $$\bigcup{X^n}$$.
Of course, this is not a proof. But it merely gives an indication why the two sets are different...
Yes, after thinking about it I came to the same conclusion. You can't put them in bijective correspondence because if you wanted to map the union onto $$X^\omega$$ you could do so injectively by adding 0s but there's no way to make this mapping surjective.
mathwonk
Homework Helper
i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.
i believe the standard cantor argument shows even a countable product of the set {0,1} is uncountable, i.e. the set of all sequences of 0's and 1's.
Exactly, hence why this was a seeming contradiction.
Exactly, hence why this was a seeming contradiction.
What's the contradiction that you're referring to?
It's already been stated that the needed bijection doesn't exist. I'm not following you're argument... | {
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# Polynomial riddle
1. Jun 8, 2014
### dislect
Hi guys,
My boss gave me a riddle.
He says that you have a "black box" with a polynomial inside it like f(x)=a0+a1x+a2x^2+a3x^3 ....
you don't know the rank of it or the coefficients a0, a1, a2 ....
You do know:
all of the coefficients are positive
you get to input two x numbers and their f(x) value
How can you find all of the coefficients a0,a1,a2 .... and the rank?
Thanks !
2. Jun 8, 2014
### hilbert2
Knowing f(x) for two values of x is not enough to determine the coefficients if the polynomial is of higher than first order.
3. Jun 8, 2014
### economicsnerd
It can be done if you know the polynomial has nonnegative integer coefficients, and if the second input is allowed to depend on the first output.
Put in $1$. The output gives you the sum $s$ of the coefficients.
- If $s=0$, you know $f=0$.
- If $s>0$, put in any integer $k>s$. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-$k$.
4. Jun 8, 2014
### dislect
Hi economicsnerd,
Couldn't follow "put in any integer k>s. Then there is a unique integer solution, which is especially easy to read off if you express the output in base-k."
Could you give me an example? lets say f(x)=5+2x+4x^2 is in the 'black box'.
i put in x=1 and receive f(x)=s=a0+a1+a2=11 > 0 so i put in x=k=15 and get f(x)=a0+15a1+900a2=935
organize:
1. a0+a1+a2=11 2. a0+15a1+900a2=935
2 equations with 3 unknowns.
Whats the rational behind putting a k > s, and how can i find all coefficients with just 2 equations ?
5. Jun 8, 2014
### economicsnerd | {
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5. Jun 8, 2014
### economicsnerd
Okay. So you know $f(1)=11$ and $f(15)=935$.
- First note that $15^3 > 935$, so that the polynomial can't be of degree $>2$. That is, $f(x)=a_0 + a_1 x + a_2 x^2$ for some $a_0,a_1, a_2\in\mathbb Z_+$.
- Next, note that $a_0,a_1\in\mathbb Z_+$ with $a_0+ a_1 \leq 11 < 15$, so that $a_0 + a_1(15)<15^2$. This means that $935 - 15^2< a_2 (15)^2 \leq 935$. It's easy to check that there's a unique integer $a_2$ that accomplishes this, namely $a_2 = 4$.
- Substituting the solution for $a_2$ in, we now know that: $a_0+a_1 < 15$ and $a_0 + a_1 (15) = 15 35$. [Of course, you have two equations and two unknowns now, but we don't actually need to know that $a_0 + a_1 =7$, because knowing they're integers gives us enough information.] Now we can do the same trick again. $35 - 15< a_1 (15) \leq 35$, which pins down $a_1=2$.
- Then substitution yields $a_0 = 5$.
We can always work backward in this way.
To get some better intuition for why, consider the following question:
Say I tell you I have a polynomial $g$ such that every coefficient of $g$ is a nonnegative integer $<10$. If I tell you $g(10)$, will you then be able to tell me what $g$ is?
For example:
- If $g(10)=5021$, then do we know $g(x) = 5x^3 + 2x + 1$?
- If $g(10)=60007$, then do we know $g(x) = 6x^4 + 7$?
Nothing is magical about $10$ that makes this trick work. It's just that we usually express numbers in a way (base-10) that makes it easy to answer this question. All the first question is needed for is to find a $k$ such that we can be sure that every coefficient of the polynomial is a nonnegative integer $<k$.
6. Jun 8, 2014
### D H | {
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6. Jun 8, 2014
### D H
Staff Emeritus
The rationale behind using asking for f(k) where k>f(1) is because otherwise the magic might not work. Suppose the function in the black box is f(x)=5+2x+4x2. The black box cranks out 11 as a response to f(1). Suppose I use 4 next. The black box churns out f(4)=77, which is 1031 when expressed base 4. That suggests the polynomial is 1+3x+5x3, which is wrong.
To see what's going on, here are f(k) from k=2 to 16, expressed in base 10 and in base k:
Code (Text):
k f(k) base k Polynomial
2 25 11001 1+x^3+x^4
3 47 1202 2+x^2+x^3
4 77 1031 1+3x+x^3
5 115 430 3x+4x^2
6 161 425 5+2x+4x^2
7 215 425 5+2x+4x^2
8 277 425 5+2x+4x^2
9 347 425 5+2x+4x^2
10 425 425 5+2x+4x^2
11 511 425 5+2x+4x^2
12 605 425 5+2x+4x^2
13 707 425 5+2x+4x^2
14 817 425 5+2x+4x^2
15 935 425 5+2x+4x^2
16 1061 425 5+2x+4x^2
Can you see the pattern? It appears that f(k) when expressed in base k is 425 for all k>5. This is indeed the case. You might want to try proving it.
Note that in this case reading off the polynomial from the base-k representation of f(k) works for all k>5. That's because the largest coefficient is five. Using k=f(1)+1 (or higher) ensures that k is larger than the largest possible coefficient.
Suppose the black box function was instead f(x)=11x3. Once again you'll get f(1)=11, but now the base k representation of f(k) keeps changing until you reach k=12, at which point it becomes b00 ('b' means 11) and stays that way for all k>11. The black box knows what the largest coefficient is. You don't, but you do know that it cannot be larger than 11.
Last edited: Jun 8, 2014
7. Jun 9, 2014
### dislect
Thank you all :) | {
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# Possible mathematical induction problem
• Dec 2nd 2013, 02:03 PM
Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
• Dec 2nd 2013, 03:09 PM
Plato
Re: Possible mathematical induction problem
Quote:
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
A strict grader may like to have seen more grouping symbols.
However, the argument is correct. | {
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Tricky Triangle Area Problem
This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.
Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$
-
Are you familiar with the Law of Cosines? – 2012ssohn Apr 28 '14 at 23:37
Additionally, it was Multiple choice, and the four answers were $2\sqrt{6}$, $5$, $3\sqrt{6}$, and $5\sqrt{6}$ – Asimov Apr 28 '14 at 23:37
@2012ssohn Of course i am, but how can i apply that formula to it? I even tried it at the time, but couldnt get anything useful about area from it – Asimov Apr 28 '14 at 23:38
Did you actually try to use Heron's formula? The formula simplifies pretty nicely. – Nate Apr 28 '14 at 23:41
When i tried it i got an ugly glob of roots and addition that just wouldn't simplify for me. Maybe i just need to practice simplification – Asimov Apr 28 '14 at 23:44
From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.
Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$
Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).
The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.
-
Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. – Asimov Apr 28 '14 at 23:48 | {
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Use the $\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula.
\begin{align} & \frac{1}{4}\sqrt{4\cdot4\cdot25-(4+25-33)^2} \\ = & \frac{1}{4}\sqrt{4^2\cdot25-4^2} \\ = & \sqrt{25-1} \\ = & 2\sqrt{6} \end{align}
-
This is the way I would have done it, nice. – Sawarnik Apr 29 '14 at 6:40
You could've used Heron's formula straight away, actually.
\begin{align} T & = \tfrac{1}{4} \sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\ & = \tfrac{1}{4} \sqrt{(2+5-\sqrt{33})(2-5+\sqrt{33})(-2+5+\sqrt{33})(2+5+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7-\sqrt{33})(-3+\sqrt{33})(3+\sqrt{33})(7+\sqrt{33})} \\ & = \tfrac{1}{4} \sqrt{(7+\sqrt{33})(7-\sqrt{33})(\sqrt{33}+3)(\sqrt{33}-3)} \\ & = \tfrac{1}{4} \sqrt{(49-33)(33-9)} \\ & = \tfrac{1}{4} \sqrt{16 \cdot 24} \\ & = \tfrac{1}{4} \sqrt{64 \cdot 6} \\ & = \tfrac{8}{4} \sqrt{6} \\ & = 2 \sqrt{6} \end{align}
-
Since this was multiple-choice, I think it's worth noting that you could have guessed the right answer without doing (much) arithmetic: the diagonal of a right triangle with sides $2$ and $5$ has length $\sqrt{2^2+5^2} = \sqrt{29}$; since $\sqrt{33}$ is fairly close to this, the answer should be close to the area of a $2-5-\sqrt{29}$ triangle, which is of course $\frac12(2)(5)=5$. If you imagine how to 'stretch out' the $\sqrt{29}$ diagonal to $\sqrt{33}$, it's clear that the right angle will have to become obtuse, and this in turn means that the area of the $2-5-\sqrt{33}$ triangle will have to be less than $5$; of the provided answers, only $2\sqrt{6}\approx4.472$ is less than $5$ (and of course very close to it). | {
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-
Where do you see the answer options (multiple choices)? Is the original problem linked to? – Jeppe Stig Nielsen Apr 29 '14 at 9:05
@JeppeStigNielsen It's now there in the question. – Ramchandra Apte Apr 29 '14 at 14:06
I like this response, and used a similar estimation method, using the areas of slightly bigger and smaller triangles that were easier to calculate – Asimov Apr 30 '14 at 0:05
I hate it when there are two good answers, and both are right and good, and you wish you could accept both as correct – Asimov Apr 30 '14 at 1:41
@JohnJPershing For what it's worth, I think 2012ssohn's answer is much better than mine; it explains how to actually derive the correct value, rather than merely how to answer the multiple-choice question. Both are valuable, but that one's likely to be more broadly applicable; I just wanted to offer an alternate approach for the test. – Steven Stadnicki Apr 30 '14 at 3:24
Let be $ABC$ the triangle. Consider the altitude $AH$ over the greatest side, the one whose length is $\sqrt{33}$. Now call $x=BH$, so $\sqrt{33}-x=CH$. Apply Pythagoras' theorem to get $$\left\{ \begin{array}{rcl} x^2+h^2&=&4\\ \left(\sqrt{33}-x\right)^2+h^2&=&25 \end{array} \right.$$
Solve for $h$ and you are (almost) done.
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# Permutation with Condition
The numbers 1-9 are drawn from a hat at random. Without repetition of numbers, how many numbers (with all digits drawn-- eg. 987654321) would be divisible by 11?
First I recalled that a number is divisible by 11 if the number's alternating digits minus the other digits are equal to a number divisible by 11.
Honestly, I did a bit of guess and check to follow and found that 948576321 was an answer divisible by 11. Therefore if I rearranged 9,8,7,3, and 1 (the leading alternating numbers) it would bring the same result. So I solved the permutation to receive 120 possible combinations. I then took the permutation of 4,5,6, and 2 and received a total of 24 combinations.
I multiplied these two answers together to get a grand total of 2880 combinations that would be divisible by 11.
-
I ran a quick python program, and the answer is 31680. I'll try to prove it in a few minutes. – Alfonso Fernandez Feb 20 '13 at 22:33
I would appreciate it :) – Hannah Feb 20 '13 at 23:06
The correct answer is 31680. Your combination of alternating numbers (9, 8, 7, 3, 1) is only one out of 11 possibilities:
1. 1, 2, 3, 4, 7
2. 1, 2, 3, 5, 6
3. 1, 3, 7, 8, 9
4. 1, 4, 6, 8, 9
5. 1, 5, 6, 7, 9
6. 2, 3, 6, 8, 9
7. 2, 4, 5, 8, 9
8. 2, 4, 6, 7, 9
9. 2, 5, 6, 7, 8
10. 3, 4, 5, 7, 9
11. 3, 4, 6, 7, 8
So we end up with $11 \cdot 5! \cdot 4! = 11 \cdot 120 \cdot 24 = 31680$. | {
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So we end up with $11 \cdot 5! \cdot 4! = 11 \cdot 120 \cdot 24 = 31680$.
-
But how do you come up with these 11 without checking all $\binom{9}{4}=126$ possibilities? – Alfonso Fernandez Feb 21 '13 at 0:21
@AlfonsoFernandez I did it by checking all of them with a script. Brian's approach is more rigorous in that regard. – ralph Feb 23 '13 at 14:40
(This kind of problem, finding a set of distinct numbers to add up to a certain number, is a key part of Kakuro puzzles. The list of valid combinations for this can be found in Kakuro tables such as brainbashers.com/combinations.asp under 5-digit combinations with a sum of either 17 or 28. Brian's answer proves why the sum must be 17 or 28.) – ralph Feb 23 '13 at 14:47
Here’s a systematic approach; it’s still a bit tedious in spots, but not unreasonably so.
The sum of all $9$ digits is $45$. You need to split the digits into a set $A$ of $5$ digits, those in the odd-numbered positions, and a set $B$ of $4$ digits, those in the even-numbered positions, whose sums differ by a multiple of $11$. If $a$ is the sum of the digits in $A$ and $b$ the sum of the digits in $B$, then $a-b$ is a multiple of $11$, and $a+b=45$.
Clearly $b=45-a$, so $a-b=a-(45-a)=2a-45$. The smallest possible value of $a$ is $1+2+3+4+5=15$, and the largest is $9+8+7+6+5=35$, so $-15\le a-b\le25$. The multiples of $11$ in that range are $-11,0,11$, and $22$. However, $2a-45$ is clearly odd, so in fact the only possible values of $a-b$ are $-11$ and $11$.
• If $a-b=2a-45=-11$, then $a=17$.
• If $a-b=2a-45=11$, then $a=28$.
What sets of $5$ digits sum to $17$? Since $1+2+3+4+5=15$, we either increase $5$ by $2$ to get the set $\{1,2,3,4,7\}$ or increase $4$ and $5$ by $1$ each to get the set $\{1,2,3,5,6\}$; anything else produces either too big a sum or duplicated digits. | {
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It’s a bit harder to enumerate the sets that sum to $28$, because there are more of them. A good starting point is the fact that $9+8+7+6+5=35$, which is $7$ too large. First note that we need to have at least one of $8$ and $9$: $7+6+5+4+3=25$ is too small. What can we do without the $9$? $8+7+6+5+4=30$, so, much as in the previous case, our only choices are to subtract $2$ from $4$ to get the set $\{8,7,6,5,2\}$ or subtract $1$ each from $4$ and $5$ to get $\{8,7,6,4,3\}$. Every other set will include the $9$.
Let’s try next for the ones that include $9$ but not $8$. $9+7+6+5+4=31$; we can subtract $3$ from the $4$, the $5$, or the $6$ to get the sets $\{9,7,6,5,1\}$, $\{9,7,6,4,2\}$, and $\{9,7,5,4,3\}$. A little experimentation shows that trying to reduce the sum by $3$ by subtracting $2$ from one digit and $1$ from another produces nothing new.
The last batch will be those containing both $9$ and $8$. If we keep the $7$ as well, we already have a total of $9+8+7=24$, and the only two digits that supply the missing $4$ are $1$ and $3$, giving us the set $\{9,8,7,3,1\}$. If we start with $9,8$, and $6$, we can get the missing $5$ from either $1$ and $4$ or $2$ and $3$, giving us the sets $\{9,8,6,4,1\}$ and $\{9,8,6,3,2\}$. If start with $9,8$, and $5$, the only way to get the missing $6$ is with $2$ and $4$, giving us the set $\{9,8,5,4,2\}$. And $9+8+4+3+2=26$ is too small, so we’ve found all of the possibilities for the set $A$.
As you already observed, we can permute $A$ and $B$ arbitrarily, so each choice of $A$ yields $5!4!$ numbers, and there are $11$ possibilities for $A$, for a grand total of $11\cdot5!4!=31,680$ numbers.
-
Every number of the form $x*11$ is divisible by 11. Now you must think about $x$.
$x$ must be such that $(x)*11$ is a number with 9 digits! If it is useful for you I can explain it more. | {
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# The Asymptotic Bode Diagram for Non-Minimum Phase Poles and Zeros
## A Real Pole with Negative ω0
Elsewhere we have discussed how to make Bode plots for a real pole. You should be familiar with that analysis. The discussion there assumed that the value of ω0 was positive; here we discuss the case if ω0 is negative. We start with
$$H(s)=\frac{1}{1+\frac{s}{\omega_0}}$$
#### Magnitude
If you carefully examine the analysis (here) of the "Magnitude" plot you'll see that the only time ω0 is used, it is squared (e.g., $\left( \frac{\omega}{\omega_0} \right)^2$). Therefore, the magnitude plot does not depend on the sign of ω0, only its absolute value, so we don't need to change anything to accomodate a negative value of ω0.
#### Phase
The phase however does change. The phase of a single real pole is given by is given by
$$\angle H\left( {j\omega } \right) = \angle \left( {{1 \over {1 + j{\omega \over {{\omega _0}}}}}} \right) = - \angle \left( {1 + j{\omega \over {{\omega _0}}}} \right) = - \arctan \left( {{\omega \over {{\omega _0}}}} \right)$$
Let us again consider three cases for the value of the frequency, but we assume ω0 is negative:
Case 1) ω<<ω0. This is the low frequency case with ω/ω0→0, and doesn't depend on the sign of ω0. At these frequencies We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx -\arctan \left( 0 \right) = 0^\circ = 0\;rad$$
Case 2) ω>>ω0. Here we will consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
This is the high frequency case with ω/ω0 → +∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left( \infty \right) = - 90^\circ$$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at -90°.
ω0 < 0 (the non-minimum phase case) | {
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ω0 < 0 (the non-minimum phase case)
This is the high frequency case with ω/ω0 → -∞. We can write an approximation for the phase of the transfer function
$$\angle H\left( {j\omega } \right) \approx - \arctan \left(- \infty \right) = + 90^\circ$$
The high frequency approximation is at shown in green on the diagram below. It is a horizontal line at +90°.
Case 3) ω=ω0. Again consider the cases of positive and negative ω0 side by side.
ω0 > 0 (the minimum phase case, discussed previously)
$$\angle H\left( {j\omega } \right) = - \arctan \left( 1 \right) = - 45^\circ$$
ω0 < 0 (the non-minimum phase case)
$$\angle H\left( {j\omega } \right) = - \arctan \left( -1 \right) =+ 45^\circ$$
From the above discussion you can see that the only effect of the pole having a negative value of ω0 is that the phase is inverted (it increases from 0 → +90° as ω increases from 0 → ∞. The images below show the Bode plots for
$$H_1(s)=\frac{1}{1+\frac{s}{10}}, \quad\quad H_2(s)=\frac{1}{1-\frac{s}{10}}$$
H1(s) has a positive ω00=+10, so H1(s)=1/(1+s/10)) and H2(s) has a negative ω00=-10, so H2(s)=1/(1-s/10)). The pole of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero)
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
The same conclusion holds for first order poles and second order poles and zeros (see below).
## A Real Zero with Negative ω0
The images below show the Bode plots for two functions, one with a positive ω00=+10) and one with a negative ω00=-10). The zero of H1(s) is at s=-10 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+10 (a positive real part, the right half of the s-plane; a non-minimum phase zero). | {
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$$H_1(s)=1+\frac{s}{10}, \quad\quad H_2(s)=1-\frac{s}{10}$$
H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites. Recall that 360° is equivalent to 0° so you can think of the plot for the angle of H2(s) as starting at 0° and dropping by 90° (though the plot shows it as starting at 360°).
## A Second Order Pole with Negative ζ
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=\frac{1}{1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}, \quad\quad H_2(s)=\frac{1}{1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2}$$
The poles of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase pole) and the pole of H2(s) is at s=+0.5±j9.987 (a positive real part, the right half of the s-plane; a non-minimum phase pole). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that, again, the magnitudes are identical, but the phases are opposites.
## A Second Order Zero with Negative ζ
The images below show the Bode plots for (note the sign of the middle term in the numerator is different)
$$H_1(s)=1+0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2, \quad\quad H_2(s)=1-0.1\cdot\frac{s}{10}+\left(\frac{s}{10}\right)^2$$
The zeros of H1(s) are at s=-0.5±j9.987 (a negative real part, the left half of the s-plane; a minimum phase zero) and the pole of H2(s) is at s=+0.5±j 9.987(a positive real part, the right half of the s-plane; a non-minimum phase zero). H1(s) is plotted as a solid blue line, and H2(s) as a dotted pink line. Note that the magnitudes are identical, but the phases are opposites.
##### Key Concept: For poles and zeros with positive real parts, phase is inverted
If a 1st order pole has a positive real part (i.e., a nonminimum phase system, pole is in right half of s-plane) at say s=+5, so
$$H(s)=\frac{1}{1 - \frac{s}{5}}$$ | {
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$$H(s)=\frac{1}{1 - \frac{s}{5}}$$
the magnitude of the Bode plot is unchanged from the case of a corresponding pole with negative value, at s=-5 (pole is in left half of s-plane)
$$H(s)=\frac{1}{1 + \frac{s}{5}}$$
but the phase of the plot is inverted.
The same rule holds Bode plots for 2nd order (complex conjugate) poles, and for 1st and 2nd order zeros.
References
Replace | {
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# Number of ways to two-color a sectored circle so that opposite sectors are different colors.
Given a circle divided into 2n equal sectors, how many ways can it be painted with two colors so that opposite sectors are not the same color? I'm interested in the number of solutions that are unique under rotation.
I've tried to enumerate the early situations and see patterns but haven't seen a path to a general solution. I think I only have three rules of thumb at the moment:
1. There will be n sectors of each color
2. An alternating paint job is only legal when n is odd.
3. I think we're bounded on top by 2^(n-2) for n>1.
I may have missed some examples but this link has my work so far: http://i.imgur.com/ZQylapu.png
If I've gotten everything, then the series goes:
• n=1, 1
• n=2, 1
• n=3, 2
• n=4, 2
• n=5, 4
• n=6, 6
n=6 is notable as two of the combinations are left- and right-handed versions of a similar arrangement. I imagine this becomes much more common beyond n=6. I counted these separately, but a general solution that wouldn't would be nice as well. | {
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• How do you define opposite sector? If each sector has just one opposite then why wouldn't they answer by $2^n$? – fleablood Oct 12 '15 at 0:31
• Oh, I guess the unique under rotation puts a damper on it. – fleablood Oct 12 '15 at 0:33
• Huh, I have difficulty putting words to what I mean by opposite. What comes to mind is two sectors are opposites if the number of sectors between them in the clockwise direction is equal to the number in the counter-clockwise direction. – Polyhog Oct 12 '15 at 0:37
• No I get your definition. As a pair of sectors has no relation to other sectors the answer would be $2^n$ as each pair of sectors has two choices. But if they are unique to rotation, we have to calculate how many are equivalent under rotation. Good question. – fleablood Oct 12 '15 at 0:54 | {
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• Here's a thought. Fix sector 1. then for sector i = 2 to n give a binary value for color and add $2^i$ for green and nothing for yellow. Thus you'll have a 1-1 correspondence between colorings and the numbers from 1 to $2^{n - 1}. Now, we have to come up with a formula for determining if two numbers represent the same coloring under rotation. – fleablood Oct 12 '15 at 1:00 ## 1 Answer This can be done using Burnside's lemma. We need to count the assignments of colors being fixed by each type of permutation from among the$2n$permutations total in the cycle index$Z(C_{2n})$of the cyclic group acting on the sectors. There are $$\varphi(d)$$ permutations having cycle structure $$a_d^{2n/d}$$ in the cycle index$Z(C_{2n})$where$d|2n.$For a permutation to fix an assignment it needs to be constant on the cycles. Suppose that the cycle length$d$is even and pick one of its elements. The slot opposing this element is also located on this cycle. But they must have different colors, so there are no valid assignments fixed by a permutation of shape$a_d^{2n/d}$when$d$is even. On the other hand when$d$is odd the$2n/d$cycles of length$d$are grouped into$n/d$pairs which are reflections of each other and hence must have opposite colors. Therefore there are two possible assignments of colors to these cycles forming a pair (as opposed to four if there were no constraint). This yields a contribution of $$\varphi(d) 2^{n/d}.$$ Averaging this over the total$2n$permutations we get $$\frac{1}{2n} \sum_{d|2n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}$$ which is $$\frac{1}{2n} \sum_{d|n,\;d\;\mathrm{odd}} \varphi(d) 2^{n/d}.$$ This yields the sequence $$1, 1, 2, 2, 4, 6, 10, 16, 30, 52, 94, 172, 316, 586, 1096, \\ 2048, 3856, 7286,\ldots$$ which is OEIS A000016 where additional material awaits. I found the OEIS entry by using a simple total enumeration algorithm (definitely not optimized) to compute the first few values (practical to about$n=10\$ which is enough to conclusively identify the | {
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compute the first few values (practical to about$n=10\$ which is enough to conclusively identify the sequence). This was the Maple code: | {
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with(numtheory);
sectors :=
proc(n)
option remember;
local d, ind, orbits, orbit, rot, pos;
orbits := {};
for ind from 2^(2*n) to 2*2^(2*n) - 1 do
d := convert(ind, base, 2);
for pos to n do
if d[pos] = d[pos+n] then
break;
fi;
od;
if pos = n+1 then
orbit := {};
for rot from 0 to 2*n-1 do
orbit := {op(orbit),
[seq(d[q], q=1+rot..2*n),
seq(d[q], q=1..rot)]};
od;
orbits := {op(orbits), orbit};
fi;
od;
nops(orbits);
end;
Q :=
proc(n)
1/2/n*
select(d->type(d,odd), divisors(n)));
end;
• That is brilliant. The equation you reached is also is the same for number of binary sequences coming from a shift register complementing the last output (equivalent to going through the circle and outputting the sector's color as binary number). Thank you! – Weaam Oct 12 '15 at 3:04
• This was not original work. The calculation is just about included in the OEIS entry, it only needed filling in the details. – Marko Riedel Oct 12 '15 at 3:06
• Still! It is a great answer. – Weaam Oct 12 '15 at 3:22 | {
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# Computing a Limit using the Limit Definition
I've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.
Compute the limit. Then, using $\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. $$\lim_{n \to \infty} \frac{1}{ln(ln(n))}$$
Firstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\infty}_{n=1}$ if for every $\epsilon > 0$ there is an integer $N=N(\epsilon)>0$ such that
$$|a_n - L|< \epsilon$$
for all $n\geq N$. Now for my informal interpretation.
I believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\epsilon$? Any clarification of the definition would be appreciated.
Now, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that
$$\left | \frac{1}{ln(ln(n))} - 0 \right | = \frac{1}{ln(ln(n))}$$
So now I need an $N$ such that, for all $n \geq N$, I get $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that
$$\frac{1}{ln(ln(N))}<10^{-6}$$
So I just solve for N in the equality, which yields
$$N > e^{e^{1000000}}$$
Hence, for any $n \geq N$,
$$\frac{1}{ln(ln(n))}<10^{-6}$$
Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated. | {
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Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated.
-
It looks to me like you are done. You could compute $N$, but keep in mind it doesn't hurt for $N$ to be higher. It doesn't have to be the minimal one over $e^{e^{1000000}}$ – rschwieb Jan 10 '13 at 21:46
You can just write your integer as $\lceil e^{e^{1000000}} \rceil$ to make it clear you're talking about an integer. Other than that you have found a value that works and so you are done. – AvatarOfChronos Jan 10 '13 at 21:49
I suspect that computing $e^{e^{1000000}}$ would be a major project, requiring quite a bit of programming skill. It doesn't make sense to try to compute the exact value in this context at all. – Henry B. Jan 10 '13 at 22:33
it's a very big number, even computers are afraid to compute it. – Santosh Linkha Jan 10 '13 at 22:43
I think your confusion arises from the phrase "there exists an $N = N(\epsilon)$. You already showed why $N$ has to be a function of $\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.
You could just say then: choose $N > e^{e^{\epsilon^{-1}}}$ for $N(\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).
But, if you want to be explicit, you could use:
• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil$ where $\lceil - \rceil$ denotes the least integer greater than its argument, as you thought
• $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil + 8434$
• $N(\epsilon) = \lceil 9434\pi e^{e^{\epsilon^{-1}}}\rceil$ | {
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See, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)
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# Why does this not seem to be random?
I was running a procedure to be like one of those games were people try to guess a number between 1 and 100 where there are 100 people guessing.I then averaged how many different guesses there are.
from random import randint
def averager(times):
tests = list()
for _ in range(times):
l = [randint(0,100) for _ in range(100)]
tests.append(len(set(l)))
return sum(tests)/len(tests)
print(averager(100))
For some reason, the number of different guesses averages out to 63.6
Why is this?Is it due to a flaw in the python random library?
In a scenario where people were guessing a number between 1 and 10
The first person has a 100% chance to guess a previously unguessed number
The second person has a 90% chance to guess a previously unguessed number
The third person has a 80% chance to guess a previously unguessed number
and so on...
The average chance of guessing a new number(by my reasoning) is 55%. But the data doesn't reflect this.
-
Shades of (1-1/e)? Note that $1 - 1/e = 0.63212$ Can you see why it has to be this? I thought I had the answer but not sure anymore. Got to look at it some more – user44197 Jan 14 '14 at 3:33
If you want more info about this problem, there's a classical one involving the same maths. It's the odds of having two people(or more) with the same birthday in a classroom. – Feu Jan 14 '14 at 3:42
@Feu an important one in cryptography too :) – Cruncher Jan 14 '14 at 14:16
A related question I asked before. Turns out $e^{-1}$ is a notable probability. Here's an older related question. – badroit Jan 14 '14 at 14:57
From the programming point of view, don't forget that python's basic random functions are not true random, or even cryptographic random. It is a seeded PRNG. If you want better random numbers, try a cryptographic number generator source such as /dev/random (full hardware generated randomness, runs out easily). – Linuxios Jan 14 '14 at 16:40 | {
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Suppose that $n$ guesses are made. For $i=1$ to $100$, let $X_i=1$ if $i$ is not guessed, and let $X_i=0$ otherwise. If $$Y=X_1+X_2+\cdots +X_{100},$$ then $Y$ is the number of numbers not guessed.
By the linearity of expectation, we have $$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_{100}).$$ The probability that $i$ is not chosen in a particular trial is $\frac{99}{100}$, and therefore the probability it is not chosen $n$ times in a row is $\left(\frac{99}{100}\right)^n$. Thus $$E(Y)=100 \left(\frac{99}{100}\right)^n.$$
In particular, let $n=100$. Note that $\left(1-\frac{1}{100}\right)^{100}\approx \frac{1}{e}$, so the expected number not guessed is approximately $\frac{100}{e}$. Thus the expected number guessed is approximately $63.2$, a result very much in line with your simulation.
In general, if $N$ people choose independently and uniformly from a set of $N$ numbers, then the expected number of distinct numbers not chosen is $$N\left(1-\frac{1}{N}\right)^N.$$ Unless $N$ is very tiny, this is approximately $\frac{N}{e}$, and therefore the expected number of distinct numbers chosen is approximately $N-\frac{N}{e}$. Note that the expected proportion of the numbers chosen is almost independent of $N$.
-
I think you are missing a "not": "then the expected number of distinct numbers [not] chosen is" – aaaaaaaaaaaa Jan 14 '14 at 19:19
The expected number not chosen is approximately $\frac{N}{e}$. Since there are $N$ numbers overall, the expected number chosen is approximately $N-\frac{N}{e}$. – André Nicolas Jan 14 '14 at 19:32
Yes, and you wrote the opposite in the second to last paragraph. – aaaaaaaaaaaa Jan 14 '14 at 19:34
Thank you, I finally found where the missing not was (not), had been looking for it towards the very end. Will fix. – André Nicolas Jan 14 '14 at 19:42 | {
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In your 1 to 10 example, it's not true in general that the third chooser has an 80% chance on choosing a new number. It's only the case if the second one has guessed a different number than the first one.
-
This is an example of the Birthday Paradox / Birthday Problem.
Birthday problem - Wikipedia, the free encyclopedia
I was just looking at my Online Cryptography class video lecture today on this very problem.
Coursera.org: crypto-009
There is an apparent paradox that there is more duplication of numbers than expected when the random numbers are supposedly independent.
But the Birthday Paradox is just one example of when our intuitive statistical sense is dead wrong.
-
Here is a cool video from Numberphile about the Birthday problem. youtube.com/watch?v=a2ey9a70yY0 – Rik Jan 14 '14 at 12:04
The birthday problem is about the probability of having at least one collision. It says very little about the number of different values found in a large number of independent samples (as in "how many different birthdays on average in a random group of $365$ people"). – Marc van Leeuwen Jan 14 '14 at 13:27 | {
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# Algorithm To Compute The Gaps Between A Set of Intervals
Problem
Given a set of intervals with possibly non-distinct start and end points, find all maximal gaps. A gap is defined as an interval that does not overlap with any given interval. All endpoints are integers and inclusive.
For example, given the following set of intervals:
$$\{[2,6], [1,9], [12,19]\}$$
The set of all maximal gaps is:
$$\{[10,11]\}$$
For the following set of intervals:
$$\{[2,6], [1,9], [3,12], [18,20]\}$$
The set of all maximal gaps is:
$$\{[13,17]\}$$
because that produces the maximal gap.
Proposed Algorithm
My proposed algorithm (modified approach taken by John L.) to compute these gaps is:
1. Order the intervals by ascending start date.
2. Initialise an empty list gaps that will store gaps
3. Initialise a variable, last_covered_point, to the end point of the first interval.
4. Iterate through all intervals in the sorted order. For each interval [start, end], do the following.
1. If start > last_covered_point + 1, add the gap, [last_covered_point + 1, start - 1] to gaps.
2. Assign max(last_covered_point, end) to last_covered_point.
5. Return gaps
I have tested my algorithm on a few cases and it produces the correct results. But I cannot say with 100% guarantee that it works for every interval permutation and combination. Is there a way to prove this handles every permutation and combination?
• @JohnL. yes the question is correct now - thanks
– Mojo
Dec 18 '20 at 21:25
The key to prove your algorithm is correct is to find enough invariants of the loop, step 4 so that we apply use mathematical induction.
Let $$I_1, I_2, \cdots, I_n$$ denote the sorted intervals. When the algorithm has just finished processing $$I_i$$, we record the values of gaps and last_covered_point as $$\text{gaps}_i$$ and $$\text{last_covered_point}_i$$ respectively.
Let us prove the following proposition, $$P(i)$$, for $$i=1, 2, \cdots, n$$. | {
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Let us prove the following proposition, $$P(i)$$, for $$i=1, 2, \cdots, n$$.
$$\text{gaps}_i$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_i$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$.
When $$i=1$$, $$\text{gaps}_1$$ is the empty set and $$\text{last_covered_point}_1$$ is the right endpoint of $$I_1$$. So $$P(1)$$ is correct.
For the sake of mathematical induction, assume $$P(i)$$ is correct, where $$1\le i\lt n$$. Let $$I_{i+1}=[s, e]$$. There are two cases.
1. If $$s\gt\text{last_covered_point}_i+1$$, then $$\text{gaps}_i\cup[\text{last_covered_point}_i +1, s-1]=\text{gaps}_{i+1}.$$
Let $$m$$ be any point between the start point of $$I_1$$ and the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$. Suppose $$m$$ not covered by any of $$I_1, I_2, \cdots, I_{i+1}$$.
• If $$m\le\text{last_covered_point}_i$$, the induction hypothesis says that $$m$$ is covered by some interval in $$\text{gap}_i$$.
• Otherwise, $$m\gt\text{last_covered_point}_i$$. Since $$m$$ is not covered by $$I_{i+1}$$, we know $$m. So $$m$$ is covered by $$[\text{last_covered_point}_i +1, s-1]$$.
In both cases, $$m$$ is covered by some interval in $$\text{gaps}_{i+1}$$. Since $$\text{last_covered_point}_i$$ is the largest point covered by one of $$I_1, I_2, \cdots, I_i$$ and $$s$$ is the smallest point covered by $$I_{i+1}$$, $$[\text{last_covered_point}_i +1, s-1]$$ is a maximal gap.
2. Otherwise, we have $$s\le\text{last_covered_point}_i$$+1. We can also verify that $$\text{gaps}_{i+1}=\text{gaps}_{i}$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_{i+1}$$.
Finally, since step 4.2 says $$\text{last_covered_point}_{i+1}=\max(\text{last_covered_point}_i, e)$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$, $$\text{last_covered_point}_{i+1}$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$. | {
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So, $$P(i+1)$$ is correct. $$\quad\checkmark$$.
• I probably didn't make my post precise enough and so I think you may have misunderstood what I meant by a gap. I have cleaned up my original post to define the problem better but I also took your example and customised it to what I think the algorithm should be. The main issue I have is that I am not sure if I have covered all possible interval permutations.
– Mojo
Dec 17 '20 at 15:19
• @Mojo, thanks for noticing that last_covered_point should be updated in step 4.2 as well. Dec 18 '20 at 16:30
• @Mojo, I just wrote a proof. The proof is not completely formal, since "maximal gap" is given by a definition (although it is easy to define) and the case 2 is not proved in detail. However, the idea should be clear enough. Dec 20 '20 at 5:24
• In my last comment, 'since "maximal gap" is given by a definition' should have been 'since "maximal gap" is not given by a definition'. Jan 4 at 1:35
I'd like to propose a quite simple algorithm as well. The idea is this: we're going to place open and close parentheses on the number line at the boundaries of each interval. For example, for the intervals $$(1,5), (2,7), (9, 10)$$, the number line would look like this:
1 2 3 4 5 6 7 8 9 10
( ( ) ) ( )
Then we'll just scan left to right, counting parentheses. When all the parentheses get closed, we start a gap. Note in particular that in the above diagram, we have lost the information about which close parenthesis is paired with which open parenthesis -- because it doesn't actually matter. So: | {
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1. Convert each interval $$(a,b)$$ to pairs $$(a,\mathtt o),(b,\mathtt c)$$. ($$\mathtt o$$ and $$\mathtt c$$ are for open and close, respectively.)
2. Sort all the pairs you get from this process. (When sorting, use lexicographic ordering and $$\mathtt o < \mathtt c$$.)
3. Iterate through them, keeping a counter that starts from $$0$$.
• When you see a pair with $$\mathtt o$$ in the second part, increment the counter.
• When you see a pair with $$\mathtt c$$ in the second part, decrement the counter.
• When you decrement the counter, if that causes it to drop to $$0$$, then look at the next element of the list to decide what to do; empty list means you're done, otherwise if the next pair's first part is just one bigger than the current one's you do nothing, and in the last case you record a maximal gap between the current end point and the next open point. | {
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# Why is $x^n\approx \left(n(x^{1/4096}-1)+1\right)^{4096}$?
There's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $\,0.751^{3.2131}$ can be calculated, which is awesome.
This provides endless possibilities, including calculating nth roots on a simple pocket calculator.
The trick goes like this:
1. Type $x$ in the calculator
2. Take the square root twelve times
3. Subtract one
4. Multiply by $n$
6. Raise the number to the 2nd power twelve times (press * and = key eleven times)
Example:
I want to calculate $\sqrt[3]{20}$ which is the same as $20^{1/3}$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:
1. $\;\;\;20$
2. $\;\;\;1.0007315$
3. $\;\;\;0.0007315$
4. $\;\;\;0.0002438$
5. $\;\;\;1.0002438$
6. $\;\;\;2.7136203$
The actual answer is $20^{1/3}\approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.
Question:
So the question is now, why does this trick work? More specifically, how do we prove that: $$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$
Note: $4096=2^{12}$.
I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.
I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:
$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$ | {
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$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$
This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $a\to\infty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.
Thank you for your time :)
• Which one would you like to prove, the approximation for $a = 12$, or just the limit expression for $a \to \infty$? – Xiangxiang Xu Sep 7 '18 at 2:01
• In my opinion the one with the limit with $a\to\infty$ is better to prove because it is more general. From there, the specific case of $a=12$ can be simpler to prove (I think) – KKZiomek Sep 7 '18 at 2:04
A standard trick is to calculate the natural logarithm first to get the exponent under control:
$$\log(\lim_{a\to\infty}(n(x^{1/a}-1)+1)^a)=\lim_{a\to\infty}a\log(nx^{1/a}-n+1)$$ Set $u=1/a$. We get $$\lim_{u\to 0}\frac{\log (nx^u-n+1)}{u}$$ Use L'Hopital: $$\lim_{u\to 0}\frac{nx^u\log x}{nx^u-n+1}=n\log x=\log x^n$$ Here we just plugged in $u=0$ to calculate the limit!
So the original limit goes to $x^n$ as desired.
• I was confused first at L'Hopital step, but then I realized that I have to differentiate with respect to $u$ not $x$ :). I'm giving this answer the best even though all other answers were equally good, because if you haven't mentioned L'Hopital, I would never think of using it. Thank you for the answer. – KKZiomek Sep 7 '18 at 2:19
• Glad to help. It's a neat problem. – Cheerful Parsnip Sep 7 '18 at 2:24 | {
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For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that $$\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n.$$ In fact, we have $$\ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] = \lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x,$$ where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.
If $x$ (actually $\ln x$) is relatively small, then $x^{1/4096}=e^{(\ln x)/4096}\approx1+(\ln x)/4096$, in which case
$$n(x^{1/4096}-1)+1)\approx1+{n\ln x\over4096}$$
If $n$ is also relatively small, then
$$(n(x^{1/4096}-1)+1)^{4096}\approx\left(1+{n\ln x\over4096}\right)^{4096}\approx e^{n\ln x}=x^n$$
Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^{1/3}=2.7144176\ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,
$$\left({1\over3}(20^{1/4096}-1)+1\right)^{4096}=2.7150785662\ldots$$
is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784\ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get
$${2.7136203+2.7152613\over2}=2.7144408$$
which is quite close to the true value! | {
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## Wednesday, 2 December 2020
### On the Colour of the Third Card
Three cards are dealt from a normal deck. You don't see them. You are told that the first two are the same colour, but not what colour they are. What is the probability that the next card is the same colour?
$\frac{1}{2},\;\frac{1}{4},\;\frac{{12}}{{25}},\;or\;\frac{4}{{17}}\;?$
This beautiful, onthefaceofit innocuous little problem comes from the great Martin Gardner's 'Modelling Mathematics with Playing Cards'. It's a problem that never fails to invoke heated discussion and vehement argument when I share it with students to play with. The four solutions and reasoning sketched out below are those typically proffered (and invariably staunchly defended) by students. Which solution would you go with, and why? Do you have a different solution? And what argument would you give to those who firmly hold the solution to be one of the others to show them that they are wrong and you are right? (Note that Gardner's solution was solution 2 [1])
## Solution 1
The last card could either be 1) the same colour by being red, or 2) a different colour by being red, or 3) the same colour by being black, or 4) a different colour by being black. So there are four possibilities and two of these result in the final card being the same colour as the previous two. So:
$\frac{2}{4} = \frac{1}{2}$
## Solution 2
There are eight ways that the arrangement of the colours of the three cards can occur, namely RRR, RRB, RBR, BRR, RBB, BRB, BBR, or BBB (where R = Red and B = Black), and two of these arrangements have the final card as the same colour as the previous two. So:
$\frac{2}{8} = \frac{1}{4}$
## Solution 3
After the first two cards, fifty cards remain in the deck. If the first two cards are both red, there remain twenty-four cards that are red. Similarly, if the first two cards are both black, there remain twenty-four cards that are black. So:
$\frac{24}{50} = \frac{12}{25}$
## Solution 4 | {
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$\frac{24}{50} = \frac{12}{25}$
## Solution 4
For the last card to be the same colour as the first two, the colours of all three cards are either RRR or BBB. You have fifty-two cards to choose from for your first card, and twenty-six of these are red and twenty-six are black. After taking the first card from the deck, you have fifty-one cards left to choose from for your second card, and assuming that the first card was red, twenty-five of the fifty-one cards left are also red. After taking the second card from the deck, you have fifty cards left to choose from for your third and final card, and assuming that the first two cards were red, twenty-four of the fifty cards left are also red. The same would be true if the cards pulled from the deck were black. So:
$\left( {\frac{{26}}{{52}} \times \frac{{25}}{{51}} \times \frac{{24}}{{50}}} \right) \times 2 = \frac{{31200}}{{132600}} = \frac{4}{{17}}$ | {
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Here we need to talk about cardinality of a set, which is basically the size of the set. there are $10$ people with white shirts and $8$ people with red shirts; $4$ people have black shoes and white shirts; $3$ people have black shoes and red shirts; the total number of people with white or red shirts or black shoes is $21$. that the cardinality of a set is the number of elements it contains. Edition 1st Edition. $$|W \cap B|=4$$ In particular, we de ned a nite set to be of size nif and only if it is in bijection with [n]. Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides of students who play both foot ball and cricket = 25, No. In Section 5.1, we defined the cardinality of a finite set $$A$$, denoted by card($$A$$), to be the number of elements in the set $$A$$. Total number of elements related to both B & C. Total number of elements related to both (B & C) only. Two sets are equal if and only if they have precisely the same elements. For example, let A = { -2, 0, 3, 7, 9, 11, 13 }, Here, n(A) stands for cardinality of the set A. To prove there exists a bijection between to sets X and Y, there are 2 ways: 1. find an explicit bijection between the two sets and prove it is bijective (prove it is injective and surjective) 2. Pages 5. eBook ISBN 9780429324819. It turns out we need to distinguish between two types of infinite sets, Any superset of an uncountable set is uncountable. then talk about infinite sets. The cardinality of a set is roughly the number of elements in a set. (Hint: Use a standard calculus function to establish a bijection with R.) 2. Math 127: In nite Cardinality Mary Radcli e 1 De nitions Recall that when we de ned niteness, we used the notion of bijection to de ne the size of a nite set. of students who play | {
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ned niteness, we used the notion of bijection to de ne the size of a nite set. of students who play cricket only = 10, No. where one type is significantly "larger" than the other. The cardinality of the set of all natural numbers is denoted by . $$\>\>\>\>\>\>\>+\sum_{i < j < k}\left|A_i\cap A_j\cap A_k\right|-\ \cdots\ + \left(-1\right)^{n+1} \left|A_1\cap\cdots\cap A_n\right|.$$, $= |W| + |R| + |B|- |W \cap R| - |W \cap B| - |R \cap B| + |W \cap R \cap B|$. ... to make the argument more concrete, here we provide some useful results that help us prove if a set is countable or not. Hence these sets have the same cardinality. The cardinality of a set is denoted by $|A|$. of students who play foot ball only = 28, No. $$|W|=10$$ The intuition behind this theorem is the following: If a set is countable, then any "smaller" set $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable. c) $(0,\infty)$, $\R$ d) $(0,1)$, $\R$ Ex 4.7.4 Show that $\Q$ is countably infinite. If $B \subset A$ and $A$ is countable, by the first part of the theorem $B$ is also a countable the idea of comparing the cardinality of sets based on the nature of functions that can be possibly de ned from one set to another. Ex 4.7.3 Show that the following sets of real numbers have the same cardinality: a) $(0,1)$, $(1, \infty)$ b) $(1,\infty)$, $(0,\infty)$. countable, we can write a proof, we can argue in the following way. So, the total number of students in the group is 100. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. If you are less interested in proofs, you may decide to skip them. The set of all real numbers in the interval (0;1). Before we start developing theorems, let’s get some examples working with the de nition of nite sets. In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. In | {
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of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. In mathematics, a set is a well-defined collection of distinct elements or members. Two finite sets are considered to be of the same size if they have equal numbers of elements. then by removing the elements in the list that are not in $B$, we can obtain a list for $B$, In addition, we say that the empty set has cardinality 0 (or cardinal number 0), and we write $$\text{card}(\emptyset) = 0$$. As far as applied probability Discrete Mathematics and Its Applications, Seventh Edition answers to Chapter 2 - Section 2.5 - Cardinality of Sets - Exercises - Page 176 12 including work step by step written by community members like you. $$|W \cup B \cup R|=21.$$ set whose elements are obtained by multiplying each element of Z by k.) The function f : N !Z de ned by f(n) = ( 1)nbn=2cis a 1-1 corre-spondence between the set of natural numbers and the set of integers (prove it!). The Math Sorcerer 19,653 views. If $A$ has only a finite number of elements, its cardinality is simply the Here is a simple guideline for deciding whether a set is countable or not. the inclusion-exclusion principle we obtain. and how to prove set S is a infinity set. you can never provide a list in the form of $\{a_1, a_2, a_3,\cdots\}$ that contains all the $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$. \mathbb {R}. thus $B$ is countable. Examples of Sets with Equal Cardinalities. This will come in handy, when we consider the cardinality of infinite sets in the next section. 4 CHAPTER 7. Cardinality of a set S, denoted by |S|, is the number of elements of the set. Since each $A_i$ is countable we can Definition. Cantor introduced a new de・］ition for the 窶徭ize窶・of a set which we call cardinality. Textbook Authors: Rosen, Kenneth, ISBN-10: 0073383090, ISBN-13: 978-0-07338-309-5, Publisher: McGraw-Hill Education The difference between the two types is We | {
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978-0-07338-309-5, Publisher: McGraw-Hill Education The difference between the two types is We will say that any sets A and B have the same cardinality, and write jAj= jBj, if A and B can be put into 1-1 correspondence. but "bigger" sets such as $\mathbb{R}$ are called uncountable. You already know how to take the induction step because you know how the case of two sets behaves. \mathbb {N} To formulate this notion of size without reference to the natural numbers, one might declare two finite sets A A A and B B B to have the same cardinality if and only if there exists a bijection A → B A \to B A → B. Cardinality Recall (from our first lecture!) (useful to prove a set is finite) • A set is infinite when there is an injection, f:AÆA, such that f(A) is … When an invertible function from a set to \Z_n where m\in\N is given the cardinality of the set immediately follows from the definition. This important fact is commonly known ... aged to prove that two very different sets are actually the same size—even though we don’t know exactly how big either one is. Thus to prove that a set is finite we have to discover a bijection between the set {0,1,2,…,n-1} to the set. Cardinality The cardinality of a set is roughly the number of elements in a set. By Gove Effinger, Gary L. Mullen. (a) Let S and T be sets. Furthermore, we designate the cardinality of countably infinite sets as ℵ0 ("aleph null"). 12:14. Here we need to talk about cardinality of a set, which is basically the size of the set. In particular, one type is called countable, If A and B are disjoint sets, n(A n B) = 0, n(A u B u C) = n(A) + n(B) + n(C) - n(A n B) - n(B n C) - n(A n C) + n(A n B n C), n(A n B) = 0, n(B n C) = 0, n(A n C) = 0, n(A n B n C) = 0, = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC). Cardinality of a set: Discrete Math: Nov 17, 2019: Proving the Cardinality of 2 finite sets: Discrete Math: Feb 16, 2017: Cardinality of a total order on an | {
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