# Convex set with empty interior is nowhere dense?

Suppose $C\subseteq\mathbb R^n$ is a convex set and $C^o=\varnothing$. Is it necessarily true that $(\overline C)^o=\varnothing$? In general, is this true if $\mathbb R^n$ is replaced by a topological vector space $X$? Or can a counterexample be found?

I know that if $C^o\neq\varnothing$, then $C^o=(\overline C)^o$, so my question is whether this result can be generalized to the case when $C^o=\varnothing$.

Update: It's not true in general topological vector spaces. Indeed, if $X$ is a topological vector space and $Y$ is a proper dense subspace (such examples can be constructed), then $Y$ is convex and has empty interior, but $(\overline Y)^o=X^o=X$ is clearly not empty. Yet, I'm still not sure if the result holds for finite-dimensional vector spaces (in which every proper subspace is closed and thus no proper subspace is dense, so the preceding counterexample doesn't work).

• Seems obvious that a convex subset of $\mathbb R^n$ with empty interior is contained in a hyperplane, which is a nowhere dense closed set. What am I missing? – bof Dec 30 '14 at 20:54

This is true in $\mathbb{R}^n$. Say $n = 3$. If $\overline{C}$ has non empty interior that $C$ is dense in some ball hence it contains the vertices of a tetrahedron (just take a tetrahedron inside the ball and move its vertices slightly to land on a point in $C$). But now by convexity, $C$ must contain the whole solid tetrahedron which contradicts the fact that $C$ has empty interior. In higher dimensions, you can start with a generalized cube and move its vertices slightly to enter $C$.

• I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! – triple_sec Dec 31 '14 at 16:03
• Maybe I'm missing something, but it seems to me that you don't need to resort to "move its vertices slightly". If $C$ is dense in some ball, then $C$ must contain $4$ non-coplanar points (because no subset of a plane can be dense in a ball). Use these points as the vertices of a tetrahedron. – Dave L. Renfro Jan 8 '15 at 17:07
• That works too. – user203787 Jan 8 '15 at 18:00

Here is a rigorous—and, accordingly, a bit lengthy—operationalization of the intuition offered by the answer by @OohAah.

$$\textbf{Proposition}\phantom{---}$$ Let $$C\subseteq \mathbb R^n$$ ($$n\in\mathbb N$$) be a convex set. If $$C$$ has no interior, then nor does $$\overline C$$.

$$\textit{Proof}\phantom{---}$$ Suppose that $$(\overline C)^o\neq\varnothing$$. I will show that $$C^o$$ is not empty, either. Let $$\mathbf x_0\in (\overline C)^o$$. Then, there exists some $$\varepsilon>0$$ such that $$\mathbf x_0\in B(2\varepsilon,\mathbf x_0)\subseteq\overline C$$, where $$B(2\varepsilon,\mathbf x_0)$$ denotes the open ball of radius $$2\varepsilon$$ around $$\mathbf x_0$$ with respect to the Euclidean norm. Let $$\{\mathbf e_i\}_{i=1}^n$$ denote the standard basis of $$\mathbb R^n$$ and define $$\mathbf x_i\equiv \mathbf x_0+\varepsilon \mathbf e_i$$ for each $$i\in\{1,\ldots,n\}$$. Clearly, $$\{\mathbf x_i\}_{i=0}^n\subseteq B(2\varepsilon,\mathbf x_0)\subseteq \overline C.$$ That is, $$\{\mathbf x_i\}_{i=0}^n$$ is included in the closure of $$C$$, so that, for each $$i\in\{0,1,\ldots,n\}$$, there exists some $$\mathbf y_i$$ such that

• $$\mathbf y_i\in C$$; and
• $$\mathbf y_i\in B((2\sqrt n)^{-1}\varepsilon,\mathbf x_i)$$.

For each $$i\in\{1,\ldots,n\}$$ define $$\mathbf b_i\equiv \mathbf y_i-\mathbf y_0$$. I claim that the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent. Indeed, suppose that $$\lambda_1,\ldots,\lambda_n\in\mathbb R$$ satisfy $$\sum_{i=1}^n\lambda_i\mathbf b_i=0.$$ Suppose, for the sake of contradiction, that not all of $$\{\lambda_i\}_{i=1}^n$$ are zero. Then, $$\sum_{i=1}^n|\lambda_i|>0\tag{1}.$$ In addition, \begin{align*} 0=&\,\sum_{i=1}^n\lambda_i\mathbf b_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_i-\mathbf x_0+\mathbf x_0-\mathbf y_0)\\=&\,\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\varepsilon\mathbf e_i+\mathbf x_0-\mathbf y_0), \end{align*} or \begin{align*} -\varepsilon\sum_{i=1}^n\lambda_i\mathbf e_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0). \end{align*} Clearly, the Euclidean norm of the left-hand side is $$\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}$$, so the following chain of inequalities holds true: \begin{align*} &\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}=\left\|\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0)\right\|\leq\sum_{i=1}^n|\lambda_i|\left(\|\mathbf y_i-\mathbf x_i\|+\|\mathbf x_0-\mathbf y_0\|\right)\\ \underset{\text{see (1)}}{<}&\,\sum_{i=1}^n|\lambda_i|\left(\frac{\varepsilon}{2\sqrt{n}}+\frac{\varepsilon}{2\sqrt{n}}\right)=\sum_{i=1}^n|\lambda_i|\frac{\varepsilon}{\sqrt{n}}\leq\sqrt{\sum_{i=1}^n|\lambda_i|^2}\sqrt{\sum_{i=1}^n\frac{\varepsilon^2}{n}}=\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}, \end{align*} where I used the Cauchy–Schwarz inequality. This is a contradiction, which implies that $$\lambda_1=\ldots=\lambda_n=0$$. Hence, the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent, and since $$\dim\mathbb R^n=n$$, it follows also that they constitute a basis.

If $$\mathbf z\in\mathbb R^n$$, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z=\sum_{i=1}^n\mu_i\mathbf b_i$$. Define $$\|\mathbf z\|_b\equiv\sum_{i=1}^n|\mu_i|$$. Then, $$\|\cdot\|_b$$ is a norm and since $$\mathbb R^n$$ is finite-dimensional, it must be equivalent to the Euclidean norm. Therefore, there exists some $$\xi_b>0$$ such that $$\|\cdot\|_b\leq \xi_b\|\cdot\|$$.

Let $$D\equiv\left\{\sum_{i=0}^n\alpha_i\mathbf y_i\,\Bigg|\,\alpha_i\geq0\text{ for all i\in\{0,1,\ldots,n\} and }\sum_{i=0}^n\alpha_i=1\right\}.$$ Since $$\{\mathbf y_i\}_{i=0}^n\subseteq C$$ and $$C$$ is convex, it follows that $$D\subseteq C$$. Define $$\mathbf w\equiv\sum_{i=0}^n\frac{1}{n+1}\mathbf y_i.$$ Clearly, $$\mathbf w\in D$$ and $$\mathbf w-\mathbf y_0=\sum_{i=1}^{n}\frac{1}{n+1}(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^{n}\frac{1}{n+1}\mathbf b_i.$$

Let $$\delta\equiv\frac{1}{n(n+1)\xi_b}>0.$$ I will show that $$B(\delta,\mathbf w)\subseteq D$$. To this end, pick any $$\mathbf z\in B(\delta,\mathbf w)$$, so that $$\|\mathbf z-\mathbf w\|<\delta.$$ Since $$\{\mathbf b_i\}_{i=1}^n$$ is a basis, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z-\mathbf y_0=\sum_{i=1}^n\mu_i\mathbf b_i$$. Then, $$\mathbf z-\mathbf w=(\mathbf z-\mathbf y_0)-(\mathbf w-\mathbf y_0)=\sum_{i=1}^n\left(\mu_i-\frac{1}{n+1}\right)\mathbf b_i.$$ Consequently, for any $$i\in\{1,\ldots,n\}$$, it follows that \begin{align*} \left|\mu_i-\frac{1}{n+1}\right|\leq\sum_{j=1}^n\left|\mu_j-\frac{1}{n+1}\right|=\|\mathbf z-\mathbf w\|_b\leq \xi_b\|\mathbf z-\mathbf w\|<\xi_b\delta=\frac{1}{n(n+1)}. \end{align*} Hence, $$0\leq\frac{(n-1)}{n(n+1)}=\frac{1}{n+1}-\frac{1}{n(n+1)}<\mu_i<\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n},$$ for each $$i\in\{1,\ldots,n\}$$ so that $$\sum_{i=1}^n\mu_i<1$$. It follows that \begin{align*} \mathbf z=\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf b_i=\mathbf y_0+\sum_{i=1}^n\mu_i(\mathbf y_i-\mathbf y_0)=\left(1-\sum_{i=1}^n\mu_i\right)\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf y_i, \end{align*} so $$\mathbf z\in D$$.

Conclusion:

• $$\mathbf w\in B(\delta,\mathbf w)\subseteq D\subseteq C$$; and
• $$B(\delta,\mathbf w)$$ is a non-empty open set; so that
• $$\mathbf w\in C^o$$.

In particular, $$C^o$$ is not empty. $$\blacksquare$$

Intuitively, the points $$\{\mathbf x_i\}_{i=0}^n$$ in $$\overline C$$ span an $$n$$-dimensional simplex, whose vertices are perturbed slightly so that the new vertices do not lie in the same hyperplane and are all in $$C$$. The convex hull of these new vertices gives rise to a “distorted simplex” fully contained in $$C$$. Then, the centroid of this distorted simplex can be surrounded by a small ball still included in the distorted simplex, and hence in $$C$$.


Suppose that ##A## is a diagonalizable ## n \times n ## matrix. Then it is similar to a diagonal matrix ##B##. My question is, is ##B## the only diagonal matrix to which ##A## is similar?

I have thought about this, but am unsure if my answer is correct. My claim is that ##B## is the only diagonal matrix to which ##A## is similar, up to a rearrangement (or permutation) of the eigenvectors of ##A## in an ordered eigenbasis for ##ℝ^{n}##. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_{A}]_{β}## but this will be merely a rearrangement of the diagonal entries of B.

Is this true? If so, how might I go about proving this conjecture of mine?

BiP

Related Linear and Abstract Algebra News on Phys.org
Yes, it's true.

A hint for the proof: the diagonal entries are the eigenvalues of ##A##.

CompuChip
Homework Helper
If ##A = X^{-1}BX = Y^{-1}B'Y## then ##B = (Y X^{-1})^{-1} B' (Y X^{-1})##.
So if A is similar to two diagonal matrices, then these matrices are similar to each other.
I guess a starting point would be to check which similarity transformations keep diagonal matrices diagonal, and see what you can say about X and Y that way.

HallsofIvy
Homework Helper
Yes, it's true.

A hint for the proof: the diagonal entries are the eigenvalues of ##A##.
Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.

Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.
That's exactly what the OP said:

Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_A]_\beta## but this will be merely a rearrangement of the diagonal entries of B.

mathwonk


# Fourier Transform Of Triangular Pulse

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The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the spectrum-shaped pulse. Amplitude-modulated optical pulse trains undergo the discrete Fourier transform (DFT) realized by the temporal Talbot effect in a dispersive fiber. Fourier series: Given a 2ˇ-periodic complex-valued function f (think of it as a function on an interval T of length 2ˇ), its Fourier (series) transform is the sequence of its Fourier coecients: (f) = fb= fck: k 2 Zg, which can be thought of as a complex-valued function of the discrete frequency variable k. The series produced is then called a half range Fourier series. A square wave is approximated by the sum of harmonics. The S-transform can be effectively utilized for the analysis of nonstationary data. Output kernel Figure 5. The Gaussian shape is often called a bell shape. I'm at a computer without MATLAB at the moment. d R pp T P, where R pp is the auto-correlation of p(t) d. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Still, many problems that could have been tackled by using Fourier transforms may have gone unsolved because they require integration that is difficult and tedious. T T 0 elsewhen. There are similar convolution theorems for inverse Fourier transforms. 2 p693 PYKC 10-Feb-08 E2. The Fourier Transform 1. That is: since the is an odd function, its contribution is zero. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. DSP First and it's accompanying digital assets are the result of more than 20 years of work that originated from, and was guided by, the premise that signal processing is the best starting point for the study of electrical and computer engineering. time signal. School of Health Information Sciences. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). Signal Distortion over a Communication Channel. The Fourier series of this general pulse train is:. Equation of the Day #11: The Fourier Transform. Except now we're going to build a composite wave form that is a triangle wave. (We shall use this in the assignment). In order for F(t) to be real, F (- t) = F* ( t) must hold, Example 9. Plotting a triangular signal and finding its Fourier transformation in MATLAB. A “Brief” Introduction to the Fourier Transform This document is an introduction to the Fourier transform. What is the FT of a triangle function? - To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. time signal. This is a good point to illustrate a property of transform pairs. 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Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Realisability of One Port Network – Consider the network function H(s) which is the ratio of Laplace transform of the output R(s) to the Laplace transform of the excitation E(s). The line spectrum, obtained from the Fourier series coefficients, indicates how the power of the signal is distributed to harmonic frequency components in the series. "Fourier Series--Triangle Wave. The Gaussian shape is often called a bell shape. Click the button "Real Transform" in the Fourier graph. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. We then generalise that discussion to consider the Fourier transform. (a) below is X(Ω) = sin Ω 2 Ω 2!2. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. A group of algorithms is presented generalizing the fast Fourier transform to the case of noninteger frequencies and nonequispaced nodes on the interval $[ - \pi ,\pi ]$. School of Health Information Sciences. Consequently, the square wave has a wider bandwidth. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. Sometimes there is a big spike at zero so try taking the log of it before plotting. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. Signals & Systems - Triangular Signal Watch more videos at https://www. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. 2 The Fourier Transform Having now presented the idea of the frequency response H(s)js=j! = H(j!) of linear systems, it is appropriate to recognise that in the special case of this Laplace transform value H(s)being evaluated at the purely imaginary value s = j!, it is known as the ‘Fourier transform’. x(t)= γ (t)-γ (t-T 0) a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T 0 that is not centered on the origin). , sin(x)/x] in the frequency domain. Consequently, the square wave has a wider bandwidth. Improving images by “ deconvolution ” of. In this chapter much of the emphasis is on Fourier Series because an understanding of the Fourier Series decomposition of a signal is important if you wish to go on and study other spectral techniques. FFT onlyneeds Nlog 2 (N). Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. The FT of a rectangle function is a sinc. tutorialspoint. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The Fourier Transform of the Box Function. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Triangle Pulse 0 0. periodic relationship, 222 discrete time Fourier series (DTFT),. Exams are approaching, and I'm working through some old assignments. The transform of a triangular pulse is a sinc2 function. Try taking the real part of it with real(). 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. If a function is defined over half the range, say 0 to L, instead of the full range from -L to L, it may be expanded in a series of sine terms only or of cosine terms only. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. In this particular SPICE simulation, I’ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series for a total of five AC voltage sources. Relation of Z transform to Laplace transform. 2005-03-15. We say that f(t) lives in the time domain, and F(ω) lives in the frequency domain. We begin by discussing Fourier series. On the other hand, as Fourier transform can be considered as a special case of Laplace transform when the real part of the complex argument is zero:. Baskakov, O I; Civis, S; Kawaguchi, K. After simplification the sinc squared function is obtained as the Fourier transform of a triangular pulse with unit area. 30) yields 2 2 ( ) ˆ p p T T. FFTs of Functions. '' Figure 8 shows an example for. Discrete Fourier Transform of Windowing Functions. )2 Solutions to Optional Problems S9. More Advanced Topics Up: Fourier Series-What, How, and Why Previous: The Fast Fourier Transform Using the Fourier Transform. Fourier Transform of Useful Functions. Since f(t) is even then g(w) is real. For any parameter, ≠. It is reversible, being able to transform from either domain to the other. You can buy my book 'ECE for GATE' here https://goo. We then generalise that discussion to consider the Fourier transform. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. The Fourier transform of the triangular pulse shown in Fig. The curve should be symmetrical with respect to the origin in 1024 points. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of a triangular pulse is a sinc 2 function. Compare the result with pan (b). Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T 0 where it has a height of one:. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. Fourier Transform of the Gaussian Konstantinos G. The reason why Fourier analysis is so important in physics is that many (although certainly. Homework Statement What is the Fourier transform of the function graphed below? According to some textbooks the Fourier transform for this function Fourier Transform (Triangular Pulse) | Physics Forums. Cvetkovic, IntechOpen, DOI: 10. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. The fast Fourier transform (FFT) is a fast algorithm for calculating the Discrete Fourier Transform (DFT). A major challenge in the data analysis process of NOM using the FT-ICR MS technique is the need to sort the entire data set and to present it in an accessible mode. Often we are confronted with the need to generate simple, standard signals ( sine, cosine , Gaussian pulse , squarewave , isolated rectangular pulse , exponential decay, chirp signal ) for. 1 The Fourier transform and series of basic signals Triangular pulse 1. The graph at the bottom of the screen is the Fourier transform of this pulse. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. (a) below is X(Ω) = sin Ω 2 Ω 2!2. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. This is the principle on which a pulse Fourier transform spectrometer operates. Discrete-Time Fourier Transform (DTFT) Chapter Intended Learning Outcomes: (i) Understanding the characteristics and properties of DTFT (ii) Ability to perform discrete-time signal conversion between the time and frequency domains using DTFT and inverse DTFT. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Sometimes there is a big spike at zero so try taking the log of it before plotting. In the synthesis, we are going to obtain a network from the given network function which may be admittance function or impedance function. Fourier Transforms For additional information, see the classic book The Fourier Transform and its Applications by Ronald N. Sometimes fft gives a complex result. rectangular pulse triangular pulse periodic time function unit impulse train (model of regular sampling). The example in this figure pertains to an output sampling rate which is times that of the input signal. Recall that normalized Fourier transform of triangular pulse is [math]sinc^{2}(f)$[math]. The Fourier transform is an integral transform widely used in physics and engineering. In this article, we will discuss the fact that choice of different window functions involves a trade-off between the main lobe width and the peak sidelobe (PSL). Improving images by “ deconvolution ” of. How to get a triangular pulse? Discrete Fourier Transform using scipy. 4 Reconstruction of a triangular pulse. What do we hope to achieve with the Fourier Transform? We desire a measure of the frequencies present in a wave. Can you explain this answer? is done on EduRev Study Group by Electrical Engineering (EE) Students. In some cases, as in this one, the property simplifies things. Triangle Pulse Sinc Pulse. What is the FT of a triangle function? – To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. DAS RISIN and MAIZAN MUHAMAD. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. A Tables of Fourier Series and Transform Properties 321 Table B. periodic relationship, 222 discrete time Fourier series (DTFT),. The Fourier transform of this convolution is the product w 1 w 2 of the two transforms, each one a series of parallel walls, and differs from zero only when both factors are different from zero. We demonstrate both amplitude and phase tailoring by generating a picosecond squarelike pulse as well as trains of femtosecond pulses with a terahertz-range repetition rate from either a. The LCFBG in the. Digital signal processing (DSP) vs. For the bottom panel, we expanded the period to T=5, keeping the pulse's duration fixed at 0. Relation of Z transform to Laplace transform. Ideal and Practical Filters. Sibbett A Wollaston prism is used in the design of a polarizing Fourier. The impulse response for the triangle in Fig. The graph at the bottom of the screen is the Fourier transform of this pulse. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. Signals & Systems - Triangular Signal Watch more videos at https://www. Exams are approaching, and I'm working through some old assignments. Signal Distortion over a Communication Channel. Fourier transform theorem table 4 1 jpg bax blog fourier transform table rh baxtyfraze blo com Pdf Fourier Transforms And Their Application To Pulse Amplitude. I One-Dimensional Fourier Transform. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Slow Fourier Transforms Consider a general 1D Fourier transform relating two vectors of length : contains the values in real-space contains the frequency components The Slow Fourier Transform : Suppose that we precompute and store the coefficients c(j,k) = exp(i 2pi jk/n). I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. I'm at a computer without MATLAB at the moment. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. Signal Energy and Energy Spectral Density. Minimalist Tall Rectangular Planter On 4 Or 6 Zebra Onion Grass. F(ω) is just another way of looking at a function or wave. A LABORATORY DEMONSTRATION OF HIGH-RESOLUTION HARD X-RAY AND GAMMA-RAY IMAGING USING FOURIER-TRANSFORM TECHNIQUES David Palmer and Thomas A. A two parts tutorial on Fourier series. For voltage signals, the power per unit frequency is proportional to |H(f)|2 and is called the power spectrum or spectral power density of h(t). I One-Dimensional Fourier Transform. Sampling theorem. 1 Fourier Series Representation of Periodic Signals 3. 900 950 1000 1050 1100 1150 0. It is reversible, being able to transform from either domain to the other. Then using the relation: one can show that. PDF | Fourier transform ultrashort optical pulse shaping using a single linearly chirped fiber Bragg grating (LCFBG) is proposed and experimentally demonstrated in this letter. Solution: g(t) is a triangular pulse of height A, width W, and is centered at t 0. Therefore the FT of a triangle function is the product of two identical sincs, or a sinc^2. 4*Pi); # The base function is f0 = f restricted to [0,T] f0:=x->x; # Then f(x)=f0(trw(x,T))=trw(x,T) # INTEGRATION STEPS FOR FOURIER COEFFICIENTS # Possible also by hand using an integral table. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Note that as long as the definition of the pulse function is only motivated by the time-domain experience of it, there is no reason to believe that the oscillatory interpretation (i. Cosine waves c. Plotting a triangular signal and finding its Fourier transformation in MATLAB. The Fourier Transform In this appendix, the most important facts pertaining to the Fourier trans form are surveyed without proof. NEW! Updated labs, visual demos, an update to the existing chapters, and hundreds of new homework problems and solutions. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. Properties of the Fourier Transform • Linearity: • Let and • then • Time Scaling: • Let • then Compression in the time domain results in expansion in the frequency domain Internet channel A can transmit 100k pulse/sec and channel B can transmit 200k pulse/sec. Signals & Systems - Triangular Signal Watch more videos at https://www. Variable position. Since the pulse is getting narrower and narrower in the limit, the multiplication with this other function has less and less effect, until it has no effect once the. This is the example given above. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform. Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. t2 exp( 2 ) 2. As is an even function, its Fourier transform is Alternatively, as the triangle function is the convolution of two square functions ( ), its Fourier transform can be more conveniently obtained according to the convolution theorem as:. Periodicity of the Fourier transform; Fourier transform as additive synthesis. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the. Note that as long as the definition of the pulse function is only motivated by its behavior in the time-domain experience, there is no reason to believe that the oscillatory interpretation (i. Another representation of signals that has been found very useful is frequency domain representation. We can sample a function and then take the FFT to see the function in the frequency domain. The triangle peak is at the integral of the signal or sum of the sequence squared. ), the frequency response of the interpolation is given by the Fourier transform, which yields a sinc function. Single-pulse, Fourier-transform spectrometer having no moving parts M. Fourier transform: F(s) = - f(x)e-'2n"dr Here, two sidebands have been introduced (in addition to the carrier frequency), and the signal of frequency s has been shifted into the region of the carrier frequency, so. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). Now, if we're given the wave function when t=0, φ(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute φ(x,t) for any t by taking the inverse Fourier transform of φ(x,0) conducting a phase shift, and then taking the Fourier transform. profile closer to Gaussian. DFT needs N2 multiplications. Of course, we must sample often enough to avoid losing content. P d P2 f , where is Fourier Transform of p(t) c. 5 The multiplication Property 4. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first row is the graph of the unit pulse function and its Fourier transform , a function of frequency. For example, a rectangular pulse in the time domain coincides with a sinc function [i. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. The actual Fourier transform are only the impulses. As in the case of ideal sampling, the spectrum contains uniformly spaced (scaled) copies of 𝑋(𝑓), with a spacing of 1⁄𝑇 Hz. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A˝sinc! ˝ 2 Thus u(t j!˝˝) !F e A˝sinc! ˝ 2 Therefore u(t a)u (t b) !F ej!a A˝sinc! ˝ 2. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Cosine waves c. Moreover, the amplitude of cosine waves of wavenumber in this superposition is the cosine Fourier transform of the pulse shape, evaluated at wavenumber. Mapping s plane semiellipse to Z plane. The corresponding intensity is proportional to this transform squared, i. To do so, you just have to divide the pulse by its norm, i. 3 Properties of the Continuous-Time Fourier Transform 4. Prince California Institute of Technology, Pasadena, CA 91125 Abstract A laboratory imaging system has been developed to study the use of Fourier-transform techniques in high-resolution hard x-ray andγ. The dotted-line is a sinc function that doesn't apply to this question, but gives the notion that this transform has something to do with the transform of a square pulse (i. Ideal and Practical Filters. F(ω) is called the Fourier Transform of f(t). Plot sine wave. 4 The Convolution Property 4. Today I want to follow up by discussing one of the ways in which reality confounds our expectations and causes confusion. Fn = 6 shows that as T/t increases the lines get closer together and the spectrum begins to look like that of the Fourier Transform. They are widely used in signal analysis and are well-equipped to solve certain partial. Signal Distortion over a Communication Channel. profile closer to Gaussian. periodic relationship, 222 discrete time Fourier series (DTFT),. Fourier series and square wave approximation Fourier series is one of the most intriguing series I have met so far in mathematics. The actual Fourier transform are only the impulses. The primary reason for it’s success is that it plots any course of constant bearing (angle w. I would guess it would be a triangle because I think the transform of a triangular pulse centered at zero is something like T sinc 2 (pi * f * T) I would restate your original question as follows: What is the transform of the following periodic rectangular function: per-rect(t) which has period T=T1+T2; is 1 during each T1 duration of its. 5 Signals & Linear Systems Lecture 10 Slide 3 Connection between Fourier Transform and Laplace. 1 The Fourier transform and series of basic signals Triangular pulse 1. 6: Roll-off of the rectangular-window Fourier transform. We describe a pulse-shaping technique that uses second-harmonic generation with Fourier synthetic quasi-phase-matching gratings. time signal. In the case of natural sampling, however, the spectral copies have different scaling factors: 𝑃(𝑘⁄𝑇)/𝑇. For above triangular wave: The square wave has much sharper transition than the triangular wave. These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. PSfrag replacements PROBLEM: Let x. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Inverse Z transform using inversion integral. Click the button "Real Transform" in the Fourier graph. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. Fourier Transforming the Triangular Pulse. Fourier Slice Theorem or central Slice Theorem. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. , use ^ M(n)/k^ M kinstead of ^ M(n). Then i deconvoluted the convoluted signal using triangular pulse with the deconvolution tool given in labview. 1-1) Since u(t) = 0 for t < 0, eq. Signal Transmission Through LTIC Systems. 2to create successive recon-structions of the pulse. Fourier transforms are used widely, and are of particular value in the analysis of single functions and combinations of functions found in radar and signal processing. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. Z transforms and difference equations. Time-harmonic impulse response calculations—The time-harmonic pressure generated by these triangular source geometries is proportional to the Fourier transform of the impulse response. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. A "Brief" Introduction to the Fourier Transform This document is an introduction to the Fourier transform. tutorialspoint. It is reversible, being able to transform from either domain to the other. Inverse Z transform using inversion integral. See Discrete Fourier Transform discrete vs. ∆(t), from Problem 3. The first example has a duty cycle of 0. The Fourier Transform: Examples, Properties, Common Pairs Square Pulse Spatial Domain Frequency Domain f(t) F (u ) 1 if a=2 t a=2 0 otherwise sinc (a u ) = sin (a u ) a u The Fourier Transform: Examples, Properties, Common Pairs Square Pulse The Fourier Transform: Examples, Properties, Common Pairs Triangle Spatial Domain Frequency Domain f(t. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Sometimes fft gives a complex result. Padgett, A. But what we're going to do in this case is we're going to add them. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. Applying the time-convolution property to y(t)=x(t) * h(t), we get: That is: the Fourier Transform of the system impulse response is the system Frequency Response L7. property, we can compute the Fourier transform of the phasor: 0 ( )0 F e f fj tω = −δ This allows us to compute the Fourier transform of periodic signals. Triangular waves d. This is the example given above. (Browser settings for best viewing results) (How to cite this work). We then generalise that discussion to consider the Fourier transform. Fourier Transforms and Theorems. A) shows the original pulse and the real part of its Fourier transform. This transform pair isn't as important as the reason it is true. " From MathWorld--A Wolfram Web Resource. t/ D X1 nD1. In mathematics, the continuous Fourier transform is one of the specific forms of Fourier analysis. 003 EECS MIT discrete operator transforms. 6 depicts a resistor and capacitor in series. Using the above, we obtain: [ ] 2 0 ( ) ( )0 j nf t n. the Fourier transform function) should be intuitive, or directly understood by humans. One imagines a delta function to be a square pulse of unit area in the limit as the base of the pulse becomes narrower and narrower and higher and higher. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 22 Summary • The Fourier Series can be formulated in terms of complex exponentials – Allows convenient mathematical form – Introduces concept of positive and negative frequencies • The Fourier Series coefficients can be expressed in terms of magnitude and phase. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Exams are approaching, and I'm working through some old assignments. The magnitude of the Fourier transform of a rectangular pulse equals the absolute value of a sinc. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. Existence of the Fourier Transform. 104 Chapter 5. jaj<1 (n+ 1)anu[n] 1 (1 ae j. In the second row is shown , a delayed unit pulse, beside the real and imaginary parts of the Fourier transform. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). The fundamental frequency is 50 Hz and each harmonic is, of course, an integer multiple of that frequency. Four chapters on analog signal processing systems, plus many updates and enhancements. This is a good point to illustrate a property of transform pairs. In other words, the input signal is upsampled by a factor of using linear interpolation.


Question

# (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.

Solution

## Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \geq 0,y \geq 0$$. We make the following table from the given data:ResourcesFoodI$$(x)$$FoodII$$(y)$$RequirementVitamins A (units/ kg)$$2$$$$1$$$$8$$Vitaminc C (units/ kg)$$1$$$$2$$$$10$$Cost (Rs/ kg)$$50$$$$70$$Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:$$2x + y \geq 8$$$$x + 2y \geq 10$$Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is$$Z = 50x 70y$$Hence, the mathematical formulation of the problem is:Minimise $$Z = 50x + 70y ... (1)$$subject to the constraints :$$2x + y \geq 8 ... (2)$$$$x + 2y \geq 10 ... (3)$$$$x, y \geq 0 ... (4)$$Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.Corner Point$$Z = 50x + 70y$$$$(0, 8)$$$$560$$$$(2, 4)$$$$380\leftarrow Minimum$$$$(10, 0)$$$$500$$In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.Mathematics

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# Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$?

Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?

• @B.Goddard $a, b$ are assumed to be relatively prime. Also, the linear combinations here are assumed to have non-negative coefficients. That changes the problem quite a bit from the one you describe. – Arthur Jun 20 '18 at 13:11
• The most explicit I can get is a summation. If $a\lt b$ then there are: $$\sum_{k=1}^{a-1}\left\lfloor\frac{bk}{a}\right\rfloor$$ such numbers. Although you should definitely double-check this result. – N. Shales Jun 20 '18 at 16:19
• Having thought a bit more I further conjecture that the result simplifies to $$\frac{1}{2}(a-1)(b-1)\, .$$ – N. Shales Jun 20 '18 at 17:16
• @N.Shales your conjecture is correct: \begin{align*} R &= \sum_{k=1}^{a-1} \left\lfloor \frac{bk}{a} \right \rfloor\\ &= \sum_{k=1}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b+\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b-1-\left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - R \end{align*} So that $2R = (a-1)(b-1)$. The key is the equality $\lfloor (-kb)/a\rfloor = -\lfloor kb/a \rfloor - 1$ which holds when $a$ does not divide $kb$. – Yong Hao Ng Jun 28 '18 at 8:20
• @N.Shales I saw it a few days ago but it took me a while before I found a not very involved solution to the floor-sum. – Yong Hao Ng Jun 29 '18 at 13:11

Let $N = ab$, we count the number of integers $0\leq t \leq N$ that can be written as $$t = ma+nb$$ such that $m,n\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them.
If $n\geq a$, then we replace it with $$ma+nb = (m+b)a + (n-a)b = m'a+n'b$$ So with repeated application we may assume that $0\leq n < a$. We require $$0 \leq ma+nb \leq N \implies 0 \leq ma \leq N-nb = (a-n)b$$ Hence we get the bounds of $m$ as $$0 \leq m \leq \left \lfloor \frac{(a-n)b}{a}\right \rfloor$$ This means the number of integers we can represent for a given $n$ is $\left \lfloor \frac{(a-n)b}{a} \right \rfloor + 1$. Now summing this over all possibilities $n=0,1,\dots a-1$, the number of representable integers $R$ is \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \left\lfloor \frac{(a)b}{a} \right\rfloor + \left\lfloor \frac{(a-1)b}{a} \right\rfloor + \cdots + \left\lfloor \frac{(2)b}{a} \right\rfloor + \left\lfloor \frac{(1)b}{a} \right\rfloor + a\\ R &= a + b + \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ \end{align*} Next we observe that if $a$ does not divide $kb$, then for some integer $t$ $$t < \frac{kb}{a} < t+1 \Longleftrightarrow -t-1 < \frac{-kb}{a} < -t$$ In particular this applies for $1\leq k\leq a-1$, since $\gcd(a,b)=1$ and $a>1$. We use the inequalities to obtain $$\left \lfloor \frac{-kb}{a}\right \rfloor = -t-1 = - \left \lfloor \frac{kb}{a}\right \rfloor -1,$$ This lets us get another equation of $R$ as \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor + b + 1 \\ &= a(b+1) + \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor\\ &= a(b+1) + \sum_{k=1}^{a-1}\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= a(b+1) + \sum_{k=1}^{a-1}-\left\lfloor \frac{kb}{a} \right\rfloor -1\\ R &= ab + 1 - \sum_{k=1}^{a-1}\left\lfloor \frac{kb}{a} \right\rfloor \end{align*} Therefore adding both equations of $R$, we have $$2R = ab + a + b + 1 = (a+1)(b+1)$$ giving us $R=(ab+a+b+1)/2$. Notice that $R$ is not an integer if and only if $a,b$ are both even, which is not possible since $\gcd(a,b)=1$.
Finally, the number of integers considered is $ab+1$, so the number of integers not representable is $$ab+1 - \frac{ab+a+b+1}{2} = \frac{ab -a - b+1}{2} = \frac{(a-1)(b-1)}{2}$$
Proposition. Let $a,b$ be relatively prime integers greater than $1,$ and let $m=ab-a-b.$
If $x\in\{0,1,2,\dots,m\},$ then exactly one of the two numbers $x$ and $m-x$ can be expressed as a linear combination of $a$ and $b$ with nonnegative integer coefficients.
It follows that exactly half the elements of $\{0,1,2,\dots,m\}$ can be expressed as nonnegative integral linear combinations of $a$ and $b;$ that is, there are exactly $\frac{m+1}2=\frac{(a-1)(b-1)}2$ numbers which cannot be so expressed.


# How many answers to $|3^x-2^y|=5$?

How many answers are there to the equation $|3^x-2^y|=5$ such that $x$ and $y$ are positive integers? Are there infinite? I've found $(2,2)$, $(3,5)$, and $(1,3)$. It seems to explode with larger values, but it's not a steady increase and there seems to be many dips. Do we KNOW that there are no large values for $x$ and $y$ where a power of 3 comes close to a power of 2?

• Its a putnam problem . – DeepSea Feb 5 '16 at 23:12
• Kf-Sansoo, had to google "putnam problem." Do you mean literally one, or one that fits the general qualifications? Do I need to add a tag? If it is one that's been mentioned before, could you give a link? – Elem-Teach-w-Bach-n-Math-Ed Feb 5 '16 at 23:18
• – Lucian Feb 6 '16 at 0:01
• @Kf-Sansoo: so what? Does that mean the question shouldn't be discussed on MSE? Does it mean that there is helpful information about the question to be found elsewhere? Or what? – Rob Arthan Feb 6 '16 at 0:01
• @Lucian: So, as it's conjectured that $Ax^n-By^m=C$ has a finite # of solutions for $(x,y,n,m)$ then $3^n-2^m=5$ can be conjectured to also have a finite # or solutions for $(n,m)$? In other words, it's unproven in a more general sense, so with the greater constraints it should be even less likely? If there are finite answers, my question could then be rephrased, "Is there a 4th answer?" – Elem-Teach-w-Bach-n-Math-Ed Feb 6 '16 at 0:18

There are only a finite number of solutions. It was proved by Pillai that $a^x - b^y = k$ where $a,b,k$ are fixed positive integers, $a > 1, b > 1, k \neq 0,$ with positive integer variables $x,y,$ has finitely many solutions. This is from page 51 in Shorey and Tijdeman, Exponential Diophantine Equations. The two papers by Pillai are 1931 and 1936. Both are in the Journal of the Indian Mathematical Society. A detail: if $k$ is larger than some bound that depends on $a,b,$ there is only one solution. Since we have more than one solution for $k = -5,$ it appears Pillai's bound is not tight enough to finish this problem. We just know one solution for $k=5.$

• I'll probably end up going with this as best answer. I've got an addendum though for your perusal, if I may. Since $|3^x-2^y|=5$ is proven to have finite solutions, I'd like to carry this knowledge to similar problems: i.e. $|(3^x)(5^y)-2^z|=7$, $|(2^x)(3^y)-5^z|=7$, $|(2^x)(3^y)-5^z|=7$. Namely working my way up to prime solutions such as $|(2^r)(7^s)(11^t)(13^u)(23^v)-(3^w)(5^x)(17^y)(19^z)|=29$ (or similarly creating primes less than the square of the largest prime used; in this case 23). Any thoughts? Probably throw this out there as a new ?. – Elem-Teach-w-Bach-n-Math-Ed Feb 8 '16 at 22:39

Here's an elementary self-contained argument that there is no solution with $y>5$.

A power of $3$ is congruent to either $1$ or $3 \bmod 8$, so once $y \geq 3$ we must have $3^x - 2^y = -5$.

Once $y \geq 6$, we then have $3^x \equiv -5 \bmod 2^6$, and thus $x \equiv 11 \bmod 16$.

But then $3^x + 5 \equiv 12 \bmod 17$, and no power of $2$ is congruent to $12 \bmod 17$ (the powers of $2 \bmod 17$ are $2,4,8,-1,-2,-4,-8,1,2,4,8,-1$ etc.), QED.

There is a large literature on such Diophantine questions. One key phrase is "$S$-unit equations". In general it has been known for some time that there are finitely many solutions, and indeed for equations of the form $$\prod_i p_i^{n_i} - \prod_j q_j^{m_j} = r$$ this already follows from Thue's theorem (1909); and by now we even have effective algorithms known to find all solutions. There's still no elementary technique known in general, but in your case (where only the primes 2,3,5) appear an elementary solution is contained in a 1976 paper
This is not a complete answer, but I'd just like to note a connection with the concept of irrationality measure. First recall that $\alpha \in \mathbf{R}$ has irrationality measure $\mu(\alpha)$ if, for all $r>\mu(\alpha)$, there are only finitely many pairs of integers $p,q$ such that $$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^r}.$$ Note that $\mu(\alpha)=\infty$ is allowed (but we have $\mu(\alpha)=2$ for all but a measure-zero set of reals $\alpha$).
With this definition out of the way, note that $$\left| 3^x - 2^y \right| = 5$$ implies $$3^x = 2^y \pm 5.$$ Take logs: $$\log(3^x) = \log(2^y) + \log(1 \pm 5 \cdot 2^{-y}).$$ So by the Taylor expansion of $\log(1+t)$, $$\left| \log(3^x) - \log(2^y) \right| < 5 \cdot 2^{-y}+\frac{5^2 \cdot 2^{-2y}}{2}+\cdots < 6 \cdot 2^{-y}$$ for $y$ sufficiently large. Hence, using $\log(3^x) = x \log 3$ and $\log(2^y) = y \log 2$ $$\left| \frac{\log 3}{\log 2} - \frac{y}{x} \right| < \frac{6}{x \log 2} \cdot 2^{-y}.$$ Since the RHS decreases much faster than anything polynomial in $x$ (at least for $y/x$ within some fixed small interval around $\log(3)/\log(2)$), the existence of infinitely many solutions to your equation would imply that $\mu(\log(3)/\log(2))$ were infinite. It is probably known that this latter statement is false, probably by methods similar to the proof in the Shorey/Tijdeman book mentioned by Will Jagy.


# istriu

Determine if matrix is upper triangular

## Syntax

• `tf = istriu(A)` example

## Description

example

````tf = istriu(A)` returns logical `1` (`true`) if `A` is an upper triangular matrix; otherwise, it returns logical `0` (`false`).```

## Examples

collapse all

### Test Upper Triangular Matrix

Create a 5-by-5 matrix.

`A = triu(magic(5))`
```A = 17 24 1 8 15 0 5 7 14 16 0 0 13 20 22 0 0 0 21 3 0 0 0 0 9```

Test `A` to see if it is upper triangular.

`istriu(A)`
```ans = 1 ```

The result is logical `1` (`true`) because all elements below the main diagonal are zero.

### Test Matrix of Zeros

Create a 5-by-5 matrix of zeros.

`Z = zeros(5);`

Test `Z` to see if it is upper triangular.

`istriu(Z)`
```ans = 1 ```

The result is logical `1` (`true`) because an upper triangular matrix can have any number of zeros on the main diagonal.

## Input Arguments

collapse all

### `A` — Input arraynumeric array

Input array, specified as a numeric array. `istriu` returns logical `0` (`false`) if `A` has more than two dimensions.

Data Types: `single` | `double`
Complex Number Support: Yes

collapse all

### Upper Triangular Matrix

A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.

For example, the matrix

$A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$

is upper triangular. A diagonal matrix is both upper and lower triangular.

### Tips

• Use the `triu` function to produce upper triangular matrices for which `istriu` returns logical `1` (`true`).

• The functions `isdiag`, `istriu`, and `istril` are special cases of the function `isbanded`, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, ```istriu(A) == isbanded(A,0,size(A,2))```.


Mathematics

# If $\displaystyle\int x^5e^{-x^2}dx=g(x)\cdot e^{-x^2}+C$ then the value of $g(-1)$ is?

$-\dfrac{5}{2}$

##### SOLUTION
Put $x^2=t$
$2xdx=dt$
$\displaystyle\int t^2e^{-t}\dfrac{dt}{2}$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}+2\displaystyle\int te^{-1}dt]+c$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}-2te^{-t}+\displaystyle\int 2e^{-t}dt]+c$
$=\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$
$=\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$
$g(x)=\dfrac{-(x^4+2x^2+2)}{2}$
$g(-1)=-\dfrac{5}{2}$.

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# Simple integral - but a question of methodology

ianb
simple integral -- but a question of methodology

$$\int \frac{(x-1)}{2} dx$$

First, try substitution. Let $$u = x - 1$$, and $$du = dx$$

$$\int \frac{u}{2} du$$

$$\frac{1}{2} \int u \ du$$

$$\frac{1}{2} \times \frac{u^{2}}{2}$$

$$\frac{1}{2} \times \frac{(x-1)^2}{2}$$

$$\frac{(x-1)^{2}}{4} + C$$

Or we could do this:

$$\int \frac{(x-1)}{2} dx$$

$$\int \frac{1}{2}(x-1)dx$$

$$\frac{1}{2} \int(x-1)dx$$

$$\frac{1}{2} (\frac{x^2}{2} - x) + C$$

$$\frac{1}{4} (x^2 - 2x) + C$$

Why should one of them be right?

Last edited:

ZioX
They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.

Homework Helper
...
$$\frac{1}{4} (x^2 - \frac{x}{2}) + C$$

Why should one of them be right?
The last line is wrong. You've pulled out the factor 1 / 4 incorrectly.
$$\frac{1}{4} \left( x ^ 2 - 2x \right)$$
Both are correct, the two results differ from each other by a constant 1/4.
You should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant.
Proof: Differentiate the LHS with respect to x, we have:
(F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0)
= F'(x) = f(x)
Can you get this? :)

ianb
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.

Thoughts?

ianb
Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.

Homework Helper
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.

Thoughts?
Nope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have:
$$\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)$$
$$\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)$$, they are still the same.
The constant will automatically vanish when you subtract the two terms. :)

ianb
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?

Last edited:
ianb
I'm sorry, I should have been clearer. Message edited accordingly.

Homework Helper
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
Uhm, well, no, they are not different.
Well, here we go:
$$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \left( \frac{(x - 1) ^ 2}{4} \right) \right|_0 ^ 1 = \frac{(1 - 1) ^ 2}{4} - \frac{(0 - 1) ^ 2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$$

$$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \frac{1}{4} \left( x ^ 2 - 2x \right) \right|_0 ^ 1 = \frac{1}{4} \left( (1 ^ 2 - 2 \times 1) - (0 ^ 2 - 0) \right) = \frac{1}{4} (-1 - 0) = -\frac{1}{4}$$
They are the same. :)

ianb
Wow. I'm embarrassed to say the least. You, sir, rock my world.


# Fewest steps to reach $200$ from $1$ using only $+1$ and $×2$

This is a problem from the AMC 8 (math contest):

A certain calculator has only two keys $$[+1]$$ and $$[\times 2]$$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$$9$$” and you pressed $$[+1]$$, it would display “$$10$$”. If you then pressed $$[\times 2]$$, it would display “$$20$$”. Starting with the display “$$1$$”, what is the fewest number of keystrokes you would need to reach “$$200$$”?

Intuitively I worked back from $$200$$, dividing by $$2$$ until I reached an odd number, subtracting $$1$$ when I did, etc..to reach the correct answer of $$9$$ steps.

However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically. The best I can come up with is that beyond the first step from $$1$$ to $$2$$, multiplication by $$2$$ is always going to yield a larger step than addition by $$1$$ and therefore I should take as many $$[\times 2]$$ steps as I can. This doesn't feel rigorous enough, though.

EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.

• Do you want the fewest steps to get to exactly $200$ or at least $200$? – John Douma Mar 17 at 19:37
• @JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice. – TonyK Mar 17 at 19:43
• @TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise. – John Douma Mar 17 at 19:44
• @JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$. – jeremy radcliff Mar 17 at 19:48

Look at what the operations $$+$$ and $$\times$$ do to the binary expansion of a number:

• $$\times$$ appends a $$0$$, and increases the length by one, leaving the total number of $$1$$'s unchanged;
• if the final digit is $$0$$, then $$+$$ increases the number of $$1$$'s by one, but doesn't change the length;
• if the final digit is $$1$$, then $$+$$ doesn't increase the total number of $$1$$'s (it may in fact decrease it), and doesn't increase the total length by more than $$1$$.

Therefore, with a single key press:

• you can increase the length by one, but this won't increase the number of $$1$$'s;
• you can increase the number of $$1$$'s by one, but this won't increase the length.

The binary expansion of $$200$$ is $$200_{10}=11001000_2$$. This has three $$1$$'s, and a length of eight. Starting from $$1$$, we must increase the length by seven, and increase the number of $$1$$'s by two. So this requires at least nine steps.

• Beautiful, this allows you to determine the optimal solution and path for an arbitrary number from it's binary representation. – Jared Goguen Mar 18 at 1:34
• There actually exist two ways of getting $200$ using nine steps. $$1+1+1\times2\times2\times2+1\times2\times2\times2=200\\ 1\times2+1\times2\times2\times2+1\times2\times2\times2=200$$ – Kay K. Mar 18 at 2:13
• @KayK.: Yes, but only because there are two ways of getting to 2. The rest is uniquely determined. – TonyK Apr 4 at 16:05

You can proceed by induction on $$n$$, showing that the quickest way to reach any even number $$2n$$ involves doubling on the last step, which is clearly true for the base case $$n=1$$ (where doubling and adding $$1$$ have a tomato-tomahto relationship).

Now if the last step to reach $$2n+2$$ isn't doubling, it can only be adding $$1$$ from $$2n+1$$. But $$2n+1$$ can only be reached by adding $$1$$ from $$2n$$, at which point the inductive hypothesis says the next previous number was $$n$$. But you can get from $$n$$ to $$2n+2$$ more quickly in two steps: add $$1$$, then double. So the last step in the quickest route to $$2n+2$$ is doubling from $$n+1$$.

• This is nice as it formalizes the OPs intuition, whereas the binary representation answer (which is super slick) might feel a little out of the blue. – jacob1729 Mar 17 at 22:55

Setting the display to binary base, $$[\times2]$$ inserts a $$0$$ to the right and $$[+1]$$ increments; if the rightmost digit is a zero, it just turns it to a $$1$$.

Using these rules you build a number of $$o$$ ones and $$z$$ zeroes in $$o-1+z$$ keystrokes, starting from $$1$$. This seems close to optimal.

• you just exactly reproduces the binary 200, why should we think it is not optimal? – dEmigOd Mar 17 at 19:54
• @dEmigOd: I didn't prove that inserting the bits one by one with $\times2$ or $\times2+1$ is optimal. – Yves Daoust Mar 17 at 19:57

I'll make a try

Since $$200=2^7+2^6+2^3$$ you will need at least $$8$$ steps to reach $$200$$ (since we start from $$1$$ and we get a number of the form $$2^a+...+2^l$$) so it remains to show that $$8$$ steps are not enough.

Maybe you could try to show that if there was a solution with $$8$$ steps then it would contain only one $$+1$$ which contradicts the fact that in $$200=2^7+2^6+2^3$$ we have two $$+$$

Working backwards, use the tree diagram and stop lagging branches (e.g. if $$\color{red}{99}$$ in step $$3$$ leads to $$1$$, then it takes one more operation than $$\color{red}{99}$$ in step $$2$$):

$$\hspace{3cm}$$


There are three formulas for calculating the area of a sector. The formula for the area of a sector is: A = r² * θ / 2. Find the central angle of a sector whose radius is 56 cm and area, is 144 cm2. Geometric skills. How to use the calculator Enter the radius and central angle in DEGREES, RADIANS or both as positive real numbers and press "calculate". Area of a sector when the central angle is given in degrees If the angle of the sector is given in degrees, then the formula for the area of a sector is given by, Area of a sector = (θ/360) πr2 A = (θ/360) πr2 Find the radius of a semi – circle with the area of 24 inches squared. Math A level Syllabus, 2016. The area of a sector of a circle with a radius of 5 cm, with an angle of 2 radians, is 25 cm 2. Then, the area of a sector of circle formula is calculated using the unitary method. 40pie units squared. Show that 2θ-3sinθ=0. The area of the minor segment as shown in Fig 5 . Area of a cyclic quadrilateral. 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We can find the area of a sector of a circle in a similar manner. Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Jul 2009 448 89. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; Displaying top 8 worksheets found for - Area Of A Sector In Radians. Both can be calculated using the angle at the centre and the diameter or radius. So in the below … The area of the circle is equal to the radius square times . as shown in Fig 4, we consider the sector as a fraction of the circle hence: Area of a segment. Introducing Radians; 9. If the diameter of a circle is 6, find the area of a sector with a sector angle of . Varsity Tutors LLC , where  is the angle measure of the sector in radians, and  is the radius of the circle. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing The full angle is 2π in radians, or 360° in degrees, the latter of which is the more common angle unit. The measure of central angle XYZ is 1.25 radians. The formulas to find the area of a sector in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr2 x θ/360 Area (Radians) = ½r2θ r, D° r, R° r, s r, A D°, s Use 3.14 for . Track your scores, create tests, and take your learning to the next level! Area of a sector given the arc length. The non-shaded area of the circle shown below is called a SECTOR. Area of an ellipse. ChillingEffects.org. The angle AOB is in radians. Area of sector. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! What is the area of Fiona's circle? 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The number inside the sector is the area. degree radian; area S . If r is in "m", the area will be in "m" 2. Now, this looks messy, but we can simplify it to get: Next, use your calculator to find a decimal answer, and then round to get our final answer. Find the area of a sector whose arc is 8 inches and radius, is 5 inches. a And then we just can solve for area of a sector by multiplying both sides by 81 pi. Find the area of the sector with radius 7\ "cm" and central angle 2.5 radians. There are two types of sectors, minor and major sector. I remember this formula as it is quite easy to remember. SECTORS . The portion of the circle's circumference bounded by the radii, the arc, is part of the sector. Perimeter of sector = r + 2r = r( + 2) Where is in radians If angle is in degrees, = Angle × π/(180°) Let us take some examples: Find perimeter of sector whose radius is 2 cm and angle is of 90° First, We need to convert angle in radians = Angle in degree × π/(180°) = 90° × π/(180° ) = π/4 Convert degrees into radians and viceversa. What is the area of the shaded sector? Fiona draws a circle with a diameter of 14 meters. A Terminal side Vertex B Initial Side C B, ABC, CBA, and are all notations for this angle. Questionnaire. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; (see diagrams below) There is a lengthy reason, but the result is a slight modification of the Sector formula: I have managed to get: 3=½r²θ and 2=½r²sinθ Therefore: ½r²θ-3=0 and ½r²sinθ-2=0 But I'm unsure where to go from there. Area of a circle. Find the area of a sector if the radius is 1 and the angle of sector is  radians. Find the area of a sector with the radius of 1 and angle of . In order to calculate the area of a sector, you need to know the following two parameters: With the above two parameters, finding the area of a circle is as easy as ABCD. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by 3. The arc length formula is used to find the length of an arc of a circle; ℓ = rθ ℓ = r θ, where θ θ is in radian. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. is obtained by the arc AB of the centre O is given by: where is in radians. Finding a Missing Angle With the Sine Rule; 13. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the Express the answer in terms of . Side of polygon given area. r O 1 radian is the size of the angle formed at the centre of a circle by 2 radii which join the ends of an arc equal in length to the radius. Some of the worksheets for this concept are Arc length and sector area, Area of a sector 1, L 2r, Find the area of the shaded sector in the following, Radians arc length and area of a sector, Radians, Mcr3ui radian work, Area and arc length of a sector. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Figure 6. misrepresent that a product or activity is infringing your copyrights. 222 r Ar r Note, to use this formula, the measure of the central angle must be given in radians. Calculate the area of the sector shown below. Calculate the area of a sector with a radius of 10 yards and an angle of 90 degrees. A-Level Maths Edexcel C2 June 2008 Q7b ExamSolutions Northeastern University, Bachelor of Science, Industrial Engineering. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. The non-shaded area of the circle shown below is called a SECTOR. If the angle of the sector is given in degrees, then the formula for the area of a sector is given by. Find the area of a sector with a radius and angle of . © 2007-2020 All Rights Reserved, Computer Science Tutors in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Francisco-Bay Area. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. In this example the sector subtends a right-angle (900) at the centre of the circle. Area of a sector of a circle. A-level : area and arc length of a sector tutorial In this tutorial you are shown how to find the area of a sector and arc length when the angle is in degrees or radians. Problem Solving With the Cosine Rule; 15. Find the angle of a sector whose arc length is 22 cm and area, is 44 cm2. When the angle at the centre is 360°, area of the sector, i.e., the complete circle = πr² When the angle at the center is 1°, area of the sector = Thus, when the angle is θ, area of sector, OPAQ = Write the formula for the area of a sector in radians. Using this formula, and approximating , the area of the circle is . Sep 2, 2009 #2 For #1. Recognize parts of a circle and use appropriate terminology. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such The circular face of a watch has an area that measures between 800 and 900 square millimeters. We know that the area of the whole circle is equal to πr². The ratio of the area of the sector to the area of the full circle will be the same as the ratio of the angle θ to the angle in a full circle. When angle of the sector is 360°, area of the sector i.e. Area of a regular polygon. Find the area of the sector. How to find the area of a sector whose central angle is in radian: formula, 1 example, and its solution. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the


1. ## Piecewise Function problem

Let f(x) =

(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2

and g(x) = [integrate] f(t)dt on [0, x]

i) Find an expression for g(x) similar to the one for f(x).

ii) Where is f differentiable? Where is g differentiable?

I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

But from that, how do I figure out an expression for g(x) from the piecewise function?

Any and all help is appreciated here.

2. Originally Posted by Tulki
Let f(x) =

(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2

and g(x) = [integrate] f(t)dt on [0, x]

i) Find an expression for g(x) similar to the one for f(x).

ii) Where is f differentiable? Where is g differentiable?

I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

But from that, how do I figure out an expression for g(x) from the piecewise function?

Any and all help is appreciated here.
i)

for $\displaystyle x < 0$, $\displaystyle f(x) = 0$ and $\displaystyle g(x) = \int f(t) ~dt = 0$

for $\displaystyle 0 \leq x \leq 1$, $\displaystyle f(x) = x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2$

for $\displaystyle 1 < x \leq 2$, $\displaystyle f(x) = 2 - x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt=$$\displaystyle \int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})$

for $\displaystyle x > 2$

3. Hello, Tulki!

Here's part (a).
Did you make a sketch?

$\displaystyle \text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm] x && 0 \leq x \leq 1 \\ \\[-4mm] 2 - x && 1 < x \leq 2 \\ \\[-4mm] 0 && x > 2 \end{array}\right\}$

and: .$\displaystyle g(x) \:=\: \int_0^x f(t)\,dt$

a) Find an expression for $\displaystyle g(x)$ similar to the one for $\displaystyle f(x).$

$\displaystyle g(x)$ gives the area under the graph from $\displaystyle 0\text{ to }x$

The graph looks like this:
Code:
|
|
1+       *
|     *:::*
|   *:::::::*
| *:::::::::| *
- * * * - - - * - + - * * * - -
0       1   x   2

Then: .$\displaystyle \displaystyle g(x) \;=\;\left\{ \begin{array}{cccc} 0 && x < 0 \\ \\[-3mm] \int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm] \frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm] 1 && x > 2 \end{array} \right\}$

4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.

For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).

For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is $\displaystyle \frac{1}{2}x^2$.

For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\displaystyle \frac{1}{2}(1)(1)= \frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\displaystyle \frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}$.
The total area is $\displaystyle \frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.

For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.

5. Thank you very much, all of you!

I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.


# Dense subset of $S^{n-1}$ with antipodal complement

Let $$n \in \mathbb{N}$$, and let $$S^{n} \subseteq \mathbb{R}^{n+1}$$ be the unit $$n$$-sphere.

Does there exist a dense subset $$D$$ of $$S^n$$ such that $$\textbf{x} \in D \implies -\textbf{x} \notin D$$?

This is true for $$n=1$$. Note that $$S^1 \cong A := [0,1]/\mathbb{Z}$$, where the equivalent question is: does there exists a dense subset $$D'$$ of $$A$$ such that $$x \in D \implies x \pm \frac{1}{2} \notin D'$$?

Consider the function $$f : [0,1) \to \{0,1\}$$,

$$f(x)= \begin{cases} 0 & x< \frac{1}{2},\text{ any decimal expansion of } x \text{ has finitely many 3s} \\ 1 & x < \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \\ 1 & x = \frac{1}{2} \\ 1 & x > \frac{1}{2}, \text{ any decimal expansion of } x \text{ has finitely many 3s}\\ 0 & x > \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \end{cases}$$

Then $$D' = f^{-1}(\{0\})$$ is a suitable dense set.

However, I am not sure how to approach the cases $$n >1$$, since $$S^n$$ is no longer homeomorphic to a very simple object and my topological toolkit is quite unsophisticated. If the question can be answered with elementary methods, I welcome any hints. I am not even sure whether or not I expect such a set $$D$$ to exist for $$n> 1$$.

• It is also equivalent to ask the analogue of the question in $\Bbb R^n$, if you’ve figured out what the antipodes look like after stereographic projection Aug 7 at 14:42
• I don't have time for a full answer, but you could consider the stereographic projection $\mathbb{S}^n\setminus\{(0,\ldots,0,1)\}\rightarrow\mathbb{R}^n$, then look at where the intersection $H_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$ with $\mathbb{S}^n$ lands, and then consider the primages of their intersection with $\mathbb{Q}^n$ and $\mathbb{R}^n\setminus\mathbb{Q}^n$ respectivly. Aug 7 at 14:58
• @SamuelAdrianAntz I almost understand your comment, but it's not quite clear. I am a little suspicious of your last step because note that $D$ quite naturally bijects with its complement, so both are uncountable. I understand the stereographic projection to be a bijection, so the preimage of $\mathbb{Q}^n$ would be countable? Or have I misunderstood? Aug 7 at 15:19
• @FShrike That's certainly a possible approach. But even if I could figure out where they antipodes go, I am not certain that this simplifies matters much (or enough for me). After all, $\mathbb{R}$ is still much simpler than $\mathbb{R}^n$, $n>1$. Aug 7 at 15:28

This answer is an expansion of my comment, but I simplified it a bit without using the stereographic projection.

First of all, the obvious way of taking the intersection of $$\mathbb{Q}^{n+1}$$ with $$\mathbb{S}^n\subset\mathbb{R}^{n+1}$$ won't work, as it can even be empty as explained in this answer here. Consider the half-spaces $$\mathbb{H}_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$$ and the (surjective) projections: $$\operatorname{pr}_\pm\colon \mathbb{H}_\pm^{n+1}\cap\mathbb{S}^n \twoheadrightarrow\mathbb{D}^n, (x_0,x_1,\ldots,x_n)\mapsto(x_1,\ldots,x_n)$$ flattening down both half-spheres.

The subsets $$A=\operatorname{pr}_-^{-1}(\mathbb{D}^n\cap\mathbb{Q}^n)$$ and $$B=\operatorname{pr}_+^{-1}(\mathbb{D}^n\cap(\mathbb{R}^n\setminus\mathbb{Q}^n))$$ are both dense (in the closure of their respective half-sphere) as preimages of dense sets under continuous maps. $$A\cup B$$ is therefore dense in $$\mathbb{S}^n$$ as the closure operator commutes with finite unions (See here).

For $$x\in A$$, we have $$x_0<0$$, therefore $$-x\notin A$$, as well as $$x_1,\ldots,x_n\in\mathbb{Q}$$, therefore $$-x\notin B$$.

For $$y\in B$$, we have $$y_0>0$$, therefore $$-y\notin B$$, as well as $$y_1,\ldots,y_n\notin\mathbb{Q}$$, therefore $$-y\notin A$$.

• So one can use effectively the same idea as for $n=1$. Fair enough. I convinced myself it would be much more difficult and didn't try it. Also, your profile description is fairly entertaining. Aug 7 at 21:26
• As a minor correction, $A$ and $B$ are images, not preimages. Aug 7 at 21:33
• Thanks a lot! :-) The mistake with $A$ and $B$ is also corrected now. Aug 7 at 21:40
• No worries, and thanks for the answer. I'm sorry, but I think I have just observed another typo. I think the projection maps surject onto $D_1(0)$ in $\mathbb{R}^{n-1}$, the interior of the $(n-1)$-sphere. Aug 7 at 21:45
• Oh, yes! Honestly, I'm really not that good in writing down my thoughts in the correct notation. Aug 7 at 21:47


# Math Help - Difficult Trigonometric Integral

1. ## Difficult Trigonometric Integral

I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

$\int {sin^2 x \over \sqrt {1+cosx}}$

Thanks for the help!

2. Originally Posted by Tylerzzz
I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

$\int {sin^2 x \over \sqrt {1+cosx}}$
Substitute $u = \cos x$. Then $du = -\sin x\,dx$ and the integral becomes

$-\int \frac{\sin x}{\sqrt {1+cosx}}(-\sin x)dx = -\int \frac{\sqrt{1-u^2}}{\sqrt{1+u}}du = -\int\sqrt{1-u}\,du = \frac23(1-u)^{3/2} = \frac23(1-\cos x)^{3/2}$ (plus a constant).

3. Wow! I never really thought about solving it that way. Thanks a lot!

4. EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]

5. Originally Posted by TheCoffeeMachine
EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
You had me worried for a moment.

6. Hello, Tylerzzz!

Another approach . . .

$\displaystyle \int \frac{\sin^2\!x}{\sqrt{1+\cos x}}\,dx$

We have: . $\dfrac{\sin^2\!x}{\sqrt{1+\cos x}}$

Multiply by $\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\!:\;\;\dfrac{\sin^2\!x}{\sqrt{1+\cos x}} \cdot\dfrac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{1-\cos^2\!x}}$

. . . $\;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{\sin^2\!x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sin x} \;=\;\sin x\sqrt{1-\cos x}$

The integral becomes: . $\displaystyle \int(1-\cos x)^{\frac{1}{2}}(\sin x\,dx)$

Let: . $u \:=\:1-\cos x \quad\Rightarrow\quad du \:=\:\sin x\,dx$

Substitute: . $\displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C$

Back-substitute: . $\frac{2}{3}(1-\cos x)^{\frac{3}{2}} + C$

7. Originally Posted by Opalg
You had me worried for a moment.
Haha! In my defence, I blame the UCAS for the plight of my eyesight today.


# Thread: Need Help for Half Life Problems using Logs

1. ## Need Help for Half Life Problems using Logs

Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.

The equation that I got was:

N = No (1/2) ^ t

N is final amount, No is the initial ammount and t is time.

I was given the data:

Time (min) 10 20 30 40
Amount (g) .83 .68 .56 .46

and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.

How do I do this? I have no clue...

2. Do you have Excel?. You can do all sorts of regressions with it. Also, many calculators do regressions. Do you have one?. If so, what is it. Maybe I can step you through it.

Anyway, using Excel I got an equation of

$y=1.0091e^{-.0196x}$

This has an R^2 of 1, so it is as good as it gets.

Enter in x=10, do you get .83?.

3. hi thanks for replying, I have a TI-83+.

4. I would have to re-familiarize myself. I suppose the 83 does exponential regressions. I am not sure.

5. it does, i did it before, I think.
I just forgot how to do it.

I think it's storing the data into 2 list and then compare them or something...

but then, how do I set the equations up?

I mean that N = No (1/2) ^ t thing , hate math final, lol.

Maybe you can show me how you did it in excel?

6. You can find the half life by using the equation $k=\frac{-1}{T}ln(2)$

We know k=-.0196

So, $-.0196=\frac{-1}{T}ln(2)$

$t=35.36$

We could also do it using the equation.

Half of 1.0091 is .50455. Because 1.0091 is the amount at t=0

So, $.50455=1.0091e^{-.0196t}$

$t=35.36$

Same as before.

To find a regression on Excel, I make the graph, then add a trendline.

Click on Insert, then Chart, then Scatter

Highlight your data which is entered in columns A and B and proceed.

It is pretty much self-explanatory. Then onece you have the graph, click on

Chart, Add Trendline, Exponential, Click on Options and check the box that says to display the equation on the chart.

There you have it.

7. Originally Posted by Billwaa
Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.

The equation that I got was:

N = No (1/2) ^ t

N is final amount, No is the initial ammount and t is time.

I was given the data:

Time (min) 10 20 30 40
Amount (g) .83 .68 .56 .46

and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.

How do I do this? I have no clue...
$N = N_0 \left ( \frac{1}{2} \right ) ^{kt}$

$ln(N) = ln(N_0) - [k~ln(2)]t$

Put t on the horizontal axis, ln(N) on the vertical axis, and do a linear regression. The slope will be $k~ln(2)$ which you can then solve for k, and the intercept will be $ln(N_0)$.

-Dan

8. thanks guys, this clear it up a little bit. ^_^

9. Originally Posted by Billwaa
hi thanks for replying, I have a TI-83+.
To do this on the 83+, input the data into two lists:

L1: {10,20,30,40}
L2: {.86,.68,.56,.46}

Now, hit [STAT] and scroll over to [CALC] Select option [0]:ExpReg.

This is an exponential regression. The solution will have the form of $y=a\times b^x$.

At the home screen, you should now see:

ExpReg

Now hit [2nd][1][,][2nd][2]

You should now have the following on the screen:

$\text{ExpReg } L_1\text{,}L_2$

Hit [ENTER]

You'll get the regression equation:

This is what the screen should say:

ExpReg
\begin{aligned}
y&=a*b^x \\
a&=1.00914514 \\
b&=.9805442185
\end{aligned}

Note that TheEmtpySet's answer was $y=1.0091e^{-.0196x}$

We can rewrite the answer that the calculator gave us to get his solution.

Note that $b^x= e^{x\ln(b)}$.

Thus, $.9805442185^x=e^{x\ln(.9805442185)}=e^{-.0196475365x}$

Rounding the regression equation, we get:

$\color{red}\boxed{y=1.0091e^{-.0196x}}$

Hope this makes sense!