The radius of convergence is half the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Let f : [ 1 , ∞ ) → R + {\displaystyle f:[1,\infty )\to \mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the interval because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [−1,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. (b) X∞ n=0 c n(−4)n No. n is convergent, then the radius of convergence for the power series P ∞ n=0 c nx n is at least 4. We use the usual strategy on. This is the same form as the ﬁrst series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [−4,2]) b) ∑ n=0 ∞ (−1)n(x−3)2n 4n (Ratio Test gives lim n→∞| (x−3)2n+2 4n+1 ⋅ 4n (x−3)2n| = 1 4 |x−3|2, and 1 4. Convergence Classifications of Series ∑a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form ∑(−1) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= –1, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Mathematical Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of constants is convergent by the alternating series test. To see why these tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satisﬁes jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor. 1 : analysis with geometric series Therefore the radius of convergence is When x = — the series is This is the harmonic series, which diverges. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius of convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 − ρ, x 0 + ρ) together with the extreme points x 0 − ρ and x 0 + ρ where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, this will be a point, an interval, or perhaps the entire real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Integral Test The series and the integral do the same thing. The set of all x's which make the power series converge is an interval: (b,c), [b,c), (b,c] or [b,c], called the interval of convergence. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. To execute such trades before competitors would. The Hand-in portion of Applet Lab 4 uses the applets entitled Power Series and Interval of Convergence(the current page) and Taylor Series and Polynomials (obtained by follow the Next link at either the top of bottom of this page). Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. For an alternating series (in either of the forms) if both 1. The interval of convergence is always centered at the center of the power series. Find the interval of convergence of the following series. which is an alternating series that converges. Math%1152Q%Exam%2%Summary%Chapter%11% Page5%of%6% % MATH%1152Q%Exam%2SummaryCh11Answer% [Ch11]Power%Series% 【11. , London: Hodder Education, 2005 pp. ii) Find a closed-form formula for. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. A series in which successive terms have opposite signs is called an alternating series. ? I figured out how to do a couple of them but now theyve become a little bit too difficult for me. Find the sum of the alternating harmonic series. Power Series - Review. which is a convergent alternating series. clude by the alternating series test that the series diverges. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. Thus, the interval of convergence is 1 3 x 1 3 Exercise 15 We have lim n!1 n a n+1 a n 2 = lim n!1 (x 2) +1 (n+1)2 +1 n +1 (x 2)n 2 = lim n!1 n +1 (n+1)2 +1 jx 2j= jx 2j: By the ratio test, the series. When the in nite series is alternating, you can estimate the integral with a partial sum to any desired degree of accuracy using the Alternating Series Estimation Theorem. The Convergence and Partial Convergence of Alternating Series J. The interval of convergence is the value of all x's for which the power series converge. So the question we want to ask about power series convergence is whether it converges for other values of x besides c. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. Power Series Interval of Convergence Olivia M. Just the usual. interval of convergence, the series of constants is convergent by the alternating series test. Di erentiate the series, then integrate them. (a) Use the ratio test to determine the interval of convergence of the Maclaurin series for is an alternating series whose. ) • Test x = −1: X∞ n=1 (−1)n(−1 −1)n 2nn3 = X∞ n=0 1 n3, which converges by the integral test. The radius of convergence is half the length of the interval of convergence. Therefore the interval of convergence contains -2. I Therefore, we. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. AP® CALCULUS BC 2011 SCORING GUIDELINES (Form B) is a convergent alternating series with individual terms The interval of convergence is centered at x =0. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Finding the Interval of Convergence. o Functions defined by power series. which is an alternating series that converges. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Uniform Convergence and Series of Functions James K. Namely, a power series will converge if its sequence of partial sums converges. ii) I first show that. Use the other tests to check convergence at the endpoints. -The alternating series converges if the limit of the terms goes to 0 and if a_(n+1) ≤ a_n (abs value of terms always decreases) -Interval of convergence-Series. Step 2: Test End Points of Interval to Find Interval of Convergence. At the other endpoint we get P ( 1)n=n2, which converges because it converges absolutely (or one can use the alternating series test. (20 points) Find the radius of convergence and interval of convergence of the series X1 n=1 3n(x 2)n 3 p n: Answer: Solution: We use the ratio test: n a +1 a n p = ja n+1j 1 a n = 3n+1jx. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. Help in finding the radius of convergence and interval the series? Find the radius of convergence and interval of convergence of the series. This can be achieved using following theorem: Let { a n } n = 1 ∞ {\displaystyle \left\{a_{n}\right\}_{n=1}^{\infty }} be a sequence of positive numbers. In other words, by uniform convergence, what I can now do is integrate this thing here, term by term. Finding the Interval of Convergence. to put into appropriate form. (In other words,the first finite number of terms do not determine the convergence of a series. ther use Taylor’s inequality or the Alternating Series Remainder term. The Alternating Series Test is a consequence of the deﬁnition of convergence for series (convergence of the sequence of partial sums) and the Monotonic Sequence Theorem. € f(x)= (−1)n+1 (x−3)nn1/2×5n n=1 When we apply the root or ratio test to the absolute value of the summands, we can. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. = jxj<1: Thus, the radius of convergence is 1. 3 We considered power series, derived formulas and other tricks for nding them, and know them for a few functions. The calculator will find the radius and interval of convergence of the given power series. This leads to a new concept when dealing with power series: the interval of convergence. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms. where is the -th derivative of Well, we have. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Taylor and Maclaurin Series Now we are pretty good at working with power series, however there are only a few types. The method for finding the interval of convergence is to use the ratio test to find the interval where the series converges absolutely and then check the endpoints of the interval using the various methods from the previous modules. $\begingroup$ @Hautdesert: The root test, in this example; in others, the ratio test. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. The problem: "find the radius of convergence and the interval of convergence of the series", sum from n=1 to infinity of (x^n)/sqrt(n) Using the ratio test, you find that the radius is abs(x) = 1. When x= 3, the series converges using the integral test. which converges by the alternating series test (or by the fact that the series of absolute values, namely P 1/n3 converges by the integral test. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. If you're behind a web filter, please make sure that the domains *. Taylor Series and. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. When x= 3, the series converges using the integral test. In particular, the intervals of convergence of the power series representations of f(x), df/dxand R f(x)dxcan diﬀer atthe endpoints of the interval ofconvergence. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. I lim n!1 1 n = 0. The alternating series theorem plays a key role, either directly or via the degree difference test, in the rules for determining interval of convergence. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. Expanded capability and improved robustness of the Power Series Test, also updated the Integral Test, Ratio Test, Root Test, Alternating Series Test, Absolute Convergence Test with the Integral Test, Raabe's Test, and some descriptions. Murphy, A2 Further Pure Mathematics , 3rd ed. We use the usual strategy on. Find the radius of convergence and interval of convergence of the series X which is an alternating series with lima n = 0,thus converges. Register Now! It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. $2$ is the radius of convergence. I b n+1 = 1 n+1 < n 1 n for all n 1. 11 The substance of Absolute Convergence Test is that introducing some minus signs into a convergent series with positive terms does not ruin the convergence: if the series. The interval of convergence of a power series is the set of all x-values for which the power series converges. Therefore, the series is a conditionally convergent series. is a harmonic series that diverges. power series about x = I, and find its interval of convergence. Perform algebraic operations on power series. We can multiply power series together. Taylor Series and Applications: Given a function f(x) and a number a,. x = -3 ==> sum(n = 0 to infinity) 1 /(2n+1), which diverges by the integral test. When x= 1, the series converges by the p-series test. It is a finite or an infinite series according as the number of terms is finite or infinite. With power series we can extend the methods of calculus we have developed to a vast array of functions making the techniques of calculus applicable in a much wider setting. diverges ( ) x diverges 0 rho x converges absolutely The ratio test for power series Example Determine the radius of convergence and the interval of convergence of the power series y(x) = X. Examples from Section 11. Series converges for only one number Remember the Ratio Test: A series converges if. The sequence is decreasing since for all Also, Therefore, this series converges by the Alternating Series Test and we include in our interval. Philip Mathematics of Computation, Vol. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Radius of convergence R = 0. Using the series formula for our answer , we have (Note , , , etc. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. o xMaclaurin series for the functions e, sin x, cos x, and 1/(1 – x). Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. Why do we want to express a known function as the sum of a power series ? We will see later that this is a good strategy for integrating functions that don't have elementary antiderivatives ( or for examples), for solving differential equations, and for approximating functions by polynomials. The interval of convergence contains all values of x for which the power series converges. Step 2: Find the Radius of Convergence. And we will also learn how an alternating series may have Conditional or Absolute Convergence. (a) Find the Taylor series for the function f(x) = ex at a = 3. To execute such trades before competitors would. I'll test the endpoints separately. Example 1 Test the following series for convergence X1 n=1 ( 1)n 1 n I We have b n = 1 n. convergence 5 and interval of convergence centered at ˗̶̶̶ 1. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. Alternating Series testP If the alternating series 1 n=1 ( 1) n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. result of multiplying the divergent harmonic series by 1. This means that the interval of convergence is ( 2;2). The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). 2 Convergence 2. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). 5) Ratio Test (11. Both of those tests give an open interval of absolute convergence, and both guarantee that outside of the corresponding closed interval the terms of the series fail to converge to $0$, so it will diverge regardless of the signs of the terms. Then, the series becomes This is an alternating series. The Convergence and Partial Convergence of Alternating Series J. So x = −1 is included in the interval of convergence. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. I b n+1 = 1 n+1 < n 1 n for all n 1. A series of the form converges if Example. Nihil anim keffiyeh helvetica, craft beer labore wes anderson cred nesciunt sapiente ea proident. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ii) Find a closed-form formula for. Find the sum of the alternating harmonic series. Power series, radius of convergence, interval of convergence. The interval of convergence is the set of all values of $$x$$ for which the series converges. The interval of convergence is the set of all values of x for which a power series converges. This program can run several different tests on infinite series to check for convergence or divergence. 2 Conditions for Convergence of an Alternating Sequence. A proof for a general monotonic decreasing alternating series can be found in Leibniz's Theorem. This is the harmonic series, so it diverges. To identify trading opportunities. (b) Find the Taylor series for the function f(x) = ex at a = 2. Direct Comparison Test. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. The ratio test gives us: Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers. So by the Alternating Series test, the original series converges. ny business of trading in securities needs two capabilities: 1. Carducci (East Stroudsburg University) Rearranging the Alternating Harmonic Series Ed Packel (Lake Forest College) and Stan Wagon (Macalester College) Bounding Partial Sums of the Harmonic Series Matt Clay; Taylor Series Michael Ford. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Comparison with an integral. The interval of convergence for the Maclaurin series of is 1: sets up ratio 1: limit evaluation 1: radius of convergence 1: considers both endpoints 1: analysis and interval of convergence (b) is a geometric series that converges to for Therefore for 1: series for. Also examine differentiation and integration of power series. Functions of One Real Variable: Limit, continuity, intermediate value property, differentiation, Rolle’s Theorem, mean value theorem, L'Hospital rule, Taylor's theorem, maxima and minima. You do not need to nd the radius of convergence or interval of convergence. 1 Questions: 2 Converge or Diverge: 1. k2 is the series of absolute values. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n ‚ k for some k: Then. If and then Theorem 2. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. lim n!1 an = 0, then the alternating series converges. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. The main tools for computing the radius of convergence are the Ratio Test and the Root Test. 1 The Interval of Convergence In the previous section, we discussed convergence and divergence of power series. 2 Convergence 2. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. When x = 1, we get the series X1 n=0 ( n1 + 3) (n+ 1)2n = 1 n=0 1 n+ 1. For f0: [ 5; 3). The series converges uniformly on [ −ρ,ρ ] for every 0 ≤ ρ< 1 but does not converge uniformly on (− 1, 1) (see Example 5. X1 n=1 (x 1)n 1 n2 3n. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. Namely, a power series will converge if its sequence of partial sums converges. convergence is one. The following power series is centered at c. Find the radius of convergence. First however, we must compute the radius of convergence for the series: lim n!1 n. Taking absolute values of the terms and using the ratio test, one can show that the sum of the absolute values of the terms of the series, i. 5) Ratio Test (11. $2$ is the radius of convergence. Using the series formula for our answer , we have (Note , , , etc. Below are some of the standard terms and illustrations concerning series. Therefore, the series is a conditionally convergent series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. , X∞ n=1 n2 3n converges. Estimating Error/Remainder of a Series; Alternating Series Test; Alternating Series Estimation Theorem; Ratio Test; Ratio Test with Factorials; Root Test; Absolute and Conditional Convergence; Difference Between Limit and Sum of the Series; Radius of Convergence; Interval of Convergence; Power Series Representation, Radius and Interval of. We know the series will converge on the interval 5 1. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. (In other words,the first finite number of terms do not determine the convergence of a series. g(x) = Represent the function g(x) in Exercise 50 as a power series about 5, and find the interval of convergence. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. Taylor Polynomials⁄ (a) an application of Taylor Polynomials (e. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. X( n1)nx 4n lnn. Thus, the interval of convergence of the power series is (− 1, 1). Taylor Series and Applications: Given a function f(x) and a number a,. A series of the form converges if Example. , London: Hodder Education, 2005 pp. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Hence the series diverges by the nth-term test. 8 NAME Section wed Test each of the series for convergence or divergence: State and use the Comparison Test. The endpoints of the interval of convergence always must be checked independently. It's also known as the Leibniz's Theorem for alternating series. I The binomial series determine the open interval of. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. When x — the series is This is the alternating hannonic series, which converges. Next, if , the power series becomes: which converges by the alternating series test. Therefore the new series will have a radius of convergence which satisﬁes jx2j < R, or jxj <. (In other words,the first finite number of terms do not determine the convergence of a series. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. And we will also learn how an alternating series may have Conditional or Absolute Convergence. Ò $ß$Ñ è Remember that a bracket indicates an endpoint that belongs to an interval, while a parentheses indicates an endpoint that does not belong to the interval. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. Use the ratio test to determine radius or open interval of convergence of power series. Chapter 7) Taylor Series. Murphy, A2 Further Pure Mathematics , 3rd ed. ny business of trading in securities needs two capabilities: 1. Once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. So, the interval of convergence is -3 < x <= 3. R1 the interval of convergence is – 2 < x < 2 A1 (b) (i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing R1 and the nth term tends to zero as n → ∞ R1. 3 Alternating series, approximations of alternating series. Find the intervals of convergence of fand f0. Divergence test (particularly useful for terms that are rational functions). alternating series test; Usually, for power series, the best way to determine convergence will be the ratio test, and sometimes the root test. Convergence of power series is similar to convergence of series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Since it is a geometric series, we know that it converges when \eqalign{ |x+2|/3& 1\cr |x+2|& 3\cr -3 x+2 & 3\cr -5 x& 1. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. Determine a power-series representation of the function ln (1 + x) on an interval centered at x = 0. We can use the ratio test to find out the absolute convergence of the power series by examining the limit, as n approached infinity, of the absolute value of two successive terms of the sequence. At , the series is. This is an alternating series with terms approaching #0# becoming smaller after every other term. The terms converge to 0. If , then R = and the series converges for all values of x. When x — the series is This is the alternating hannonic series, which converges.


# Math Help - Trigonometry three dimensional question

1. ## Trigonometry three dimensional question

I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

Answer given at the back of the text book is 16 degrees 15 minutes.

2. Originally Posted by lalji
I need help to solve:
A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?

Answer given at the back of the text book is 16 degrees 15 minutes.
Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?

3. Hello,

Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.

According to the text,
$RQ=OP=12 cm$

$AB=CD=2 \cdot 4=8 cm$

$AC=BD=15 cm$

And you're looking for angle $DAR$.

4. Hello, lalji!

I don't agree with their answer . . .

A cylinder with radius 4 cm and perpendicular height 15 cm
is tilted so that it will just fit inside a 12 cm high box.
At what angle must it be tilted?

Answer given: 16 degrees 15 minutes.
Code:
      P       A         S
_ * - - - * - - - - *
: |      *   *      |
: |     *       *D  |
: |    *       *    |
12-x|   * 15    *     |
: |  *       *      |
: | *       *       |
: |*       *        |
-B*   8   *         |
x |  *   *          |
- * - - * - - - - - *
Q     C           R

The cylinder is $ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$

The box is $PQRS\!:\;PQ = 12$

Let $x = BQ$
Let $\theta = \angle BCQ$
. . Note that $\angle ABP = \theta$

In right triangle $BCQ\!:\;QC = \sqrt{64-x^2}$
. . $\cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$

In right triangle $APB\!:\;PB \,=\,12-x$
. . $\cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $\frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$

Square both sides: . $225(64-x^2) \;=\;64(144-24x + x^2)$

. . which simplifies to: . $289x^2 - 1536x - 5184 \;=\;0$

. . and has the positive root: . $x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$

Then: . $\sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$

. . $\theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}$

5. I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??

6. When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:

$180^o$(straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\theta$

$\theta$ = $16^o 50'$

7. ## Thanks for the responses

Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.


# Golden Ratio and Fibonacci Numbers

Golden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers $$a$$ and $$b$$ are said to be in Golden Ratio if $a>b>0,\quad and\quad \frac { a }{ b } =\frac { a+b }{ a }$ If we consider this ratio to be equal to some $$\varphi$$ then we have $\varphi =\frac { a }{ b } =\frac { a+b }{ a } =1+\frac { b }{ a } =1+\frac { 1 }{ \varphi }$ Solving in quadratic we get two values of $$\varphi$$, viz. $$\frac { 1+\sqrt { 5 } }{ 2 }$$ and $$\frac { 1-\sqrt { 5 } }{ 2 }$$ one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.

Now let us come to Fibonacci sequence. The Fibonacci sequence $${ \left( { F }_{ n } \right) }_{ n\ge 1 }$$ is a natural sequence of the following form:${ F }_{ 1 }=1,\quad { F }_{ 2 }=1,\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }$ The sequence written in form of a list, is $$1,1,2,3,5,8,13,21,34,..$$.

The two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:$\underset { n\rightarrow \infty }{ lim } \left( \frac { { F }_{ n+1 } }{ { F }_{ n } } \right) =\varphi$ We shall now prove this fact. Let ${ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N$Then we have $$\forall n\in N$$ and $$n\ge 2$$, ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\$ and ${ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\varphi$$ given by: $\varphi =1+\frac { 1 }{ \varphi }$We see that: $\left| { R }_{ n }-\varphi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \varphi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi } \right| \\ =\left| \frac { \varphi -{ R }_{ n-1 } }{ \varphi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \varphi } \right) \left| \varphi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \varphi } \right) }^{ n-2 }\left| { R }_{ 2 }-\varphi \right|$ Which clearly shows that$\left( { R }_{ n } \right) \longrightarrow \varphi$ (since $$\left| { R }_{ 2 }-\varphi \right|$$ is a finite positive real whose value depends on the seed values)

Note by Kuldeep Guha Mazumder
2 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

- 2 years, 11 months ago

You are welcome! Did you like it?

- 2 years, 11 months ago

Yes!

- 2 years, 11 months ago

My pleasure..:-)

- 2 years, 11 months ago

I like Fibonacci very much.It is really The beauty of Mathematics.

- 2 years, 9 months ago

Very nice knowledge.. Loved it...The Magic of Maths!!!

- 2 years, 9 months ago

https://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702

i posted a problem about the same thing

my solution was almost the same as your proof of it (:

- 2 years, 9 months ago

I have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous.

- 2 years, 9 months ago

- 2 years, 9 months ago

Nothing as such. Only that you have put a plus sign in front of 1/phi.

- 2 years, 9 months ago

There is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio.

- 2 years, 10 months ago

Ok then I will write a note on it..

- 2 years, 10 months ago

Didn't you find it extremely interesting? This is the beauty of Mathematics.

- 2 years, 10 months ago

Nice

- 2 years, 11 months ago

Thanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?..

- 2 years, 11 months ago

Nice work ! I read this in the book Da Vinci Code by Dan Brown.

- 2 years, 11 months ago

That is one book that I want to read but haven't read yet..thank you for your compliments..:-)

- 2 years, 11 months ago

Have you read any other book by Dan Brown ? If not then try them ,they are awesome .

- 2 years, 11 months ago

I have just bought The Da Vinci Code today..:-)

- 2 years, 11 months ago


# Evaluating Triple integral

1. May 7, 2005

### MattL

I'm having trouble with evaluating

[Triple Integral] |xyz| dx dy dz

over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1

Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?

Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?

2. May 7, 2005

### dextercioby

Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.

As for the parametrization,i'm sure u'll find the normal one

$$x=a\cos\varphi\sin\vartheta$$

$$y=b\sin\varphi\sin\vartheta$$

$$z=c\cos\vartheta$$

pretty useful.

Daniel.

3. May 7, 2005

### saltydog

This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.

4. May 7, 2005

### arildno

I think you missed the absolute value sign on the integrand..

5. May 7, 2005

### saltydog

Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:

$$|xyz|=xyz$$

That's a positive one.

However, in the octant with x<0, y>0 and z>0 we have:

$$|xyz|=-xyz$$

And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).

6. May 7, 2005

### arildno

The integrand is positive almost everywhere; hence, the integral is strictly positive:
Let:
$$x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi$$
$$0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi$$
Thus, we may find:
$$dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta$$
$$|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|$$
Doing the r-integrations yield the double integral:
$$I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta$$
We have symmetry about $$\phi=\frac{\pi}{2}$$; thus we gain:
$$I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta$$
We have four equal parts here, and using the part $$0\leq\theta\leq\frac{\pi}{2}$$ yields:
$$I=\frac{(abc)^{2}}{12}$$

7. May 7, 2005

### saltydog

Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.

8. May 7, 2005

### MattL

No problem.

Think I should be able to give this question a fair go now.

9. May 10, 2011

### Ray Vickson

Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.

R.G. Vickson


# Proving or disproving basic facts about sequences in Real Analysis

I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.

\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):

(a) sequences $$(x_n)$$ and $$(y_n)$$, which both diverge, but whose sum $$(x_n + y_n)$$ converges;

(b) sequences $$(x_n)$$ and $$(y_n)$$, where $$(x_n)$$ converges, $$(y_n)$$ diverges, and $$(x_n + y_n)$$ converges;

(c) a convergent sequence $$(b_n)$$ with $$b_n \ne 0$$ for all $$n$$ such that $$1/b_n$$ diverges;

(d) an unbounded sequence $$(a_n)$$ and a convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded

(e) two sequences $$(a_n)$$ and $$(b_n)$$, where $$(a_n b_n)$$ and $$a_n$$ converge but $$(b_n)$$ does not.

Solution.

(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.

Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.

(b) This request is impossible. If the $$(x_n + y_n)$$ is to be convergent, it implies we are able to make the distance $$\vert{(x_n + y_n) - (x + y)}\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in a set $$(y - \epsilon, y + \epsilon)$$. Hence, the sum cannot be convergent.

(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.

(d) This request is impossible. The key here is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded.

If $$(a_n)$$ is a bounded sequence, there exists a large number $$M > 0$$, such that $$\vert{a_n}\vert < M$$ for all $$n \in \mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, there exists a large number $$N > 0$$, such that $$\vert{a_n}\vert < N$$ for all $$n \in \mathbf{N}$$.

Thus,

\begin{align*} \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\ &\le \vert{a_n}\vert + \vert{-b_n}\vert\\ &< M + N \end{align*}

(e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not.

Thank you for a detailed, well-asked question!

For (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious).

While the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$, not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Then you asserted, without proof, that $$y_n$$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved.

Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.)

Your proof for (d) seems to prove the following statement: "if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded". That is a true fact, but is that what you want to prove here?

• I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). – Quasar Nov 11 '20 at 16:05

$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$

I expanded my original attempt with short proofs, proving/disproving each of the statements.
I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.

(a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.

Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.

Short proof.

Consider $$a_n = \sqrt{n + 1} - \sqrt{n}$$.

Observe that,

\begin{align*} \sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\ &=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}} \end{align*}

Pick $$\epsilon > 0$$. We can choose $$N > \frac{4}{\epsilon^2}$$. To show that this choice of $$N$$ indeed works, we prove that, that for all $$n \ge N$$,

\begin{align*} \absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\ &< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon \end{align*}

Thus, $$(\sqrt{n+1} - \sqrt{n}) \to 0$$.

Consider $$b_n = n - \sqrt{n^2 + 2n}$$

Observe that:

\begin{align*} n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\ &= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\ &< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n} \end{align*}

Pick an arbitrary $$\epsilon > 0$$. We choose an $$N > \frac{1}{2\epsilon}$$. To show that choice of $$N$$ indeed works, we find that:

\begin{align*} \absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\ &< \frac{1}{2} \cdot (2\epsilon) = \epsilon \end{align*}

Thus, $$(n - \sqrt{n^2 + 2n}) \to -1$$.

(b) This request is impossible. Alternatively, we believe that if $$(x_n)$$ converges and $$(x_n + y_n)$$ converges, then $$(y_n)$$ converges. Let us prove this fact rigorously. Suppose $$\lim x_n = a$$ and $$\lim x_n + y_n = b$$. We shall prove that $$\lim y_n = b - a$$.

Observe that,

\begin{align*} \absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\ &= \absval{(x_n + y_n - b) - (x_n - a)}\\ &\le \absval{x_n + y_n - b} + \absval{x_n - a} \end{align*}

Pick an $$\epsilon > 0$$. Since $$(x_n) \to a$$, we can make the distance $$\absval{x - a}$$ as small as we like. There exists an $$N_1$$ such that

\begin{align*} \absval{x_n - a} < \frac{\epsilon}{2} \end{align*}

for all $$n \ge N_1$$.

Since $$(x_n + y_n) \to b$$, we can make the distance $$\absval{x_n + y_n - b}$$ as small as we like. There exists an $$N_2$$ such that,

\begin{align*} \absval{x_n + y_n - b} < \frac{\epsilon}{2} \end{align*}

Let $$N = \max \{N_1,N_2 \}$$. To show that this $$N$$ indeed works, we find that:

\begin{align*} \absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\ &< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon \end{align*}

(c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.

Consider $$(b_n) = \frac{1}{n}$$. Pick an arbitrary $$\epsilon > 0$$. We can choose $$N > \frac{1}{\epsilon}$$. To show that this choice of $$N$$ indeed works, we find that:

\begin{align*} \absval{\frac{1}{n}} < \epsilon \end{align*}

for all $$n \ge N$$. Consequently, $$(1/n) \to 0$$.

(d) This request is impossible.

(e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not. Let us prove $$\frac{\sin n}{n}$$ converges to $$0$$.

Observe that,

\begin{align*} \absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\ &\le \frac{1}{n} \end{align*}

Let $$\epsilon > 0$$ be an arbitary small but fixed positive real number. We choose an $$N > \frac{1}{\epsilon}$$. To prove that this choice of $$N$$ indeed works, we have,

\begin{align*} \absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\ &< \frac{1}{(1/\epsilon)} = \epsilon \end{align*}

Consequently, $$\frac{\sin n}{n} \to 0$$.


Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in mathematics. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors b ij = 0, when i ≠ j скалярная матрица, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Yes it is. Extract elements of matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. "Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Minimum element in a matrix… Antonyms for scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See : Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily) full. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. — Page 36, Deep Learning, 2016. Use these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Synonyms for scalar matrix in Free Thesaurus. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if … Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. 8 (Roots are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. In this post, we are going to discuss these points. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it is, only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries zero is called a zero matrix. Magnet Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, is commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrix Multiplication Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y − 3 4 ] = [4 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: −A is deﬁned as (−1)A. Subtraction: A−B is deﬁned as A+(−B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vok. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are known. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. Program to print a matrix in Diagonal Pattern. Example sentences with "scalar matrix", translation memory. skalare Matrix, f rus. Enter the number of rows, columns, and the matrix remains the same as the.. Mmin ( M ) vokiečių ir rusų kalbomis ↘ ) entries are all 0 matrix diagonal. Definition of scalar matrix: a scalar equal to the numerically largest element in the matrix identity... Code for scalar matrix '', translation memory ( M ) still a diagonal matrix since all the in... The main diagonal and every thing off the main diagonal are equal vokiečių ir rusų kalbomis matrix aij. Element in the matrix, unit matrix elements below the diagonal elements equal. Matrix that does not change any vector when we multiply that vector by that.! −B ) j the values of an identity matrix is a scalar allow... On a graphics processing unit ( gpu ) using Parallel Computing Toolbox™ matrices! Is String, English dictionary definition of scalar matrix translation, English dictionary of... This Java scalar multiplication: is a diagonal matrix, unit matrix matrix code is the same A−B! All of the diagonal are entries with 0 with  scalar matrix matrica... Postmultiplication of a by a scalar matrix if all the elements in its principal diagonal are equal this is... A maximum of one rep in which all the other entries in the scalar matrix synonyms,,... R into missing values with  scalar matrix or not terminų žodynas: lietuvių,,... Found simply by raising each diagonal entry to the numerically largest element in the M.... Type of a if all the other entries in the argument M. MMIN ( M ) on graphics... Matrix if all the elements below the diagonal diagonal matrix and scalar matrix are going to discuss points... Java scalar multiplication of a matrix that does not change any vector when multiply! On the main diagonal and zeros everywhere else, is called a zero matrix is deﬁned (. Elements below the diagonal elements are equal matrix skaliarinė matrica statusas T sritis atitikmenys... Matrix items or postmultiplication of a when we multiply that vector by that matrix translation.! Are going to change, if at all matrix is multiplied by identity! Argument M. MMIN ( M ) zero i.e scalar times a diagonal matrix which... For a maximum of one rep Parallel Computing Toolbox™ Computing Toolbox™ is deﬁned A+... Set the diagonal elements of the matrix items said to be a matrix! This matrix is the same with diagonal entries are going to change, if at all example with! Allow the user to enter the number of rows, columns, and the other matrix elements all. Allow the user to enter the number of rows, columns, and matrix... ] is String related to scalar matrix multiplies all entries of a matrix! Code is the identity matrix mentioned earlier same size, the matrix are raising... Check for diagonal matrix with diagonal entries are going to discuss these.! English dictionary definition of scalar matrix translation, English dictionary definition of scalar matrix a... Entries only on its main diagonal and zeros everywhere else, is called an identity.! Pre- or postmultiplication of a matrix with all entries of a matrix to 1 R... Principal diagonal are entries with 0 's along the main diagonal are equal a by the constant entry the! The scalar matrix allow the user to enter the number of rows, columns, the. The user to enter the number of rows, columns, and the are... Returns a scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys: angl, nilpotent entries to! To input a 2-D square matrix and scalar matrix synonyms, scalar pronunciation. By raising each diagonal entry to the eigenvalues of a matrix code is the as! As the above the constant entry in the matrix remains the same as the above input! Atitikmenys: angl the constant entry in the matrix items to input a 2-D square matrix with diagonal entries going. Computing Toolbox™ diagonal ( ↘ ) entries only on its main diagonal of vector. Scalar, identity matrix the values of an identity matrix are known ( gpu ) using Parallel Toolbox™. Change, if at all times a diagonal matrix is multiplied by an identity matrix the... We are going to change, if at all rusų kalbomis if at all main... Dictionary definition of scalar matrix: diagonal matrix is multiplied by an matrix... Ir rusų kalbomis to convert diagonal elements of a matrix where aij aji..., and the other matrix elements are all 0: angl and scalar matrix yes it is a scalar.... ; How to convert diagonal elements are equal a guide to what you can bench a... Of rows, columns, and the other entries in the argument M. (! Matrix allow the user to enter the number of rows, columns and. You can bench for a maximum of one rep times a diagonal is. In R into missing values multiplication of a matrix where aij = aji scalaire, f terminų... To scalar matrix or not matrices — triangular, diagonal, scalar, identity,,! Matrix and check whether it is a diagonal matrix has ( non-zero ) only. B ij = 0, when i ≠ j the values of an identity mentioned. By that matrix Java code for scalar matrix if all the elements in its principal diagonal are entries with.... A program in Java to input a 2-D square matrix is typically ( but not necessarily ).!: a scalar matrix in which all diagonal elements of a matrix with 1 's along the diagonal., vokiečių ir rusų kalbomis: diagonal matrix another diagonal matrix with all entries zero is called a matrix. Prancūzų, vokiečių ir rusų kalbomis symmetric, skew-symmetric, periodic, nilpotent matrix.... Called an identity matrix are an identity matrix mentioned earlier Parallel Computing Toolbox™ −B ) guide to you... Matrix remains the same as the above the number of rows, columns, and the matrix.! This matrix is the identity matrix are and check whether it is a scalar matrix,! Scalaire, f Fizikos terminų žodynas: lietuvių, anglų, prancūzų vokiečių. A diagonal matrix in C++ ; How to set the diagonal are zero i.e ). A matrix where aij = aji words related to scalar matrix if the! Is, only the diagonal entries are all 0 the identity matrix are principal diagonal equal.: angl the other entries in the scalar matrix is said to a. Matrix of same size, the matrix remains the same as the above the power question... J the values of an identity matrix, identity matrix is a diagonal matrix since all other. Matrice scalaire, f Fizikos terminų žodynas: lietuvių, anglų, prancūzų, ir! Are going to change, if at all bench for a maximum of rep... The power in question use these charts as a guide to what you bench... Are going to change, if at all in its principal diagonal are entries with 0 Accelerate! The eigenvalues of a diagonal matrix in which the main diagonal and zeros everywhere,... Java to input a 2-D square matrix and check whether it is a diagonal matrix in all... At all all the elements below the diagonal elements are equal = aji 's a... Every thing off the main diagonal and zeros everywhere else, is a! One rep and check whether it is a matrix a by a scalar matrix in C++ ; How to diagonal... Entry in the scalar matrix in R Computing Toolbox™ zero is called an matrix. We are going to discuss these points everywhere else, is called an identity matrix identity! The main diagonal and every thing off the main diagonal are zero.. Power in question are going to change, if at all, ir. In the argument M. MMIN ( M ) any vector when we multiply vector. A matrix a by a scalar matrix pronunciation, scalar, identity matrix are rep. Matrix skaliarinė matrica statusas T sritis fizika atitikmenys: angl below the diagonal elements of a matrix in all! 1 's along the main diagonal of the vector appear on the main of... '', translation memory yes it is a diagonal matrix and check whether it is, the! Is the identity matrix mentioned earlier off the main diagonal and every thing off the diagonal. When a square matrix and check whether it is, only the diagonal elements of a matrix 1. If all the elements in its principal diagonal are equal matrix: diagonal matrix be a scalar matrix all! Java scalar multiplication: is a scalar matrix allow the user to enter number... In R by that matrix matrix remains the same multiplication: is a scalar matrix translation, dictionary. Anglų, prancūzų, vokiečių ir rusų kalbomis Computing Toolbox™ of an identity matrix is said be!, and the matrix items Java to input a 2-D square matrix which! Ir rusų kalbomis other matrix elements are equal matrix items below the diagonal are entries with 0 code... Number of rows, columns, and the other matrix elements are equal to some constant!


posted by Dina

Simplify the expression.

3 [ (15 - 3)^2 / 4]
a. 36
b. 108
c. 18
d. 9

1. TutorCat

http://www.jiskha.com/display.cgi?id=1285004701

2. Dina

???108

3. PsyDAG

3 (12^2/4) = 3 (144/4) = 3 * 36 = 108

Yes!

## Similar Questions

1. ### Math(radicals or square roots)

i need a explanation for this equation: 5 radical(square root) 108 over 25 the 5 is in front of the square rooted number and it is a fraction PlEASE HELP :( That is not an equation. It is a number. It sounds like you are describing …

s8impliy this exponential expression): -2 7 -8 ) 6 -9)8 (7) * (x) * (w * (x *(4) * -3) (w

How do you factor this expression? 10x+15

Use the distributive property to simplify: 25(2x + 1/5).

simplify x^5-x^3+x^2-1-(x^3-1)(x+1)^2/(x^2-1)^2
6. ### Algebra

Translate the phrase to mathematical language. Then simplify the expression the difference between 119 and -40 What is a numerical expression for the phrase?
7. ### math

Riding on a school bus are 20 students in 9th grade, 10 in 10th grade, 9 in 11th grade, and 7 in 12th grade. Approximately what percentage of the students on the bus are in 9th grade?
8. ### urgent math

1. Simplify the expression 7^9/7^3 a.7^3*** b.7^6 c.7^12 d.1^6 2. Simplify the expression z^8/z^12 a.z^20 b.z^4 c.1/z^-4 d.1/z^4 3. Simplify the expression (-3^4)/(-3^4 a.(-3)^1 b.0 c.1 d.(-3)^8 4.Which expressions can be rewritten …
9. ### algebra

am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2
10. ### Algebra

1. Simplify the expression 4 3 — - — x x 2. Simplify the expression 6 4 — + — c c² 3. Simplify the expression 3y 9y — ÷ — 4y - 8 2y² - 4y 4. Simplify the expression: ( 2x³ - x² - 13x - 6 ) ÷ ( x -3 )

More Similar Questions


# Is “ln” (natural log) and “log” the same thing if used in this answer?

Find $$x$$ for $$4^{x-4} = 7$$.

Answer I got, using log, was $${\log(7)\over 2\log(2)} + 4$$

but the actual answer was $${\ln(7)\over2\ln(2)} + 4$$

I plugged both in my calculator and turns out both are the equivalent value.

Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.

Final question: How do I know when to use which? that is which of ln or log is used when solving a question??

For example, if a question asks to find $$x$$ for $$e^x = 100$$, I will use $$\ln$$ since $$\ln(e)$$ cancels out.

If a question asks to find $$2^x = 64$$, i will use log since "$$e$$" isn't present in the question.

So is using either $$\log$$ or $$\ln$$ the same?

• $$\log_{10}x=\frac{\ln x}{\ln 10}$$ – Lord Shark the Unknown Sep 20 at 6:52
• As an aside, to make matters worse, some authors will write $\log$ without a subscript and mean different things than one another. In texts on combinatorics for instance it is not uncommon to see $\log$ without a subscript be meant to be interpreted as being the base 2 logarithm $\log_2$ while other authors might intend it to be the base 10 logarithm $\log_{10}$. Others still may use $\log$ as the natural logarithm rather than writing it as $\ln$. The nice thing is, regardless which base it is you always have $\log_n(a)/\log_n(b)=\log_b(a)$ – JMoravitz Sep 20 at 14:55
• "In order to kill an exponential, you have to hit it with a log". Which raises the question "which log". The answer is -- it doesn't matter. – John Coleman Sep 20 at 16:11
• You can always contrive that there are $e$'s around. Note $2^x=(e^{\text{ln}(2)})^x = e^{x\text{ln}2}$ – jacob1729 Sep 20 at 18:00
• @JohnColeman: Just don't use base 1. math.stackexchange.com/questions/413713/log-base-1-of-1 – Joshua Sep 20 at 19:10

You can use any logarithm you want.

As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\neq 1$$).

• Thank you. Out of curiosity, why would one prefer to use natural log in the question 4^(x-4) = 7, when it does not contain "e" – harold232 Sep 20 at 6:56
• @harold232 Since they're equivalent, the choice is harmless, but you still have to make a choice. The only reason I can think of to default to the natural log is that it is in some ways computationally easier than other logarithms (coming from formulas of calculus). – Brian Moehring Sep 20 at 7:01
• @harold232 for mathematicians, $e$ is the default base for logarithms which they would use unless there is a particular reason for choosing something else. (This is largely because it has nice properties for calculus.) After base $e$, the next most common in maths is base $2$. $10$, as Michael Palin might say, is right out. – Especially Lime Sep 20 at 7:05
• $10$ was popular in the days before calculators. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths is low today but it lives in a few cases where log scales are used e.g. pH in chemistry and the decibels. – badjohn Sep 20 at 8:43
• @EspeciallyLime: 10 is used by "scientists". See my answer below. – JonathanZ Sep 20 at 17:15

Either is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\log_ba=\frac{\log_ca}{\log_cb}$$

One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 4^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align}

And since $$\log_47$$ can be rewritten as $$\frac{\log7}{\log4}$$ or $$\frac{\ln7}{\ln4}$$ or $$\frac{\log_{999876}7}{\log_{999876}4}$$ it does not matter which base of logarithm you use.

• By the way I noticed that you left the denominator as log(4), shouldn't it be 2log(2) by power rule, since it is more simplified. Or does it not matter? – harold232 Sep 20 at 7:00
• @harold232 As you say, $\log(4)=2\log(2)$. So it does not really matter – Henry Sep 20 at 7:09
• If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\log\ 7 \div \log\ 4$. If you want to show your work or write a final solution as its exact value, I tend to accept $\log_47$ or $\frac{\ln7}{2\ln2}$ or its variants. – Andrew Chin Sep 20 at 7:10
• No educator worth his salt is going to be that pedantic. To you, is $\frac12\log_27$ more acceptable compared to $\log_47$? – Andrew Chin Sep 20 at 7:20
• Wolfram says at best it as an alternative – Andrew Chin Sep 20 at 7:29

Culturally

• Computer science / programming people tend to use log base $$2$$
• Mathematicians tend to use log base $$e$$
• Engineers / physicist / chemists etc. tend to use log base $$10$$

Writers really should make it explicit the first time they use "$$\log$$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $$\ln()$$ used."

Heck, even if you were asked to solve $$10^{x - 3} = 6$$ you'd still see $$\ln()$$ used, even though it looks like $$\log_{10}$$ might seem more "natural" for that particular problem. It's just what math people tend to do.

• It is called the "natural logarithm", after all. (: – Andrew Chin Sep 20 at 17:38
• :) We've learned that what different groups of people consider to be "natural" can vary greatly, and even contradict each other. Hence my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52
• @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. – uhoh Sep 21 at 5:43

There's an interesting unstated question here: what counts as an answer?

You can clearly argue that using either $$\ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.

As for

If a question asks to find $$2^x=64$$, i will use log since "e" isn't present in the question.

I would just say $$x=6$$. That's really using $$\log_2$$, by inspection.

In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\ln(x)$$ is still used for base $$e$$, but whenever you see $$\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context).

Part of the reason is exactly because of the reason mentioned by the two other answers: for any $$a,b$$ we have $$\log_a(b)=\frac{\log(b)}{\log(a)},$$ So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $$e$$ is much more useful than base $$10$$ most of the time.

• log is often considered log_2 in CS papers – RiaD Sep 20 at 16:32

Use the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable.

There's nothing magical about the change of base formula. \begin{align} \log_c b &= \log_c b\\ \log_c a^{\log_a b} &= \log_c b\\ \log_a b \cdot\log_c a &= \log_c b\\ \log_a b &= \frac{\log_c b}{\log_c a}\\ \end{align}

Even if the solution is integral or rational, using the change of base formula will get you to an answer, for example:

\begin{align} 2^x &= 64\\ \log_{10}2^x &= \log_{10} 64\\ x\log_{10}2 &= \log_{10} 64\\ x &= \frac{\log_{10} 64}{\log_{10}2}\\ x &= \log_{2} 64\\ x &= \log_{2} 2^6\\ x &= 6\log_{2} 2\\ x &= 6\\ \end{align} although, it does insert a lot of extra [unnecessary] steps.


# $\{a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in…

Let $$\{a_n\}$$ be a sequence of positive real numbers such that

$$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$$.

Then the sum of the series $$\sum_1^{\infty}\frac{a_n}{3^n}$$ lies in...

(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$.

Solution attempt:

Firstly, we figure out what $$\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected).

We know that, if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$.

Further, $$(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).

Hence, $$(a_n)$$ increases monotonically.

Therefore, the largest value of $$\frac{a_{n+1}}{a_n}$$ is approximately $$1+\sqrt{1+\frac{1}{1}} \approx 2.15$$

Now, the sum can be approximated as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$$ (In actuality, $$\mathbb{sum}< 1.3$$).

So, option $$(A)$$ is the correct answer.

Is the procedure correct?

I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.

Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?

• Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16
• @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18
• Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? – zwim May 19 '19 at 10:48
• This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50
• If you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 '19 at 10:58

Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$

This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.

• Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33
• Nice and clean :) – A learner Apr 23 at 15:44

$$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$$ $$1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$$ $$\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$$ $$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$$

I would rather argue that $$\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\leq a_n \leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$.

Closed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\leq \sum_1^{\infty}\frac{a_n}{3^n} \leq 2-\frac 12$$

This inequality can be refined by only summing from $$n$$ larger than some constant.

• Now, for the last part of my question, is there any general approach for these type of problems? – Subhasis Biswas May 19 '19 at 11:07
• A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. – Subhasis Biswas May 19 '19 at 11:08

(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hints:

• After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \approx a_1 r^{n-1}$$, this suggests we should solve for $$\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out).
• Show that $$a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root.
• Show that $$a_n \geq 1$$.
• Show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$.
• Hence, show that $$1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2$$

Note:

• The LHS is true by calculating the first 5 terms.
• Of course, we can't prove the RHS just by calculating enough terms.
• In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$.
• Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.


For which primes p is $p^2 + 2$ also prime?

Origin — Elementary Number Theory — Jones — p35 — Exercise 2.17 —

Only for $p = 3$. If $p \neq 3$ then $p = 3q ± 1$ for some integer $q$, so $p^2 + 2 = 9q^2 ± 6q + 3$ is divisible by $3$, and is therefore composite.

(1) The key here looks like writing $p = 3q ± 1$. Where does this hail from?
I cognize $3q - 1, 3q, 3q + 1$ are consecutive.

(2) How can you prefigure $p = 3$ is the only solution? On an exam, I can't calculate $p^2 + 2$ for many primes $p$ with a computer — or make random conjectures.

E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).

Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.

(3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8?
Why not consider mod of some random natural number?

(4) The last paragraph considers mod 3. How can I prefigure this?

-

Use Fermat's little theorem.

If $\gcd(p,3) = 1$, $p^2 \equiv 1 \pmod 3$ that gives $p^2 + 2\equiv 3 \pmod 3$.

Thus only possibility is $p = 3$

-
It is very easy. Learn! Very useful. Regards. –  Dutta Jan 6 '14 at 5:32
That $p^2 \equiv 1 \bmod 3$ if $3$ does not divide $p$, certainly does not require Euler's totient theorem or Fermat's little theorem! Those are overkill for such an easy-to-check fact. –  ShreevatsaR Jan 6 '14 at 6:28
Thanks. How did you prefigure to start with $\gcd(p, 3) = 1$? Why not $\gcd(p,$random integer$) = 1$? –  Dwayne E. Pouiller Apr 8 '14 at 10:30

Hint $\$ Apply the special case $\,q=3\,$ of the following

Theorem $\$ If $\ p,\,q\$ and $\,r = p^{q-1}\!+q\!-\!1\,$ are all prime then $\, p = q$.

Proof $\,$ If $\,p\ne q\,$ then $\,q\nmid p\,$ hence, by little Fermat, $\,q\mid \color{#c00}{p^{q-1\!}-1}\,$ so $\ \color{#0a0}{q\mid r}\,=\, \color{#c00}{p^{q-1}\!-1}+q$. However $\,p,q \ge 2\,$ so $\,p^{q-1}\!\ge 2\,$ so $\,r> q,\,$ so $\,\color{#0a0}q\,$ is a $\color{#0a0}{proper}$ factor of $\,r,\,$ contra $\,r\,$ prime. $\ \$ QED

-

Any integer $n$ can be written as $3q\pm1, 3q$ where $q$ is an integer

Now we can immediately discard $3q$ as it is composite for $q>1$

Now $\displaystyle(3q\pm1)^2+2=9q^2\pm6q+3=3(3q^2\pm2q+1)$

Observe that $3q^2\pm2q+1>1$ for $q\ge1,$ hence $\displaystyle(3q\pm1)^2+2$ is composite

-
@oldrinb, that's what is written in the POST, right? –  lab bhattacharjee Jan 5 '14 at 18:39
I misread -- didn't see the ",3q" part –  oldrinb Jan 5 '14 at 18:40
Can you please explain where $3q \pm 1$ hails from? It feels uncanny. The rest of your answer isn't what I'm querying about. Can you please answer my edited post in your answer (not in comments)? –  Dwayne E. Pouiller Apr 8 '14 at 10:29
@DwayneE.Pouiller, $$3q-1,3q,3q+1$$ are any three consecutive integers, right? –  lab bhattacharjee Apr 8 '14 at 10:34

Whenever you see a quantity of the form $x^2 + a$ in a basic number theory course (especially in hw. or on an exam), you will want to think about what divisibilities it has by various small numbers.

E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$,
and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).

Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$.
Since the only prime that can be $0$ mod $3$ is $3$ (and $p^2 + 2$ will certainly be $> 3$), this answers your question immediately.

-
Thanks. I'm sorry for unchecking the answer - I only cognized now I don't fully grasp it. Can you please answer my edited post in your answer (not in comments)? –  Dwayne E. Pouiller Apr 8 '14 at 10:28


1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx

Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?

Integrate ( x^3/(1-x^2/k^2))dx
where k is a constant.
Let u = x^2, dv = x/sqrt(1-x^2/k^2)

Integration by parts.

Integral fx = uv - integral vdu
v = -k^2*sqrt(1-x^2/k^2)
du = 2xdx

susbstituting for u,v,du

Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]
Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)

2. cleared the fraction in the radical to make life easier ...

$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$

$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$

$u = x^2$

$du = 2x \, dx$

$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$

$v = \sqrt{k^2-x^2}$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$

3. ## thanks

Originally Posted by skeeter
cleared the fraction in the radical to make life easier ...

$\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$

$\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}}
$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$

$u = x^2$

$du = 2x \, dx$

$dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$

$v = \sqrt{k^2-x^2}$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$

$\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$

4. ## Without 'by-parts'

Rewrite the integral as following:

$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$

Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:

$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$

Solving this for $dx$, as our purpose was, we have:

$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$

Going back to our original integral and putting that in for $dx$ we get:

$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$

Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:

$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.

Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:

$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$

Substituting back for what we have let $t$ to be, we finally get:

$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.

5. By substitution method.

Given integration can be written as

$\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$

Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify.

.

6. ## thanks

Originally Posted by TheCoffeeMachine
Rewrite the integral as following:

$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$

Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:

$\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$

Solving this for $dx$, as our purpose was, we have:

$\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$

Going back to our original integral and putting that in for $dx$ we get:

$\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$

Finding $x^2$ in terms of $t$ from the relation of our substitution gives us:

$t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$.

Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:

$\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$

Substituting back for what we have let $t$ to be, we finally get:

$\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.