# Calculation Of Taylor Series

Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Then, for every x in the interval, where R n(x) is the remainder (or error). We prove a general Steady-state theorem for Volterra series operators, and then establish a general formula for the spectrum of the output of a Volterra series operator in terms of the spectrum of a periodic input. Actually, this is now much easier, as we can use Mapleor Mathematica. Annette Pilkington Lecture 33 Applications of Taylor Series. A minimum of 1 double value to hold the total and 2 integer values, 1 to hold the current term number and 1 to hold the number of terms to evaluate. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. Part 2) After completing part 1, modify the series for faster convergence. Other articles where Taylor Standard Series Method is discussed: David Watson Taylor: …known since 1910 as the Taylor Standard Series Method, he determined the actual effect of changing those characteristics, making it possible to estimate in advance the resistance of a ship of given proportions. In mathematics, the Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. Series can expand about the point x = ∞. Series can construct standard Taylor series, as well as certain expansions involving negative powers, fractional powers, and logarithms. Sign up to read all wikis and quizzes in math, science, and engineering topics. The Taylor series is a power series that approximates the function f near x = a. Concrete examples in the physical science division and various engineering ﬁelds are used to paint the applications. TAYLOR AND MACLAURIN SERIES 102 4. That is, every reasonable function can be written as This module describes how to compute the coefficients for a given function. Binomial Theorem Calculator Binomial Theorem Calculator This calculators lets you calculate __expansion__ (also: series) of a binomial. Math 142 Taylor/Maclaurin Polynomials and Series Prof. And by knowing these basic rules and formulas, we can learn to use them in generating other functions as well as how to apply them to Taylor Series that are not centered at zero. The first is the power series expansion and its two important generalizations, the Laurent series and the Puiseux series. The taylor series expansion of f(x) with respect to xo is given by: Generalization to multivariable function: (5) Using similar method as described above, using partial derivatives this time, (Note: the procedure above does not guarantee that the infinite series converges. Miyagawa 4-12-11-628 Nishiogu, Arakawa-ku, Tokyo 116-0011, Japan. Definition: The Taylor series is a representation or approximation of a function as a sum. Taylor’s Series. The Taylor Series is also referred to as Maclaurin (Power) Series. is determined by using multivariate Taylor series expansion. The power series converges globally to the function. Sum of Series Programs / Examples in C programming language. for any x in the series' interval of convergence. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied. Maclaurin's formula or Maclaurin's theorem: The formula obtained from Taylor's formula by setting x 0 = 0 that holds in an open neighborhood of the origin, is called Maclaurin's formula or Maclaurin's theorem. The Taylor series for ex based at b = 0is ex = X∞ n=0 xn n! so we have e3x = X∞ n=0 (3x)n n! and x2e3x = X∞ n=0 3nxn+2 n! =. OBTAINING TAYLOR FORMULAS Most Taylor polynomials have been bound by other than using the formula pn(x)=f(a)+(x−a)f0(a)+ 1 2! (x−a)2f00(a) +···+ 1 n! (x−a)nf(n)(a) because of the diﬃculty of obtaining the derivatives f(k)(x) for larger values of k. Here are a few examples. Suppose we wish to find the Taylor series of sin( x ) at x = c , where c is any real number that is not zero. Computing Taylor Series Lecture Notes As we have seen, many different functions can be expressed as power series. The Taylor Series Calculator an online tool which shows Taylor Series for the given input. Derivative calculator Integral calculator Definite integrator Limit calculator Series calculator Equation solver Expression simplifier Factoring calculator Expression calculator Inverse function Taylor series Matrix calculator Matrix arithmetic Graphing calculator. A minimum of 1 double value to hold the total and 2 integer values, 1 to hold the current term number and 1 to hold the number of terms to evaluate. Series Converges Series Diverges Diverges Series r Series may converge OR diverge-r x x x0 x +r 0 at |x-x |= 0 0 Figure 1: Radius of. Let G = g(R;S) = R=S. To estimate the square root of a number using only simple arithmetic, the first-order Taylor series of the square root function provides a convenient method. Taylor_series_expansion online. Other Power Series Representing Functions as Power Series Functions as Power Series Derivatives and Integrals of Power Series Applications and Examples Taylor and Maclaurin Series The Formula for Taylor Series Taylor Series for Common Functions Adding, Multiplying, and Dividing Power Series Miscellaneous Useful Facts Applications of Taylor. As such, he found that by calculating the time needed for the various elements of a task, he could develop the "best" way to complete that task. Part 2) After completing part 1, modify the series for faster convergence. For this I need to calculate the taylor series expansion of the function. the series for , , and ), and/ B BB sin cos. , x 0 2I : Next consider a function, whose domain is I,. The Taylor package was written to provide REDUCE with some of the facilities that MACSYMA's TAYLOR function oﬀers, but most of all I needed it to be faster and more space-eﬃcient. Taylor Series. So I want a Taylor polynomial centered around there. Taylor’s series is an essential theoretical tool in computational science and approximation. In this video I'm going to show you how you can find a Taylor series. (The formula used is shown on page 100 of the text. It is nothing but the representation of a function as an infinite sum of terms. If only concerned about the neighborhood very close to the origin, the n = 2 n=2 n = 2 approximation represents the sine wave sufficiently, and no. The nonlinear restoring forces are given in R(x,x˙) and fext(t) is a vector of external dynamic loads. The Taylor series obtained when we let c = 0 is referred to a Maclaurin series. This script lets you input (almost) any function, provided that it can be represented using Sympy and output the Taylor series of that function up to the nth term centred at x0. How can I get a Taylor expansion of the Sin[x] function by the power series? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. these developments was the recursive calculation of the coeﬃcients of the Taylor series. TAYLOR and MACLAURIN SERIES TAYLOR SERIES Recall our discussion of the power series, the power series will converge absolutely for every value of x in the interval of convergence. About the calculator: This super useful calculator is a product of wolfram alpha, one of. via the usual Taylor series, we get the same result as above without using Taylor’s mul-tivariable formula. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. I'm currently in an introductory course of MATLAB and one of my assignments is to calculate the Taylor series of a given formula, without using the available taylor(f,x) function. 1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. Browse other questions tagged power-series functional-equations taylor-series or ask your own question. This calculator turns your data into a Mathematical formula by generating a Fourier Series of sines and cosines. This is clearly not the case. Compute the interval of convergence for each series on the previous page. Several methods exist for the calculation of Taylor series of a large number of functions. Solution: Since 1 1 −𝑥 = 𝑥𝑘 ∞ 𝑘=0 we get 𝑥2 1 −𝑥 = 𝑥2 𝑥𝑘 ∞ 𝑘=0 = 𝑥𝑘+2 ∞ 𝑘=0 J. Euler's Method, Taylor Series Method, Runge Kutta Methods, Multi-Step Methods and Stability. taylor_series is a univariate Taylor series. These are called the Taylor coefficients of f, and the resulting power series is called the Taylor series of the function f. 2 Calculating a Maclaurin series Use Maxima to calculate the terms in the Maclaurin series up to and including x7 for the following functions; (1) g(x) = arctanx (2) G(x) = arctanhx (the inverse of the hyperbolic function tanh) What do you notice about the terms and signs in the series (1) and (2) here?. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. Taylor Series • The Taylor Theorem from calculus says that the value of a function can be approximated near a given point using its “Taylor series” around that point. Expansions of e. May 20, 2015 firstly we look. x, sin x, and cos x, and related series. See Examples. the Taylor expansion of 1 1−x) • the Taylor expansions of the functions ex,sinx,cosx,ln(1 + x) and range of va-lidity. I have to do it using taylor series using iterations, but only for first 13 nominators&;denominators. The Taylor expansion of a function at a point is a polynomial approximation of the function near that point. 1 Introduction This section focuses on deriving a Maclaurin series for functions of the form f(x) = (1 + x)k for any number k. n=0 for some constant C depending on the choice of antiderivative of f. This series — known as a "power series" — can be written in closed. 1 Introduction This chapter has several important and challenging goals. It is nothing but the representation of a function as an infinite sum of terms. A term that is often heard is that of a “Taylor expansion”; depending on the circumstance, this may mean either the Taylor series or the n th degree Taylor polynomial. Taylor's formula and Taylor series can be defined for functions of more than one variable in a similar way. Change the function definition 2. A consequence of this is that a Laurent series may be used in cases where a Taylor. Say you are asked to find the Taylor Series centered at a=0 up to degree n=3 (really a MacLaurin series as the center is at 0 ) So plug into Calculus Made Easy option 7 D as follows : The derivatives are taken in order to compute the coefficients for each term up to degree 3. Find the interval of convergence for ∞ n=0 (x−3)n n. We know that the th Taylor polynomial is , and we have spent a lot of time in this chapter calculating Taylor polynomials and Taylor Series. In other words, in this particular instance, from the fact that the series 1+ 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + converges, one is likely to erroneously infer thatall in nite series converge. In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. program for calculation of the sine of an angle using the sine series to radians because the Taylor series method for calculating sine uses radians and not. REVIEW: We start with the diﬀerential equation dy(t) dt = f (t,y(t)) (1. For sin function. For this reason, we often call the Taylor sum the Taylor approximation of degree n. Binomial Theorem A-Level Mathematics revision section of Revision Maths looking at Binomial Theorem and Pascals Triangle. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. 7:34 AM - 14 May 2018. As an example, let’s use the Maclaurin polynomial (with just four terms in the series) for the function f(x) = sin(x) to approximate sin(0. These notes discuss three important applications of Taylor series: 1. It explains how to derive power series of composite functions. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. Calculation of Taylor series Several methods exist for the calculation of Taylor series of a large number of functions. And by knowing these basic rules and formulas, we can learn to use them in generating other functions as well as how to apply them to Taylor Series that are not centered at zero. The proofs are complete, and use only the basic facts of analysis. Convergence of Taylor Series SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. List of Maclaurin Series of Some Common Functions / Stevens Institute of Technology / MA 123: Calculus IIA / List of Maclaurin Series of Some Common Functions / 9 | Sequences and Series. What is the Taylor series representation of f(x + delta(x)) and how is it arrived at? I am trying to understand the derivation of continuous compounding rate of interest calculation. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. Cook great food for your customers. Find the Taylor expansion series of any function and see how it's done! Up to ten Taylor-polynomials can be calculated at a time. TI-89 - Vol 2 - Sect 08 - Calculator Taylor and Maclaurin Polynomials. Of course, the polynomial function will not have the same shape for all values of "x". For instance, in Example 4 in Section 9. These notes discuss three important applications of Taylor series: 1. For this I need to calculate the taylor series expansion of the function. Taylor Series Convergence (1/n!) f (n) (c) (x - c) n = f(x) if and only if lim (n-->) R n = 0 for all x in I. This variable is first initialized to 0. Rapid and Accurate Calculation of Water and Steam Properties Using the Tabular Taylor Series Expansion Method K. Modern numerical algorithms for the solution of ordinary diﬀerential equations are also based on the method of the Taylor series. This smart calculator is provided by wolfram alpha. 0 Calculator question. It is nothing but the representation of a function as an infinite sum of terms. Added Nov 4, 2011 by sceadwe in Mathematics. We first write the terms of the series from n = 0 to n = 3. If you are interested in seeing how that works you. A Taylor series isn't really a good way to compute this function unless you're looking for asymptotic accuracy around a particular point, rather than general accuracy along the whole thing. Sequences and series • A sequence is a (possibly infinite) collection of numbers lined up in some order • A series is a (possibly infinite) sum – Example: Taylor’s series k ¦ 2 n o o n o o o o o n k o o k n f x n fx k T c ( ) 1 ( )! 1 2 1! 1 Example: sine function approximated by Taylor series expansion The approximation of f(x)=sin. If you are interested in seeing how that works you. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. The Delta Method gives a technique for doing this and is based on using a Taylor series approxi-mation. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. Statistical moments of X , such as the variance, are then computed in terms of the Taylor coefficients and the moments of x [ 3 , ]. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. Specifically, the Taylor series of an infinitely differentiable real function f, defined on an open interval (a − r, a + r), is :. Example: The Taylor Series for e x e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! +. As you have noticed, the Taylor series has infinite terms. Taylor’s Series of sin x In order to use Taylor’s formula to ﬁnd the power series expansion of sin x we have to compute the derivatives of sin(x):. The more terms you use, however, the better your approximation will be. It seems that, instead of Taylor series for int_0^x, it calculates Taylor series for int_x^{2x} I have only found this bug when calculating Taylor series for functions f defined as in the example (by means of an integral). MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. sin x=x-1/6x^3 +1/120x^5 -1/5040x^7 The calculator substitutes into as many terms of the polynomial that it needs to in order to get the required number of decimal places. About the calculator: This super useful calculator is a product of wolfram alpha, one of. Actually, this is now much easier, as we can use Mapleor Mathematica. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. Solve for g(pi/3) using 5, 10, 20 and 100 terms in the Taylor series (use a loop) So I tried the following in the script editor:. MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. The result is in its most. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. , x 0 2I : Next consider a function, whose domain is I,. Part 1) Given a list of basic taylor series, find a way to approximate the value of pi. As any calculus student knows, the first-order Taylor expansion around x 2 is given by sqrt(x 2 + a) ~ x + a / 2x. A Maclaurin series is a special case of a Taylor series, where “a” is centered around x = 0. The Taylor Series Calculator an online tool which shows Taylor Series for the given input. A summary of The Remainder Term in 's The Taylor Series. Every Maclaurin series, including those studied in Lesson 22. Taylor Series SingleVariable and Multi-Variable • Single variable Taylor series: Let f be an inﬁnitely diﬀerentiable function in some open interval around x= a. Especially I wanted procedures that would return the logarithm or arc tangent of a Taylor series, again as a Taylor series. The crudest approximation was just a constant. The first is to calculate any random element in the sequence (which mathematicians like to call the "nth" element), and the second is to find the sum of the geometric sequence up to the nth element. See Examples. sin(x) of java. The answer lies in estimating the function f(x) by its Taylor expansion. When p = 1, the p-series is the harmonic series, which diverges. Textbook solution for Engineering Fundamentals: An Introduction to… 5th Edition Saeed Moaveni Chapter 18 Problem 44P. , the difference between the highest and lowest power in the expansion is 4. The use of duration, in the second term of the Taylor series, to determine the change in the instrument value is only an approximation. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. 2 The variance of g ( y n) is then approximated. Thenlet x= 1 in the earlier formulas to get Most Taylor polynomials have been bound by other than using the formula pn. * The more simple the expression, the better range of accuracy with less terms. The question is, for a specific value of , how badly does a Taylor polynomial represent its function?. We want a power series in factors of (x-a), where we can easily obtain the coefficients of (x-a)^k. Simple components for Ada The Simple components for Ada library provides implementations of smart pointers for automatically c taylor series calculator free download - SourceForge. The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0. Find the Taylor series expansion of any function around a point using this online calculator. TAYLOR AND MACLAURIN SERIES 102 4. These "time and motion" studies also led Taylor to conclude that certain people could work more efficiently than others. Each term of the Taylor polynomial comes from the function's derivatives at a single point. For the square roots of a negative or complex number, see below. Specifically, the Taylor series of an infinitely differentiable real function f, defined on an open interval (a − r, a + r), is :. Taylor Series. This non-linear mapping (on inhomogeneous coordinates) can be expanded in a Taylor series. Things you should memorize: • the formula of the Taylor series of a given function f(x) • geometric series (i. Requires a Wolfram Notebook System. This is the program to calculate the value of 'e', the base of natural logarithms without using "math. Sum of Taylor Series Program. It is a series expansion around a point. Math 142 Taylor/Maclaurin Polynomials and Series Prof. 1 What is a Laurent series? The Laurent series is a representation of a complex function f(z) as a series. Questions: 1. An Easy Way to Remember the Taylor Series Expansion which is technically known as a Maclaurin rather than a Taylor). C / C++ Forums on Bytes. Then, for every x in the interval, where R n(x) is the remainder (or error). These risk statistics are also known as greeks. If the Taylor series is centered at zero, then that series is also called a Maclaurin series, after the Scottish mathematician Colin Maclaurin, who made extensive. Taylor Series Cos x Calculator. This is an infini te sum, but your function should stop the summation when the addition of a successive term makes a negligible change in e. Miyagawa 4-12-11-628 Nishiogu, Arakawa-ku, Tokyo 116-0011, Japan. TI-89 - Vol 2 - Sect 08 - Calculator Taylor and Maclaurin Polynomials. I'm trying to write a program to find values for arctan of x by using taylor series. Cascade IP3 calculation formula for 3 or four stages. Things you should memorize: • the formula of the Taylor series of a given function f(x) • geometric series (i. ADVERTISEMENTS: Read this article to learn how to Calculate Standard Deviation in 3 different Series! A. A Quick Note on Calculating the Radius of Convergence The radius of convergence is a number ˆsuch that the series X1 n=0 a n(x x 0)n converges absolutely for jx x 0j<ˆ, and diverges for jx x 0j>0 (see Fig. Once you get the basic idea it will be very easy to do. The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. List of Maclaurin Series of Some Common Functions / Stevens Institute of Technology / MA 123: Calculus IIA / List of Maclaurin Series of Some Common Functions / 9 | Sequences and Series. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. After all, you are finding the nth derivative of f, when all you need is the nth derivative of f , evaluated at a. C++ templates for sin, cos, tan taylor series. This appendix derives the Taylor series approximation informally, then introduces the remainder term and a formal statement of Taylor's theorem. Featured on Meta Congratulations to our 29 oldest beta sites - They're now no longer beta!. Questions: 1. It is nothing but the representation of a function as an infinite sum of terms. Taylor Series Calculator. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. Program to Calculate the Exponential Series in C | C Program Posted by Tanmay Jhawar at 9:12 PM - 9 comments Here's a C program to calculate the exponential series using For loop with output. Example: Calculation of Pi to 707-digit accuracy (like William Shanks):. See how it's done with this free video algebra lesson. Please see Jenson and. Sequences and series • A sequence is a (possibly infinite) collection of numbers lined up in some order • A series is a (possibly infinite) sum – Example: Taylor’s series k ¦ 2 n o o n o o o o o n k o o k n f x n fx k T c ( ) 1 ( )! 1 2 1! 1 Example: sine function approximated by Taylor series expansion The approximation of f(x)=sin. Find f11(0). Gas Turbines Power (July, 2001) Supplementary Backward Equations for Pressure as a Function of Enthalpy and Entropy p(h,s) to the Industrial Formulation IAPWS-IF97 for Water and Steam. The Taylor Series is also referred to as Maclaurin (Power) Series. The free tool below will allow you to calculate the summation of an expression. The more terms you use, however, the better your approximation will be. For any f(x;y), the bivariate ﬁrst order Taylor expansion about any = (. 1) and its associated formula, the Taylor series, is of great value in the study of numerical methods. Then combine the terms with the same exponent. What makes these important is that they can often be used in place of other, more complicated functions. Since this series expansion for tan-1x allows us to reach higher accuracy much faster, it is the more efficient of the two series. Using a popular tool of monetary policy analysis, our Taylor Rule Calculator lets you estimate where short-term interest rates should move as economic conditions change. A Maclaurin series is a specific type of Taylor series that's evaluated at x o = 0. An example of a Taylor Series that approximates $e^x$ is below. Taylor Series Calculator with Steps Taylor Series, Laurent Series, Maclaurin Series. (c) Check that the Taylor series for ex, sinhxand coshxsatisfy the same equation. Maclaurin Taylor Series for Transcendental Functions: A Graphing-Calculator View of Convergence Marvin Stick Most calculus students can perform the manipulation necessary for a polynomial approximation of a transcendental function. Find approximations for EGand Var(G) using Taylor expansions of g(). So I want a Taylor polynomial centered around there. Glenn Research Center, Cleveland, Ohio More efficient versions of an interpo-lation method, called kriging, have been introduced in order to reduce its tradi-. Find the Taylor series for f(x) centered at the given value of 'a'. Calculate g(x) = sin(x) using the Taylor series expansion for a given value of x. The Taylor package was written to provide REDUCE with some of the facilities that MACSYMA's TAYLOR function oﬀers, but most of all I needed it to be faster and more space-eﬃcient. Concrete examples in the physical science division and various engineering ﬁelds are used to paint the applications. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. It is a series expansion around a point. 1730's) with using the series expansion of the arcsine function,,. Wyrick family find themselves Cialis 20 Mg Paypal to spot fakes per GiB of RAM I hope to hear adjustment and metal ball. Belbas Mathematics Department University of Alabama Tuscaloosa, AL. What is the difference between Power series and Taylor series? 1. Therefore, ex= X1 k=0 f(k)(0) k! xk= X1 k=0 xk k! as we already know. e-mail: [email protected] Evaluate all these at. Taking the first two terms of the series gives a very good approximation for low speeds. The SURVEYLOGISTIC Procedure. Due to the increasing quartic moveout term (which is negative for the example in Figure 4. java from §9. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. We want a power series in factors of (x-a), where we can easily obtain the coefficients of (x-a)^k. Simple Calculator to find the trigonometric cos x function using cosine taylor series formula. Drek intends to pollute into my fifties my irony is sarcastic and to thin more and. Calculating the sine of a number from the Taylor sum I'm trying to write code to basically mimic the sum to infinity expression, Sine - Wikipedia, the free encyclopedia , but with a finite number of terms, and have the user input the number of which the sine is to be calculated. Taylor and Maclaurin Series If a function $$f\left( x \right)$$ has continuous derivatives up to $$\left( {n + 1} \right)$$th order, then this function can be expanded in the following way:. On the other hand, this shows that you can regard a Taylor expansion as an extension of the Mean Value Theorem. 1 Introduction This chapter has several important and challenging goals. With modern calculators and computing software it may not appear necessary to use linear approximations. , the Riemann zeta function evaluated at p. Maclaurin Series Calculator. The taylor command computes the order n Taylor series expansion of expression, with respect to the variable x, about the point a. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. Constructing a Taylor Series [ edit ] Several methods exist for the calculation of Taylor series of a large number of functions. Each term of the Taylor polynomial comes from the function's derivatives at a single point. They suggest using T(x) = 1+ax^2+bx^4+cx^6 and (cosx)(secx)=1. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. We use the power series for the sine function (see sine function#Computation of power series): Dividing both sides by (valid when ), we get: We note that the power series also works at (because ), hence it works globally, and is the power series for the sinc function. (Assume that 'f' has a power series expansion. 1) y(0) = y0 This equation can be nonlinear, or even a system of nonlinear equations (in which case y is a vector and f is a vector of n diﬀerent functions). 2, is a Taylor series centered at zero. ADVERTISEMENTS: Read this article to learn how to Calculate Standard Deviation in 3 different Series! A. Taylor and Laurent Series We think in generalities, but we live in details. TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. 10 Taylor and Maclaurin Series 677 If you know the pattern for the coefficients of the Taylor polynomials for a function, you can extend the pattern easily to form the corresponding Taylor series. When a = 0, Taylor’s Series reduces, as a special case, to Maclaurin’s Series. F(t0 + ∆t) ≈ F(t0) +F′(t0)∆t. To find the value of sin 1 (in radians), a calculator will use the Maclaurin Series expansion for sin x, that we found earlier. i+1 = (i+ 1)h, we may solve for the accelerations in terms of the displacements, velocities, and the applied forces. How can I get a Taylor expansion of the Sin[x] function by the power series? Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. While all the previous writeups have concentrated on the fact that you can calculate a Taylor series by calculating the appropriate derivatives, this is often not the best solution, especially if you just need a few terms. Example 1 Taylor Polynomial Expand f(x) = 1 1–x – 1 around a = 0, to get linear, quadratic and cubic approximations. Arribas, A. Rapid Calculation of Spacecraft Trajectories Using Efficient Taylor Series Integration Software greatly accelerates the calculation of spacecraft trajectories. Use Taylor's formula to estimated the accuracy En(x) of the approximation Tn(x) to f(x) when for -4 < x < 4 and n=4. factorial(i) sign = -sign return cosx x. Expansions of e. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Chapter 4: Taylor Series 19 1 + x + x2 2! + x3 3! +··· xn n! +···= ∞ i=0 xi i! As another example we calculate the Taylor Series of 1 x. Once you get the basic idea it will be very easy to do. e-mail: [email protected] We find the desired polynomial approximation using the Taylor Series. In this section we present numerous examples that provide a number of useful procedures to find new Taylor series from Taylor series that we already know. Solution We will be using the formula for the nth Taylor sum with a = 0. A simple example is if we scale a function, say g(t) = 5f(t), the the Fourier series for g(t) is 5 times the Fourier series of f(t). It seems that, instead of Taylor series for int_0^x, it calculates Taylor series for int_x^{2x} I have only found this bug when calculating Taylor series for functions f defined as in the example (by means of an integral). For example, the derivative f (x) =lim h→0 f (x +h)−f (x) h is the limit of the difference quotient where both the numerator and the denominator go to zero. Part 2) After completing part 1, modify the series for faster convergence. sin(x) of java. The Taylor Series is also referred to as Maclaurin (Power) Series. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X∞ n=0 f(n)(b) n! (x−b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. Taking the first two terms of the series gives a very good approximation for low speeds. For instance, in Example 4 in Section 9. Annette Pilkington Lecture 33 Applications of Taylor Series. Why don't you code the formula to calculate Taylor series? The more terms you add, the better precision you get I suggest you see these articles about it:. Taylor and Maclaurin Series (27 minutes, SV3 >> 84 MB, H. In this video I'm going to show you how you can find a Taylor series. Solution: Since 1 1 −𝑥 = 𝑥𝑘 ∞ 𝑘=0 we get 𝑥2 1 −𝑥 = 𝑥2 𝑥𝑘 ∞ 𝑘=0 = 𝑥𝑘+2 ∞ 𝑘=0 J. Rapid Calculation of Spacecraft Trajectories Using Efficient Taylor Series Integration Software greatly accelerates the calculation of spacecraft trajectories. Created a rational approximation method like the Abramowitz & Stegun with 13 terms which I think is good to 10**(-12). As you increase the degree of the Taylor polynomial of a function, the approximation of the function by its Taylor polynomial becomes more and more accurate. Let us start with the formula 1 1¡x = X1 n=0. As an example, let’s use the Maclaurin polynomial (with just four terms in the series) for the function f(x) = sin(x) to approximate sin(0. Enter the x value and find the sin x value in fraction of seconds. This is not always a good value of a to pick. Either the integral test or the Cauchy condensation test shows that the p-series converges for all p > 1 (in which case it is called the over-harmonic series) and diverges for all p ≤ 1. "Write a program consisting of only the main function, called piApproximator. There is a lot of good information available on line on the theory and applications of using Padé approximants, but I had trouble finding a good example explaining just how to calculate the co-efficients. For example, we know from calculus that es+t = eset when s and t are numbers. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Build your own widget. Maclaurin series are simpler than Taylor's, but Maclaurin's are, by definition, centered at x = 0. The fact-checkers, whose work is more and more important for those who prefer facts over lies, police the line between fact and falsehood on a day-to-day basis, and do a great job. Today, my small contribution is to pass along a very good overview that reflects on one of Trump’s favorite overarching falsehoods. Namely: Trump describes an America in which everything was going down the tubes under  Obama, which is why we needed Trump to make America great again. And he claims that this project has come to fruition, with America setting records for prosperity under his leadership and guidance. “Obama bad; Trump good” is pretty much his analysis in all areas and measurement of U.S. activity, especially economically. Even if this were true, it would reflect poorly on Trump’s character, but it has the added problem of being false, a big lie made up of many small ones. Personally, I don’t assume that all economic measurements directly reflect the leadership of whoever occupies the Oval Office, nor am I smart enough to figure out what causes what in the economy. But the idea that presidents get the credit or the blame for the economy during their tenure is a political fact of life. Trump, in his adorable, immodest mendacity, not only claims credit for everything good that happens in the economy, but tells people, literally and specifically, that they have to vote for him even if they hate him, because without his guidance, their 401(k) accounts “will go down the tubes.” That would be offensive even if it were true, but it is utterly false. The stock market has been on a 10-year run of steady gains that began in 2009, the year Barack Obama was inaugurated. But why would anyone care about that? It’s only an unarguable, stubborn fact. Still, speaking of facts, there are so many measurements and indicators of how the economy is doing, that those not committed to an honest investigation can find evidence for whatever they want to believe. Trump and his most committed followers want to believe that everything was terrible under Barack Obama and great under Trump. That’s baloney. Anyone who believes that believes something false. And a series of charts and graphs published Monday in the Washington Post and explained by Economics Correspondent Heather Long provides the data that tells the tale. The details are complicated. Click through to the link above and you’ll learn much. But the overview is pretty simply this: The U.S. economy had a major meltdown in the last year of the George W. Bush presidency. Again, I’m not smart enough to know how much of this was Bush’s “fault.” But he had been in office for six years when the trouble started. So, if it’s ever reasonable to hold a president accountable for the performance of the economy, the timeline is bad for Bush. GDP growth went negative. Job growth fell sharply and then went negative. Median household income shrank. The Dow Jones Industrial Average dropped by more than 5,000 points! U.S. manufacturing output plunged, as did average home values, as did average hourly wages, as did measures of consumer confidence and most other indicators of economic health. (Backup for that is contained in the Post piece I linked to above.) Barack Obama inherited that mess of falling numbers, which continued during his first year in office, 2009, as he put in place policies designed to turn it around. By 2010, Obama’s second year, pretty much all of the negative numbers had turned positive. By the time Obama was up for reelection in 2012, all of them were headed in the right direction, which is certainly among the reasons voters gave him a second term by a solid (not landslide) margin. Basically, all of those good numbers continued throughout the second Obama term. The U.S. GDP, probably the single best measure of how the economy is doing, grew by 2.9 percent in 2015, which was Obama’s seventh year in office and was the best GDP growth number since before the crash of the late Bush years. GDP growth slowed to 1.6 percent in 2016, which may have been among the indicators that supported Trump’s campaign-year argument that everything was going to hell and only he could fix it. During the first year of Trump, GDP growth grew to 2.4 percent, which is decent but not great and anyway, a reasonable person would acknowledge that — to the degree that economic performance is to the credit or blame of the president — the performance in the first year of a new president is a mixture of the old and new policies. In Trump’s second year, 2018, the GDP grew 2.9 percent, equaling Obama’s best year, and so far in 2019, the growth rate has fallen to 2.1 percent, a mediocre number and a decline for which Trump presumably accepts no responsibility and blames either Nancy Pelosi, Ilhan Omar or, if he can swing it, Barack Obama. I suppose it’s natural for a president to want to take credit for everything good that happens on his (or someday her) watch, but not the blame for anything bad. Trump is more blatant about this than most. If we judge by his bad but remarkably steady approval ratings (today, according to the average maintained by 538.com, it’s 41.9 approval/ 53.7 disapproval) the pretty-good economy is not winning him new supporters, nor is his constant exaggeration of his accomplishments costing him many old ones). I already offered it above, but the full Washington Post workup of these numbers, and commentary/explanation by economics correspondent Heather Long, are here. On a related matter, if you care about what used to be called fiscal conservatism, which is the belief that federal debt and deficit matter, here’s a New York Times analysis, based on Congressional Budget Office data, suggesting that the annual budget deficit (that’s the amount the government borrows every year reflecting that amount by which federal spending exceeds revenues) which fell steadily during the Obama years, from a peak of $1.4 trillion at the beginning of the Obama administration, to$585 billion in 2016 (Obama’s last year in office), will be back up to $960 billion this fiscal year, and back over$1 trillion in 2020. (Here’s the New York Times piece detailing those numbers.) Trump is currently floating various tax cuts for the rich and the poor that will presumably worsen those projections, if passed. As the Times piece reported:


# Theorems on Rationals

We present three theorems involving rational numbers.

### $$n$$th Power Theorem. Let $$n$$ be a positive integer. If $$a$$ is a positive integer such that $$a = r^n$$ for some rational number $$r$$, then $$r$$ must be an integer.

Proof: We apply the integer root theorem to the polynomial $$x^n - a$$. Since this polynomial has a rational root $$r$$, this root must be an integer.$$_\square$$

### Theorem 2. If the sum and product of two rational numbers are both integers, then the two rational numbers must be integers.

Proof: We have $$r_1 + r_2 = b$$ and $$r_1 \cdot r_2 = c$$, hence $$r_1, r_2$$ are rational roots of the polynomial $$x^2 - bx + c$$. By the integer root theorem, $$r_1$$ and $$r_2$$ are both integers. $$_\square$$

### 2. $$a$$ divides both $$b$$ and $$c$$.

Proof: By the Remainder Factor Theorem, if the roots are integers $$n$$ and $$m$$, then $$f(x) = a (x-n)(x-m)$$. By comparing terms, we obtain $$b = -a(n+m)$$ and $$c = anm$$. Hence, (2) is satisfied. Furthermore, $b^2 - 4ac = a^2 (n+m)^2 - 4 a \times anm = a^2 (n-m)^2,$ so (1) is also satisfied.

Conversely, by the quadratic formula, the roots of $$f(x)$$ are $$\frac {-b \pm \sqrt{b^2- 4ac} }{2a}$$. By condition (1), it follows that the roots are rational. By condition (2) and Vieta's formula, both the sum and product of the roots are integers. Hence by Theorem 2, the roots are integers. $$_\square$$

Note by Calvin Lin
4 years ago

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what's &nbsp?

- 3 years, 8 months ago

It was meant to be a non-breaking space, to separate out indented paragraphs. I wanted the theorems to be displayed individually, instead of in a long chunk of text.

Staff - 3 years, 8 months ago

Something like a non-breaking space.

Edit: Actually, it's

- 3 years, 8 months ago


# A bird flies between two cars infinite times [duplicate]

The following question and answer is taken from careerbless

A naughty bird is sitting on top of the car. It sees another car approaching it at a distance of 12 km. The speed of the two cars is 60 kmph each. The bird starts flying from the first car and moves towards the second car,reaches the second car and come back to the first car and so on. If the speed at which bird flies is 120kmph then

1)the total distance travelled by the bird before the crash is?

2)the total distance travelled by the bird before it reaches the second car for the second time is?

3)the total number of times that the bird reaches the bonnet of the second car is(theoretically)?

I am clear with the explanation available in the mentioned site for 1 and 2.

For the third part, i.e., the total number of times that the bird reaches the bonnet of the second car is(theoretically), answer given is infinite times with the explanation proving the infinite sequence.

But we know it cannot go on infinite times. It has to be a finite number if we look at it practically. How to explain this?

I am puzzled because the explanation looks convincing (infinite times) whereas we know it cannot be infinite times. Please help in understanding this contradiction.

EDIT: Adding the explanation(below) from careerbless as advised by @DylanSp for clarify and my aim is to understand why it looks as infinite times(theoretically) whereas we know it is finite practically or where my understanding is wrong. (it was a detailed explanation and that is why I have not added initially)

(3) infinite times

As explained for the previous case,

The bird reaches the second car in $\dfrac{12}{180}=\dfrac{1}{15}$ hour for the first time.

In this time, the cars together covers a distance of $\dfrac{1}{15}×120=8$ km and therefore the distance between the cars becomes 12-8=4 km.

The bird reaches back the first car in $\dfrac{4}{180}=\dfrac{1}{45}$ hour.

In this time, the cars together covers a distance of $\dfrac{1}{45}×120=\dfrac{8}{3}$ km and therefore the distance between the cars becomes $4-\dfrac{8}{3}=\dfrac{4}{3}$ km.

Now the bird flies to the second car for the second time. It takes $\dfrac{\left(\dfrac{4}{3}\right)}{180}=\dfrac{1}{135}$ hour for this.

In this time, the cars together covers a distance of $\dfrac{1}{135}×120=\dfrac{8}{ 9}$ km and therefore the distance between the cars becomes $\dfrac{4}{3}-\dfrac{8}{9}=\dfrac{4}{9}$ km.

The bird reaches back the first car in $\dfrac{\left(\dfrac{4}{9}\right)}{180}=\dfrac{1}{405}$ hour.

In this time, the cars together covers a distance of $\dfrac{1}{405}×120=\dfrac{8}{ 27}$ km and therefore the distance between the cars becomes $\dfrac{4}{9}-\dfrac{8}{27}=\dfrac{4}{27}$ km.

Now the bird flies to the second car for the third time. It takes $\dfrac{\left(\dfrac{4}{27}\right)}{180}=\dfrac{1}{1215}$ hour for this.

so on.

Sine this goes on repeatedly, the bird reaches the bonnet of the second car infinite times(theoretically)

## marked as duplicate by Ron Gordon sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 18 '16 at 18:56

• Can you add the explanation from that site to your question here? Questions should be self-contained as much as possible. – DylanSp Feb 18 '16 at 18:39
• It is not much different from Zeno's Paradoxes. Theoretically we can talk about an infinite number of iterations of still smaller distances adding up to a finite distance as a total. It makes sense theoretically. – String Feb 18 '16 at 18:43
• @ DylanSp, added – Kiran Feb 18 '16 at 18:47
• @String, never thought this will open up a completely new area for me which I was not aware. But, we can find sum of an infinite geometric progression(when r<1) to a finite number, right? so it could be the mathematical explanation? – Kiran Feb 18 '16 at 18:49
• This is why the question asks for a theoretical answer rather than a practical one. Also, I think you can give a much simpler explanation of why the answer is infinity: in short, the bird flies faster than the cars, and at every step the distance between the cars is positive. Finally, if you wanted to make the scenario slightly more realistic, you could suppose that it takes the bird one second, say, to turn around. There will be a point where the cars are less than one second apart, so the bird can only change direction finitely many times. – Théophile Feb 18 '16 at 18:50


# Lecture 022

## Number Theory

Example:

• Euclid: there exists infinite many positive integer s.t. x^2+y^2=z^2 and grd(x, y, z) = 1

• Fermat's Last Theorem: there don't exist any positive integer x, y, z s.t. x^n+y^n=z^n for n>=3

• Lagrange's Four Square Theorem: Every non-negative integer can be written as the sum of 4 perfect squares

• Golbach's Conjecture: every even integer > 2 can be expressed as the sum of 2 primes (e.g. 14=3+11=7+7)

• Twin Prime conjecture: there are infinite many pairs of primes of the form (p, p+2)

Prime: let $n\in \mathbb{Z}$ s.t $n\geq 2$. n is prime iff its only positive divisors are 1 and itself.

• Theorem: for every integer > 2, is either a prime or a product of primes (every integer has at least one prime factorization)

• Lemma: for non zero m, n, if m|n, then $|m|\leq|n|$

Composite: n is not prime (ie. $(\exists a,b \in \mathbb{Z})(1< a \leq b)

Theorem: if n is composite, then n has a prime factor $p<=\sqrt{n}$

• prove by definition of composite and contradiction and prime factorization (either prime or product of prime)

• Lemma: if $n \in \mathbb{Z} \ {0}$, then n has finitely many divisors (so we always have finite many common divisors) TODO: why

Common Divisor: for $a, b \in \mathbb{Z}$ not both 0, integer d is called a common divisor iff $d|a$ and $d|b$. d is called the greatest common divisor(d=gcd(a, b)) iff d is the greatest of the common divisor

• $d=gcd(a,b) \iff d|a \land d|b \land (\forall x \in \mathbb{Z})(x|a \land x|b \implies x \leq d)$

• note: negative number has the same set of divisors as its positive

• note: gcd(a, b) = gcd(|a|, |b|)

• note: gcd(0, n) = n

• note: gcd(0, 0) undefined

• note: gcd = product of common prime factor

coprime(relatively prime): iff gcd(a,b) = 1

• divide a and b by gcd(a, b) to ensure they are relatively prime

• means no common divisor other than 1.

Theorem: for none zero a,b. d=gcd(a,b) then gcd(a/d, b/d) = 1

• proof: let n be common divisor of a/d and b/d. then |nd| is common divisor of a and b by algebra. |n|d <= d because d=gcd(a,b), n=+-1, so the greatest(among -1 and 1) common divisor is 1.

Euclidean Algorithm:

• gcd(x, y) = gcd(x+my, y) where x can be seen as remainder and m is divisor.

Division Algorithm:

• if a, b \in \mathbb{Z} with b > 0, then there exists unique q,r \in \mathbb{Z} s.t. a=bq+r and 0<=r<b

• proof: consider the set of remainder that divide a by 1~b for a,b: $S_{(a,b)} = \{a-bk | k \in \mathbb{Z} a-bk \geq 0 \}$, set is non-empty, has least element (by WOP-well-ordering-property) then we find the lowest reminder. WTS q,r pair unique, and r<b.

• r=0, then r-b \in S because a-b(g+1) > 0. r-b<r and r is the least element contradicts.
• q,r unique: a=bq_1 + r_1, b=bq_2 + r_2 (r1, r2 between 0 and b), WTS r1=r2,q1=q2. then -b<-r1<=0, -b<b(q1-q2)<b, -1<(q1-q2)<1, q1=q2, r1=r2

Corollary to Division Algorithm (DA): let a, b, in \mathbb{Z} with b!=0. if a=bq+r, then gcd(a,b) = gcd(b,r)

• note that q and r doesn't have to be correct divisor or remainder

• proof: let d1=gdc(a,b), d2=gcd(b,r). $d1|a \land d1|b \implies (\exists k,l \in \mathbb{Z})(a=ka_1 \land b=ld_1)$...

TODO: don't understand

Proposition: $a|b \land a|c \implies a|bx+cy$

Euclidean Algorithm (formal)

Bezout's Lemma

1. $(\exists x, y \in \mathbb{Z})(ax+by = gcd(a,b)))$
2. $(x,y \in \mathbb{Z} \land ax+by>0 \implies ax+by \geq gcd(a,b))$ TODO: proof ignored

3. corollaries: let a,b \in \mathbb{Z}, not both 0. let d=gcd(a,b) then

1. if $t \in \mathbb{Z}$ s.t. $t|a$ and $t|b$, then $t|d$. (greatest common divisor divides any common divisor)
2. $gcd(a,b) | c \iff (\forall c \in \mathbb{Z})(\exists m, n \in \mathbb{Z})(c=am + bn)$
3. if a and b are relatively prime, then $(\exists m, n \in \mathbb{Z})(am+bn=1)$
4. $(\forall m \in \mathbb{Z}^+)(gcd(ma, mb)= m \times gcd(a,b))$

TODO: proof ignored

Table of Content


# Can you determine a set of values from a set of distinct sums

Consider an array of positive integers $$A$$ of length $$n$$. Now consider the set of sums of all the contiguously indexed subarrays of $$A$$. For example if $$A = (1,3,5,6)$$ then the set would be $$S_A = \{1,3,5,6,4,8,11,9,14,15\}$$.

If the sums of all contiguously indexed subarrays are distinct (as they are in the example above), does the set of these sums uniquely specify the set of integers in the array the sums were calculated from?

We can certainly compute the smallest element in the original array as it is the smallest element in $$S_A$$. Similarly there must be a value in the original array which is the largest value in $$S_A$$ minus the second largest.

To show one of the subtleties of this problem, consider $$A = (1, 6, 2, 3)$$ and $$S_A = \{1, 6, 2, 3, 7, 8, 5, 9, 11, 12\}$$. We can immediately tell from $$S_A$$ that $$1$$ occurs somewhere in $$A$$. Similarly we can tell that $$2$$ occurs somewhere in $$A$$. But what can we tell about $$3$$? If $$1$$ and $$2$$ were next to each then as $$1+2=3$$ we would know that $$3$$ can't be in $$A$$. But if $$1$$ and $$2$$ are not next to each in $$A$$ then we know $$3$$ must be in $$A$$. How do we tell which case we are in?

The answer turns out to be NO. Take $$A = (4, 6, 5, 2, 1)$$ and $$B = (3, 8, 2, 4, 1)$$. We have that $$S_A = S_B$$ but the set of elements in $$A$$ and $$B$$ are distinct.

• yeah, there have been a few questions recently (in the last few months) dealing with subsequence sums of an array. so i was just wondering where all the common interest comes from. :) OK, lets clarify a few things: (1) $S_A$ is a SET, so you should use curly braces like $S_A = \{1,3,5,...\}$. (2) your highlighted question asks does $S_A$ uniquely specify the SET of integers, e.g. $\{1, 3, 5, 6\}$, but is that what you mean? Or do you mean to ask if $S_A$ uniquely specify the SEQUENCE (ARRAY) of integers e.g. $(1,3,5,6)$ -- which may still be true, modulo sequence reversal $(6,5,3,1)$? – antkam Nov 9 '18 at 20:26
• @antkam I asked about the set only because it’s a weaker claim. That is it is more likely to be true. Of course if it’s also true for the array, modulo reversal, that’s even better. (Fixed bracket error now) – felipa Nov 9 '18 at 20:51
• If $S_A$ is a set, shouldn't it be $\{1,3,4,\cdots\}$, i.e. an ordered list without repetitions ? – G Cab Nov 10 '18 at 19:48
• This is a version of the “turnpike” problem, or equivalently the “partial digest” problem. It is known that there can be very many distinct arrays with the same set of substring sums for large $n$, however I am not aware of any results with the restriction that all the sums are distinct. Nice question. – Erick Wong Nov 10 '18 at 21:13
• @mathlove No sorry that isn't right. The question is if for all arrays A (which are ordered) does $S_A$ (unordered) uniquely determine the set (unordered) of integers in A under the constraint that the subarray sums of A are all distinct. In your example the set $\{1,2,3\}$ does indeed tell us that A contains exactly the integers 1 and 2 (but not their order). – Anush Nov 14 '18 at 10:47

The answer turns out to be NO. Take $$A = (4, 6, 5, 2, 1)$$ and $$B = (3, 8, 2, 4, 1)$$. We have that $$S_A = S_B$$ but the set of elements in $$A$$ and $$B$$ are distinct.
• To clarify, there are no counter examples at all for length < 5 since it is known that every array of length $\le 4$ is uniquely determined (up to reversal) by its multi set of substring sums. – Erick Wong Nov 15 '18 at 21:07
• There is a decent amount known about this actually. I recommend reading “Reconstructing Sets from Interpoint Distances” by Skiena, Smith, Lemke, which you should be able to find a PDF for. They give a nice argument that the set of solutions for a given set of sums has a hypercube structure, in particular it is always a power of $2$ (if non-zero). – Erick Wong Nov 15 '18 at 22:18


# Polynomials sprint: How to solve a seemingly disgusting polynomial

On page 248, it is shown how to solve the polynomial $$x^4+x^3+x^2+x+1=0$$. In this note, I will explain how to solve the polynomial $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ First we multiply both sides by $$(x-1)$$ to get $x^9-1=0\implies x^9=1$ Then since $$x^9=1$$, we can divide some terms by $$x^9$$ $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=$ $\dfrac{x^8}{x^9}+\dfrac{x^7}{x^9}+\dfrac{x^6}{x^9}+\dfrac{x^5}{x^9}+x^4+x^3+x^2+x+1=$ $\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+\dfrac{1}{x^4}+x^4+x^3+x^2+x+1$ $\left(x+\dfrac{1}{x}\right)+\left(x^2+\dfrac{1}{x^2}\right)+\left(x^3+\dfrac{1}{x^3}\right)+\left(x^4+\dfrac{1}{x^4}\right)+1=0$ Now we let $$y=x+\dfrac{1}{x}$$ so we have $(y)+(y^2-2)+(y^3-3y)+((y^2-2)^2-2)+1=0$ which after simplification becomes $y^4+y^3-3y^2-2y+1=0$ Using the Rational Root Theorem (discussed on page 246) we quickly find that $$y=-1$$ is a root and factor the polynomial as $(y+1)(y^3-3y+1)$ Now the rest is simple! We can use the cubic solving method discussed on pages 247-248 to find the roots of $$y^3-3y+1$$ Then simple quadratic bashing gives us our roots!

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4 years, 2 months ago

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\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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Or, we can use the 9th roots of unity.

- 4 years, 2 months ago

Yes, what happened to the classic roots of unity? Roots are just $$e^{2ki\pi/9}$$with $$k=1\to 8$$.

- 4 years, 2 months ago

Except all roots of unity give you is $$e^{\frac{2\pi i}{9}}$$ and not a numerical answer with radicals. The method here allows you to compute $$\sin {40^{\circ}}$$

- 4 years, 2 months ago

Or you can also just do some Euler's formula with $$e^{\dfrac{2\pi i}{9}}$$.

- 4 years, 2 months ago

ummmm which euler's formula?

- 4 years, 2 months ago

$$e^{i\phi}$$ = cos$$\phi$$ + $${i}$$sin$$\phi$$ ... substituting $$\pi$$ gives $$e^{i\pi}$$ = -1, where $${i}$$ = $$\sqrt{-1}$$

- 3 years, 8 months ago

Where i can find this book

- 3 years, 8 months ago

$$e^{i\theta}=\text{cis}(\theta)$$

- 4 years, 2 months ago

How does that let you solve for $$\sin{40^{\circ}}$$

- 4 years, 2 months ago

wow

- 4 years, 2 months ago

:o

- 4 years, 2 months ago

2 Daniels

- 4 years, 2 months ago

Oh God, this is not good. One Daniel is enough for me, but two? Double Trouble.

- 4 years, 2 months ago

2 Lius

- 4 years, 2 months ago

2 14-year-old-Daniel-Lius. Enough to get a conversation off-topic.

- 4 years, 2 months ago

2 14-year-old-Daniel-Lius-residing-in-USA. conversation closed.

- 4 years, 2 months ago

Where are all other pages?

- 3 years, 8 months ago

Yes I used the same to expand cosnx

- 3 years, 8 months ago

@Nathan Ramesh Can you add this to the Roots of Unity Applications Wiki page?

Select "Write a summary", and then copy-paste your text into it (with minor formatting adjustments if relevant. Thanks!

Staff - 3 years, 11 months ago

Done! I put it at the bottom. Let me know if it is bugged (I posted it from my phone). Thanks!

- 3 years, 11 months ago


# A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as log e ). Natural logarithms have the same properties.

## Presentation on theme: "A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as log e ). Natural logarithms have the same properties."— Presentation transcript:

A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as log e ). Natural logarithms have the same properties as log base 10 and logarithms with other bases. The natural logarithmic function f(x) = ln x is the inverse of the natural exponential function f(x) = e x.

The domain of f(x) = ln x is {x|x > 0}. The range of f(x) = ln x is all real numbers. All of the properties of logarithms from Lesson 7-4 also apply to natural logarithms.

Simplify. A. ln e 0.15t B. e 3ln(x +1) ln e 0.15t = 0.15te 3ln(x +1) = (x + 1) 3 ln e 2x + ln e x = 2x + x = 3x Example 2: Simplifying Expression with e or ln C. ln e 2x + ln e x

Simplify. a. ln e 3.2 b. e 2lnx c. ln e x +4y ln e 3.2 = 3.2 e 2lnx = x 2 ln e x + 4y = x + 4y Check It Out! Example 2

The formula for continuously compounded interest is A = Pe rt, where A is the total amount, P is the principal, r is the annual interest rate, and t is the time in years.

What is the total amount for an investment of \$500 invested at 5.25% for 40 years and compounded continuously? Example 3: Economics Application The total amount is \$4083.08. A = Pe rt Substitute 500 for P, 0.0525 for r, and 40 for t. A = 500e 0.0525(40) Use the e x key on a calculator. A ≈ 4083.08

The half-life of a substance is the time it takes for half of the substance to breakdown or convert to another substance during the process of decay. Natural decay is modeled by the function below.

Download ppt "A logarithm with a base of e is called a natural logarithm and is abbreviated as “ln” (rather than as log e ). Natural logarithms have the same properties."

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# Validity of the change-of-base formula for all bases

Suppose I want to solve $3^x=10$. I convert it to logarithmic form $$\log_{3}10=x$$ then change bases $$\frac{\log10}{\log3}=x$$ and this will yield the same answer as if I had written $$\frac{\ln10}{\ln3}=x$$ But log and ln have different bases. Why does this work both ways?

• Your first "inserted" equation is wrong. It should be $x = \log_3 10$, not $x=\log_{10} 3$. Generally, $a^b=c \iff b=\log_a c$.
– MPW
Sep 1 '16 at 16:54
• @MPW Right, I had to edit the question. And the second half was just rambling about the first half. Sep 1 '16 at 16:59
• @ParclyTaxel Thank you very much for the concise summary. Sep 1 '16 at 17:08
• Mathematicians writing "$\log$" usually mean the same thing as "$\ln$". The convention used by many engineers, biologists, astronomers, and others, that $\log$ means the base-$10$ logarithm, was to a large extent rendered obsolete by in advent of calculators around 1970 or so, but it is perpetuated by calculators. Base-10 logarithms are useful in computations by hand because you only need a table going from $1$ to $10$: If you want the logarithm of $145$, you find the logarithm of $1.45$ in the table and then add $2$ to it since the decimal point is $2$ places to the right of there. $\qquad$ Sep 1 '16 at 17:20
• It used to be that if you wanted to tell students how to find things like $\log_3 20$ with a calculator, you told them it's $\dfrac{\log_{10} 20}{\log_{10} 3}$ or else it's $\dfrac{\ln 20}{\ln 3}$, and they got base-$10$ or base-$e$ logarithms from the calculator. But today you find that some of them have calculators on which you enter both the base and the argument. (I haven't yet encountered on that gives GCDs. Maybe that means I'm not up to date?) $\qquad$ Sep 1 '16 at 17:22

You can use whatever basis: the equality $3^x=10$ is equivalent to $$\log_a 3^x=\log_a 10$$ for any $a>0$, $a\ne1$. Since $\log_a 3^x=x\log_a 3$, we get $$x=\frac{\log_a 10}{\log_a 3}$$ so you see that the final result is independent of the base.

You can get the change of base formula by considering $b^c=k$ that says $c=\log_b k$; but we also have $c\log_a b=\log_a k$ and therefore $$\log_b k=\frac{\log_a k}{\log_a b}$$ In the case of $b=3$, $k=10$ and $a=e$, you get $$\log_3 10=\frac{\ln 10}{\ln 3}$$

• Perhaps more to the point is the fact that $$\frac{\log_a 10}{\log_a 3}= \frac{\frac{\log_a 10}{\log_a b}}{\frac{\log_a 3}{\log_a b}} = \frac{\log_b 10}{\log_b 3}$$ so the choice of base doesn't affect the result.
– MPW
Sep 1 '16 at 17:07
• @MPW Since $a$ is completely arbitrary (provided $a>0$ and $a\ne1$), there's no need to do that computation. Sep 1 '16 at 17:12
• I know you know that. I just meant to demonstrate to OP explicitly that they are the same.
– MPW
Sep 1 '16 at 18:02

Possibly the best way to see this is to write a change to an arbitrary base instead of choosing either $\log_{10}$ or $\ln$ initially. (There is already an answer that does that.) But I think it's also noteworthy that the change-of-base formula itself provides an easy derivation of the fact that $$\frac{\log10}{\log3}=\frac{\ln10}{\ln3}.$$

Starting with $\frac{\ln10}{\ln3}$ (for example), simply apply the formula to change the base from $e$ to $10$ on both the numerator and denominator: \begin{align} \ln10 &= \log_e 10 = \frac{\log_{10} 10}{\log_{10} e}, \\ \ln3 &= \log_e 3 = \frac{\log_{10} 3}{\log_{10} e}, \\ \frac{\ln10}{\ln3} &= \frac{\left( \frac{\log_{10} 10}{\log_{10} e} \right)} {\left(\frac{\log_{10} 3}{\log_{10} e} \right)} \end{align} Multiply both the numerator and denominator of the right-hand side of the last equation by $log_{10} e$, and we find that $$\frac{\ln10}{\ln3} = \frac{\log_{10} 10}{\log_{10} 3}$$

In other words, every time we change the base of both the numerator and denominator from base $b$ to base $c$ simultaneously, we divide both the numerator and denominator by $\log_c b$, and those two operations cancel each other out.

In your question note that the definition of the logarithm implies that $$\log_310=x \iff 3^x=10 \tag 1.$$ In general $$\log_bw=x \iff b ^x=w \tag 2$$ where $b>0, b\neq 1$.

Now take the logarithm of both sides to some base $c>0$, $c\neq 1$ and use the fact that $\log_c(b^x)=x\log_cb$ to get that $$x\log_cb=\log_cw\implies x=\frac{\log_cw}{\log_cb}.\tag 3$$ Equations $(2)$ and $(3)$ imply that$$\log_bw=\frac{\log_cw}{\log_cb}. \tag 4$$ Take $c=10$ and $c=e$ in equation $(4)$ to get the equality $$\log_{3}10=\frac{\log_{10}10}{\log_{10}3}=\frac{\ln10}{\ln3}. \tag 5$$


# Kernel and Image of a Linear Transformation

I have the following linear transformation

$$L:\Bbb R^3\rightarrow \Bbb R^2, (x_1,x_2,x_3)\mapsto(x_3+x_1,x_2-x_1)$$

And I want to determine the kernel and the image of $L$.

$\text{ker}(L):=v\in V \space | \space L(v)=0$

Is it accurate to say that I want to find the set of vectors that will map to the zero-vector when plugged into the linear transformation?

In matrix form:

$$\begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$

$$\implies x_1+x_3=0 \iff x_1=-x_3 \space \space \space \text{and} \space \space \space x_2-x_1=0 \iff x_2=x_1$$

Therefore:

$$ker(L)=\begin{pmatrix}t\\t\\-t\end{pmatrix}$$

Is this correct?

What is the image of a linear transformation? Is it the subspace of the co-domain that the linear transformation actually maps to? So for example, if I had the transformation:

$M:\Bbb R^2 \rightarrow \Bbb R, (x_1,x_2) \mapsto (2)$

The Image of $Q$ would be {$2$}? How do I determine the image for much more complicated linear transformations?

• Small detail: In your last example, $M$ is not linear ($M(\lambda X) \neq \lambda M(X)$) – Clement C. Jun 13 '15 at 14:25
• @ClementC. You're right. I will fix that. – qmd Jun 13 '15 at 14:27

As soon as you have the matrix of a linear map $f$, you have a system of generators of the image of $f$.

Indeed the column-vectors of the matrix are the images of the vectors of the basis in the source-space, say $v_1,v_2,v_3$. Hence for any vector $v$ in $\mathbf R^3$, $v=\lambda_1 v_1+\lambda_2v_2+\lambda_3v_3$, we have: $$f(v)=\lambda_1 f(v_1)+\lambda_2f(v_2)+\lambda_3f(v_3).$$

Naturally, this system of generators is not minimal, i. e. it is not a basis of the image in general. But from this system, you can deduce a basis by column reduction.

It will not be necessary here, because the rank-nullity theorem ensures $\dim\operatorname{Im}f=2$ since you've found that $\dim\ker f=1$. Thus, $f$ is surjective, i. e. the image of $f$ is the whole of $\mathbf R^2$.

• So in this case I have the matrix $\begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}$ with columns $\begin{pmatrix}1\\1\end{pmatrix}$,$\begin{pmatrix}0\\1\end{pmatrix}$, $\begin{pmatrix}1\\0\end{pmatrix}$. Are these column-vectors the image of $L$? Is it always the case that the column-vectors are the image? – qmd Jun 13 '15 at 14:51
• The column-vectors are not the image of $L$: they span the image of $L$. But $\begin{pmatrix}1\\-1\end{pmatrix}$ andt $\begin{pmatrix}0\\1\end{pmatrix}$ also generate the image. – Bernard Jun 13 '15 at 14:55
• I see. So I have to look at that column vectors and calculate the span. In this case $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ span $\Bbb R^2$ and therefore Im($L$)=$\Bbb R^2$? – qmd Jun 13 '15 at 14:58
• Yes. You also know that here without any calculation thanks to the rank-nullity theorem. – Bernard Jun 13 '15 at 15:16

An option to characterize all $\vec{y}\in\mathrm{im}(L)$ is to express that there must be a $\vec{x}$ such that $L( \vec{x}) = \vec{y}$. That is, write $$\begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}y_1\\y_2\end{pmatrix}$$ and solve for $\begin{pmatrix}y_1\\y_2\end{pmatrix}$ as a function of of $x_1,x_2,x_3$. This will give you the set (when $x_1,x_2,x_3$ vary in $\mathbb{R}$ of all $\vec{y}$'s in the image.

• Thanks. Can you give me an intuitive explanation of what the image is? Is it, as I asked in my question, "the subspace of the co-domain that the linear transformation actually maps to?" or am I confusing something here? Solving the linear system in your answer I get: $x_1+x_3=y_1$ and $x_2-x_1=y_2$. I don't understand how that helps me. – qmd Jun 13 '15 at 14:43
• The image is the set of all poin ts that are "reached" by the linear transformation. In your case, it's basically the set of all points that can be written $(a+b,c-a)$ for some $a,b,c\in\mathbb{R}$. (Also, one can show that here, this is actually the whole set $\mathbb{R}^2$: for any choice of $y_1,y_2$, you can solve the system for $x_1,x_2,x_3$ and find solutions: that is, any point in $\mathbb{R}^2$ is reached by $L$. – Clement C. Jun 13 '15 at 14:45


# Does $\varphi(1)=1$ if $\varphi$ is a field homomorphism?

Is it by definition that $\varphi(1)=1$ if $\varphi$ is a field homomorphism ?

My field theory lecture said that yes, but now $\varphi\equiv0$ is not a field homomorphism...

What is the 'standard' in mathematics ? (i.e. if someone says that $\varphi$ is a field homomorphism does it imply $\varphi(1)=1$ ?)

-
It is standard to assume $\varphi(1) = 1$. It isn't hard to show that if $\varphi(1)\neq 1$, then $\varphi\equiv 0$. Thus assuming $\varphi(1) = 1$ is just to rule out this case. –  froggie May 6 '12 at 14:25
Why would you want $0$ to be a homomorphism? It's not like you can add field (or ring) homomorphisms. –  Zhen Lin May 6 '12 at 14:39
Hi, it seems that my answer is somewhat misleading and possibly even wrong. Please unaccept it so I can delete the answer. Pete Clark has agreed to write an answer of his own, which I am sure will be much better than mine. –  Asaf Karagila May 6 '12 at 17:40
@AsafKaragila - done. –  Belgi May 6 '12 at 18:17

Upon request, I am leaving an answer.

First, let's acknowledge that this is a question about conventions first and only secondarily about mathematics. Your lecturer's definition of a field homomorphism is whatever she says it is: that's what a definition means. ("When I use a word, it means just what I choose it to mean -- neither more nor less." -- Humpty Dumpty) On the other hand, definitions, while in some formal sense arbitrary, can certainly be good or bad, for both mathematical and sociological reasons.

Your lecturer's definition is a good one both because of its mathematical consequences and because, as she claimed, this is very much the "standard" definition, i.e., the one you will find in the vast majority of contemporary texts and courses, the one which is being used by practitioners both in field theory itself and in areas of mathematics where field theory is applied, and so forth.

The notion of "homomorphism" is so important in mathematics that there has been a lot of effort to make it as systematic and "un-(ad hoc)" as possible. Two very general frameworks for discussing homomorphisms are universal algebra and category theory. The basic objects of study in universal algebra are relational structures, namely sets equipped with various relations (and, what is a special case of that, functions), of various "arities".

From this perspective, the relational structure of a ring is a set $R$, two binary operations called $+$ and $\cdot$ and two "nullary functions" -- i.e., constants $0$ and $1$. There is more to the definition of a ring, namely the axioms that these relations must satisfy. In this sense, model theory is being overlayed on top of universal algebra. However, the definition of a homomorphism of relational structures doesn't depend on the axioms: a homomorphism of relational structures is just a map of the underlying sets which preserves, in a rather straightforward sense which I won't completely write out here (one can easily look it up online) all the relations. In the case of rings, this means that a homomorphism $f: R \rightarrow S$ is a map of the underlying sets such that

$\bullet$ For all $x,y \in R$, $f(x+y) = f(x) + f(y)$,
$\bullet$ For all $x,y \in R$, $f(x \cdot y) = f(x) \cdot f(y)$,
$\bullet$ $f(0) =0$, and
$\bullet$ $f(1) = 1$.

Again, I stress that this does not depend on the axioms (or, in more model-theoretic terminology, the theory, of the structure): a homomorphism of non-associative rings (with distinguished elements $0$ and $1$) is the same definition as a homomorphism of not necessarily commutative rings, which is the same definition as a homomorphism of fields. This uniformity is useful and powerful.

Now let $f: R \rightarrow S$ be a homomorphism of rings. An element $x \in R$ is a unit if there exists $y \in R$ such that $xy = yx = 1$. Two easy facts:

(i) For all $x \in R$, there is at most one $y \in R$ such that $xy = yx = 1$.
(Proof: If also $xz = zx = 1$, then $y = y \cdot 1 = y(xz) = (yx)z = 1 \cdot z = z$.)

Thus if $x$ is a unit, it has a well-defined inverse which we may denote $x^{-1}$. However, $x \mapsto x^{-1}$ is not a completely kosher function from the perspective of universal algebra, since it is not a function on $R$ but only on $R^{\times}$, the set of units of $R$. There are ways to get around this (a keyword is sorts) but the easiest solution is simply not to regard $x \mapsto x^{-1}$ is being part of the relational structure defining a commutative ring.

(ii) If $f: R \rightarrow S$ is a homomorphism of rings and $x \in R^{\times}$, then $f(x) \in S^{\times}$ and $f(x^{-1}) = f(x)^{-1}$.
(Proof: $f(x^{-1}) f(x) = f(x^{-1} x) = f(1) = 1 = f(1) = f(x x^{-1}) = f(x) f(x^{-1})$.)

Thus the preservation of inverses comes for free from this definition of ring homomorphisms.

There is also the categorical perspective, which is more general and thus more permissive. It has the bright idea that in the same breath as we define a mathematical object of a certain kind ("category"!), we should define the notion of homomorphism between objects of that category. In order to be a homomorphism, certain very mild axioms of composition and identities must be satisfied. So in particular, there is a category $\operatorname{Ring}$ in which the objects are rings and the morphisms are defined exactly as above. But there is also a category $\operatorname{Rng}$ in which the objects are "rngs", i.e., what you get by taking the constant $1$ out of the relational structure and taking out all the parts of the axioms for rings which refer to it. And there is even the category -- call it $\operatorname{Ring}'$ -- in which the objects are rings -- i.e., have $1$ -- but in which homomorphisms are not required to carry $1$ to $1$. In particular, for any rings $R$ and $S$, the identically zero map from $R$ to $S$ is a homomorphism in $\operatorname{Ring}'$ but not in $\operatorname{Ring}$ (except in the trivial case where $0 = 1$ in $S$).

It is worth spending a little time getting experience with the category $\operatorname{Ring}'$, both because (i) it is not simply a pathology: these sorts of maps between rings do come up sometimes and (ii) it gives you insight about the much more standard category $\operatorname{Ring}$. In particular, for a homomorphism $f: R \rightarrow S$ in $\operatorname{Ring}'$ $f(1)$ cannot be just any old element of $S$ but must be an idempotent element:

$f(1) = f(1 \cdot 1) = f(1) \cdot f(1)$.

Many rings have only $0$ and $1$ as idempotents -- in particular, this holds for fields -- so either $f(1) = 1$, in which case we're back to the standard case, or $f(1) = 0$, and in this case for all $x \in R$, $f(x) = f(x \cdot 1) = f(x) \cdot f(1) = f(x) \cdot 0 = 0$, so $f$ is identically zero.

So now we know exactly what we're excluding by defining a homomorphism of fields to satisfy $f(1) = 1$: we're excluding the identically zero map between fields. So we can be rather confident that we're not excluding anything important: certainly the zero map is not interesting in its own right, just as the identity element of a group or zero element of a ring is not really interesting in its own right. However, the identity element of a group and zero element of a ring are there for another very important reason: they enable the appropriate algebraic structure of group or ring to exist!

In contrast, if $R$ and $S$ are rings, then I see no intrinsic algebraic structure on the set of homomorphisms from $R$ to $S$: we cannot add them, multiply them (so as to get another homomorphism from $R$ to $S$) and so forth. So adding the zero element is not helping us in any evident way. In fact, if we don't believe this at first, no matter -- keep it in and see what happens. We'll find that in our further study of field theory, we'll keep having to say "except for the zero homomorphism" in results and arguments. For instance, most of field theory is based upon the fact that absolutely every field homomorphism $\iota: E \rightarrow F$ is injective and makes $F$ into an $E$-vector space...except for the zero homomorphism.

-


# How to find if this function is always zero?

The question I faced was:

Let $f(x)$ be a non-negative continuous and bounded function for all $x \ge 0$. If
$$(\cos x)f'(x) \le (\sin x - \cos x)f(x), \; \forall \; x \ge 0$$ then which of the following is/are correct?

(A) $f(6) + f(5) > 0$

(B) $x^2 - 3x + 2 + f(9) = 0$ has two distinct solutions

(C) $f(5)f(7) - f(6)f(5) = 0$

(D) $\lim\limits_{x \to 4} \dfrac{f(x) - \sin(\pi x)}{x-4} = 1$

(B), (C)

By observation, $f(x)=0$ satisfies the given conditions. But is it the only solution? If so, how to prove it is?

After rearranging the terms and combining them, I converted the inequality to this form: $$\left( f(x)\,\cos x \right)' + f(x)\,\cos x \le 0$$ Despite its allure, this inequality isn't getting me anywhere! It doesn't seem to have any information about $f(x)$, since it is stuck with a "$\cos x$". Even then, I don't see where I can go with it.

So how to solve this problem? Thank you.

• Your condition is equivalent to $$\frac{\mathrm{d}}{\mathrm{d}x} \ f(x) \cos x \ e^x \le 0$$ – Crostul Apr 28 '17 at 10:56
• @BST Actually the discriminant is $(-3)^2-4(2+f(9)) = 1-4f(9)$ which may be negative – Crostul Apr 28 '17 at 11:00
• Sorry, you're right. My mistake. I'll remove the comment. I miscalculated the discriminant. – be5tan Apr 28 '17 at 11:02

We assume that $f(x)$ is bounded and non-negative, for $x \geqslant 0$. This means that there is a non-negative function $g(x)$ with the property that $f(x) = g(x)e^{-x}$ for $x \geqslant 0$.

Plugging this into the inequality you found gives, $$0 \geqslant (f(x)\cos (x))' + f(x) \cos (x) = e^{-x} (g(x) \cos(x))'.$$ And since $e^{-x} > 0$, we have $$0 \geqslant (g(x) \cos(x))'.$$ This means that the function $g(x) \cos(x)$ is weakly decreasing. Because any point $x \in \mathbb{R}$ is between two zeroes of $\cos(x)$, we have that $g(x) \cos(x) = 0$ for all $x$.

• Instead of integrating the inequality, it gives more information simply to note that the derivative of $g(x)\cos x$ can never be positive. Since every $x$ is between two zeroes of $g(x)\cos x$, the mean value theorem forces it to be $0$ everywhere, not just where the cosine is positive. – Henning Makholm Apr 28 '17 at 11:07
• Ah, I missed that! Thank you, will edit my answer. – Peter Apr 28 '17 at 11:08
• Also, I think it would be slicker simply to define $g(x)=f(x)e^x$ at the beginning. Then we don't need to appeal to "bounded and nonnegative", and the first inequality is still true. – Henning Makholm Apr 28 '17 at 11:22
• This is brilliant! How did you figure out the substitution $f=ge^{-x}$? – FreezingFire Apr 28 '17 at 11:28
• I asked myself what function $h(x)$ would satisfy the strict version of your bound, i.e.~$h'(x) + h(x) = 0$. Evidently the function $h(x) = e^{-x}$ solves this equation. The rest was the result of a bit of experimentation with this function. – Peter Apr 28 '17 at 11:56

If you set $g(x)=f(x)\cos x$, your rearrangement tells you that $$\tag{1} g'(x) \le -g(x)$$

Consider an interval $[2\pi k-\frac12\pi , 2\pi k +\frac12\pi]$ where $\cos x$ is $\ge 0$. Then you know that $g(x)\ge 0$ on this interval, and $g(x)=0$ at the start of the interval. Then, because of (1), $g(x)$ must be identically zero on that interval, and so must $f(x)$.

This settles at least (A) and (C), because $5$, $6$ and $7$ are all within $\pi/2$ of $2\pi$.

For (B) and (D) this argument doesn't tell you enough; I recommend Peter's slicker answer instead.

Notice that whenever some $f$ satisfies the hypothesis, then so does any positive multiple of $f$. In particular:

Note that (B) is equivalent to $$0<(-3)^2-4(2+f(9)) = 1-4f(9)$$ i.e. $f(9)<1/4$. Hence (B) can hold if and only if $f(9) =0$ for any $f$ satisfying the hypothesis.


# Explain proof that any positive definite matrix is invertible

If an $n \times n$ symmetric A is positive definite, then all of its eigenvalues are positive, so $0$ is not an eigenvalue of $A$. Therefore, the system of equations $A\mathbf{x}=\mathbf{0}$ has no non-trivial solution, and so A is invertible.

I don't get how knowing that $0$ is not an eigenvalue of $A$ enables us to conclude that $A\mathbf{x}=\mathbf{0}$ has the trivial solution only. In other words, how do we exclude the possibility that for all $\mathbf{x}$ that is not an eigenvector of $A$, $A\mathbf{x}=\mathbf{0}$ will not have a non-trivial solution?

Note that if $Ax=0=0\cdot x$ for some $x\ne 0$ then by definition of eigenvalues, $x$ is an eigenvector with eigenvalue $\lambda = 0$, contradicting that $0$ is not an eigenvalue of $A$. $$Ax=\lambda x$$
$$\det A = \prod_{j=1}^n \lambda_j \implies \det A = 0 \Leftrightarrow \exists\ i \in \{1,2,\ldots, n\}:\lambda_i = 0$$
Because if $Ax=0$ for some nonzero $x$, then $0$ would be an eigenvalue.
• Yes. You have that $0$ is an eigenvalue of $A$ $\iff$ $A$ is not invertible $\iff$ there exists nonzero $x$ with $Ax=0$. – Martin Argerami Sep 11 '19 at 11:57


# Prove $\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$

Let $f:[a,b]\rightarrow \mathbb{R}$ be a positive and differentiable function. Prove: $$\frac{f(b)}{f(a)}=e^{(b-a)\frac{f'(c)}{f(c)}}$$

I tried to apply the MVT and make some algebraic manipulations, but got stuck.

Any help appreciated.

• Do you want to prove that the equality holds for a certain $c$, a certain $a<c<b$, or any $c$? – toliveira Mar 8 '17 at 19:42
• Did you forget an exponential somewhere, e.g. in $e^{\frac{f'(c)}{f(c)}}$? – Clement C. Mar 8 '17 at 19:42
• @ClementC. Sorry for that, editing. – Itay4 Mar 8 '17 at 19:44
• Guys, he means $e^{(b-a)\frac{f'(c)}{f(c)}}$ – Gabriel Romon Mar 8 '17 at 19:44
• @LeGrandDODOM Yet, that is not what (s)he wrote initially. Hence my comment. – Clement C. Mar 8 '17 at 19:46

High-level idea: The reflex to use the MVT is a good one. Now, the issue is that you do not want to apply it directly to $f$, but to some other function $g$ related to $f$. In particular, the term $e^{b-a}$ strongly hints that you're going to take an exponential after that; and seeing $\frac{f'(c)}{f(c)}$ should prompt you to think of $(\ln f)' = \frac{f'}{f}$. These two together suggest we set $g=\ln f$, and apply the MVT to $g$: good thing, this is legitimate, as $f$ is assumed positive. So let's try that...

Let $g\colon [a,b]\to \mathbb{R}$ be defined by $g\stackrel{\rm def}{=} \ln f$. $g$ is well-defined, and differentiable, as $f$ is positive and differentiable.

Applying the Mean Value Theorem on $g$, we get that there exists $c\in(a,b)$ such that $$g(b)-g(a) = g'(c) (b-a)$$ and, exponentiating both sides, we obtain $$e^{g(b)-g(a)} = e^{g'(c)(b-a)}.$$ Recalling that $e^{g(b)-g(a)} = \frac{e^{g(b)}}{e^{g(a)}} = \frac{f(b)}{f(a)}$ and $g'(c) = \frac{f'(c)}{f(c)}$ concludes the proof.

• The right-hand side of your last displayed equation should be $\exp(g^{\prime}(c)(b-a))$. Great answer otherwise though... – carmichael561 Mar 8 '17 at 20:18
• @carmichael561 Thanks for spotting this! – Clement C. Mar 8 '17 at 20:19

Let $g(x)=\ln(f(x))$. According to the mean value theorem, $$\frac{g(b)-g(a)}{b-a}=g^{\prime}(c)=\frac{f^{\prime}(c)}{f(c)}$$ for some $c\in(a,b)$. This implies that $$\ln\Big(\frac{f(b)}{f(a)}\Big)=(b-a)\frac{f^{\prime}(c)}{f(c)}$$ hence $$\frac{f(b)}{f(a)}=\exp\Big((b-a)\frac{f^{\prime}(c)}{f(c)}\Big)$$

• Darn, you beat me by a few seconds. (+1) – Clement C. Mar 8 '17 at 19:47


+0

# How to convert polar to cartesian and vice versa?

0
507
2

how to convert polar coordinates to cartesian and vice versa?

Guest Sep 9, 2014

### Best Answer

#2
+20040
+5

How to convert polar to cartesian and vice versa ?

I. polar to cartesian : $$\boxed{\begin{array}{rcl} x=r*\cos{(\alpha)} \\ y=r*\sin{(\alpha)} \end{array} }$$

II. cartesian  to polar : $$r: \begin{array}{rcl} x^2+y^2&=&r^2\cos^2{(\alpha)}+r^2\sin^2{(\alpha)} \\ x^2+y^2&=& r^2( \underbrace{ \sin^2{(\alpha)}+\cos^2{(\alpha)} }_{=1} ) \\ x^2+y^2&=&r^2 \\ \end{array} \boxed{r=\sqrt{x^2+y^2}}$$

$$\alpha: \begin{array}{rcl} \frac{y}{x}&=& \frac { r\sin{(\alpha)} } { r\cos{(\alpha)} }\\ \frac{y}{x}&=& \tan{(\alpha)} \end{array} \boxed{ \alpha=\tan^{-1}\left( \frac{y}{x} \right) }$$

$$\begin{array}{rcll} y>0 &and& x>0 & \quad \alpha \\ y>0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x>0 & \quad \alpha +360\ensurement{^{\circ}}\\ \hline y=0 &and& x>0 & \quad \alpha = 0\ensurement{^{\circ}}\\ y>0 &and& x=0 & \quad \alpha = 90\ensurement{^{\circ}}\\ y=0 &and& x<0 & \quad \alpha = 180\ensurement{^{\circ}}\\ y<0 &and& x=0 & \quad \alpha = 270\ensurement{^{\circ}} \end{array}$$

y = 0   and  x = 0   $$\alpha$$  undefined!

heureka  Sep 9, 2014
#1
+27062
+5

cartesian to polar:

$$r=\sqrt{x^2+y^2}$$

$$\theta=\tan^{-1}(\frac{y}{x})$$

polar to cartesian:

$$x=r\cos{\theta}$$

$$y=r\sin{\theta}$$

Alan  Sep 9, 2014
#2
+20040
+5
Best Answer

How to convert polar to cartesian and vice versa ?

I. polar to cartesian : $$\boxed{\begin{array}{rcl} x=r*\cos{(\alpha)} \\ y=r*\sin{(\alpha)} \end{array} }$$

II. cartesian  to polar : $$r: \begin{array}{rcl} x^2+y^2&=&r^2\cos^2{(\alpha)}+r^2\sin^2{(\alpha)} \\ x^2+y^2&=& r^2( \underbrace{ \sin^2{(\alpha)}+\cos^2{(\alpha)} }_{=1} ) \\ x^2+y^2&=&r^2 \\ \end{array} \boxed{r=\sqrt{x^2+y^2}}$$

$$\alpha: \begin{array}{rcl} \frac{y}{x}&=& \frac { r\sin{(\alpha)} } { r\cos{(\alpha)} }\\ \frac{y}{x}&=& \tan{(\alpha)} \end{array} \boxed{ \alpha=\tan^{-1}\left( \frac{y}{x} \right) }$$

$$\begin{array}{rcll} y>0 &and& x>0 & \quad \alpha \\ y>0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\ y<0 &and& x>0 & \quad \alpha +360\ensurement{^{\circ}}\\ \hline y=0 &and& x>0 & \quad \alpha = 0\ensurement{^{\circ}}\\ y>0 &and& x=0 & \quad \alpha = 90\ensurement{^{\circ}}\\ y=0 &and& x<0 & \quad \alpha = 180\ensurement{^{\circ}}\\ y<0 &and& x=0 & \quad \alpha = 270\ensurement{^{\circ}} \end{array}$$

y = 0   and  x = 0   $$\alpha$$  undefined!

heureka  Sep 9, 2014

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