}(xa)^2\ \\[4pt]p_3(x) &=f(a)+f(a)(xa)+\dfrac{f''(a)}{2!}(xa)^2+\dfrac{f'''(a)}{3! }-\frac{x^{6}}{6 ! ln ) Use these polynomials to estimate \(\sqrt{6}\). Find the Maclaurin series for \(f(x)=\cos x\). From this fact, it follows that if there exists M such that |f(n+1)(x)|M|f(n+1)(x)|M for all x in I, then. let me paste that. Therefore, we have, \(|R_2(11)|\dfrac{0.0011468}{3!}(118)^30.0065104.\). \nonumber \]. x^{3}+\cdots\), \(= 1+ x+ \dfrac 1 {2!} = 0 Here, f (x=0) = e2(0) = 1. I think its something like pi, which you see in a circle. Write out the Taylor expansion for e^x = 1 + x + x^2/2! \[\begin{gathered} f\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f\left( 0 \right) = {e^0} = 1 \\ f\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f\left( 0 \right) = {e^0} = 1 \\ f\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f\left( 0 \right) = {e^0} = 1 \\ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = {e^0} = 1 \\ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{gathered} \], Now using Maclaurins series expansion function, we have We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence. 1 Find the Maclaurin series representation of functions step-by-step. 0 Question. These are the same terms as in the Maclaurin series I wrote above. That is, \(f^{(2m)}(0)=0\) and \(f^{(2m+1)}(0)=(1)^m\) for \(m0.\) Thus, we have, \(\begin{align*} p_0(x)&=0,\\[5pt] n x &=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0)}{2 !} the second derivative is the degree that the tangent line of one point differs from the tangent line of a point next to it. Notice that there is a telescoping effect. xn = f(0) + f (0)x + f (0) 2! &=\sum_{k=0}^m(1)^k\dfrac{x^{2k+1}}{(2k+1)!}.\end{align*}\]. Thus, the Maclaurin series formula is, \(\begin{align} Consider the simplest case: n=0.n=0. Our mission is to improve educational access and learning for everyone. 2 Given a function f(x), the Maclaurin series of f(x) is given by: The value of the first derivative of xe^x is 1 This is the second derivative for xe^x This is the third derivative This is the fourth derivative x x x The slope at any Therefore, if a function ff has a power series at a,a, then it must be the Taylor series for ff at a.a. In this tutorial we shall derive the series expansion of e x by using Maclaurin's series expansion function. Since, Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by for all n0.n0. \(R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\). },\\[5pt] (if-1 f'(0)=1 # Yes, we can use 0! cos }|xa|^{n+1} \nonumber \]. [T] In the following exercises, identify the value of x such that the given series n=0ann=0an is the value of the Maclaurin series of f(x)f(x) at x.x. + f (3)(0) x3 3! Let p0 be the 0th Taylor polynomial at a for a function f.f. ? here is that cosine of x looks a lot like Therefore, the interval of convergence is \((0,2)\). And in the next 0 This is . x3 + e So let me use a p_5(x)&=0+x+0\dfrac{1}{3!}x^3+0+\dfrac{1}{5!}x^5=x\dfrac{x^3}{3!}+\dfrac{x^5}{5! f^{(4)}(x)&=\sin x & f^{(4)}(0)&=0.\end{align*}\], Since the fourth derivative is \(\sin x,\) the pattern repeats. x2 + f (0) 3! Find formulas for the Maclaurin polynomials p0,p1,p2p0,p1,p2 and p3p3 for f(x)=11+x.f(x)=11+x. 1 Therefore, all coefficients of the series are equal to 1. Thanks for the feedback. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. ) Conclude that \(R_n(1)0\), and therefore, \(sn!R_n(1)0\). }(xa)^3+ \nonumber \]. 2 Required fields are marked *. this term, plus this term, plus a negative version would have guessed maybe there's some relationship To find the interval of convergence, we use the ratio test. \(\displaystyle \sum_{n=0}^\dfrac{x^n}{n!}\). Therefore, in the following steps, we suppose e=r/se=r/s for some integers r and s where s0.s0. Posted 12 years ago. Here, we state an important result. at x is equal to 1. Use the fifth Maclaurin polynomial for \(\sin x\) to approximate \(\sin\left(\dfrac{}{18}\right)\) and bound the error. Have questions on basic mathematical concepts? n This formula allows us to get a bound on the remainder \(R_n\). ( Using the product rule, we note that, \[\dfrac{d}{dt}\left[\dfrac{f^{(n)}(t)}{n!}(xt)^n\right]=\dfrac{f^{(n)}(t)}{(n1)!}(xt)^{n1}+\dfrac{f^{(n+1)}(t)}{n!}(xt)^n. When x = 0, xe^x = 0 because anything multiplied by 0 is 0. And in clearly: For the sequence of Taylor polynomials to converge to f,f, we need the remainder Rn to converge to zero. 4 Then the Taylor series, \[\sum_{n=0}^\dfrac{f^{(n)}(a)}{n! x Find the Taylor polynomials \(p_0,p_1,p_2\) and \(p_3\) for \(f(x)=\dfrac{1}{x^2}\) at \(x=1\). How do you find the maclaurin series expansion of f (x) = (ex) + (e2x)? ln f'''(x)&=\dfrac{32}{x^4} & f'''(1)&=3!\\[5pt] ) 2 = Write your answer using sigma notation. Copy function Maclaurin n=5; x=input ('enter the x value: ') epsilon_s= (0.5*10^ (2-n)) epsilon=100 maclaurin (1)=1 iteration=1 while epsilon>epsilon_s iteration=iteration+1 maclaurin (iteration)=maclaurin (iteration-1)+ (x)^ (iteration-1)+x^ (2+ (iteration-1))/factorial (2+ (iteration-1)) + x^3/(2^3)3! p_3(x)&=f(0)+f(0)x+\dfrac{f''(0)}{2}x^2+\dfrac{f'''(0)}{3!}x^3=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3,\end{align*}\). We choose a series that is close to the one we want and then modify it accordingly. 0 x, f We now show how to use this definition to find several Taylor polynomials for f(x)=lnxf(x)=lnx at x=1.x=1. 1 + 1x^2/(2!) consent of Rice University. And what you see \(\dfrac{1}{2}\displaystyle \sum_{n=0}^\left(\dfrac{2x}{2}\right)^n\). And that's going Using the \(n^{\text{th}}\)-degree Maclaurin polynomial for \(\sin x\) found in Example b., we find that the Maclaurin series for \(\sin x\) is given by. between cosine and sine, but what about e to the x? Direct link to sjfeg.barthe's post Quite complex ! Answer link In the following exercises, find the Taylor series of the given function centered at the indicated point. At \(x=2\), we see that, \(\displaystyle \sum_{n=0}^(1)^n(21)^n=\sum_{n=0}^(1)^n\), diverges by the divergence test. This is equal to the For f(x)=1x,f(x)=1x, the values of the function and its first four derivatives at x=1x=1 are, That is, we have f(n)(1)=(1)nn!f(n)(1)=(1)nn! [T] Compare S5(x)C4(x)S5(x)C4(x) on [1,1][1,1] to tanx.tanx. really cool-- that e to the x can be approximated by We now discuss issues of convergence for Taylor series. = = So plus x plus, this \nonumber \]. Therefore, Using the Mean Value Theorem in a similar argument, we can show that if \(f\) is \(n\) times differentiable on an interval \(I\) containing \(a\) and \(x\), then the \(n^{\text{th}}\) remainder \(R_n\) satisfies, \[R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)! Therefore, for any real number \(b\), the maximum value of \(e^x\) for all \(|x|b\) is \(e^b\). 1 OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. We now show how to find Maclaurin polynomials for ex, sinx,sinx, and cosx.cosx. For each of the following functions, find formulas for the Maclaurin polynomials \(p_0,p_1,p_2\) and \(p_3\). \[g'(t)=\dfrac{f^{(n+1)}(t)}{n! ) Recall that Newtons method xn+1=xnf(xn)f(xn)xn+1=xnf(xn)f(xn) approximates solutions of f(x)=0f(x)=0 near the input x0.x0. From Example 6.12b., the Maclaurin polynomials for sinxsinx are given by. It is important to note that the value \(c\) in the numerator above is not the center \(a\), but rather an unknown value \(c\) between \(a\) and \(x\). You just have x to the f(x) &=\sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n ! The following exercises make use of the functions S5(x)=xx36+x5120S5(x)=xx36+x5120 and C4(x)=1x22+x424C4(x)=1x22+x424 on [,].[,]. x x^{2}+\dfrac{f^{\prime \prime \prime}(0)}{3 !} 0 for each step. f(x)&=\cos x & f(0)&=1\\[5pt] x \(\dfrac{|a_{n+1}|}{|a_n|}=\dfrac{(1)^{n+1}(x1)n^{+1}}{|(1)^n(x1)^n|}=|x1|\). x = approximately over here. Use a graphing utility to compare the graph of \(f\) with the graphs of \(p_0,p_1,p_2\) and \(p_3\). + x^3/(3!) = And especially if we had an ) ) equal to f prime of 0. I want until the term. x 2 The nth Taylor polynomial for ff at 0 is known as the nth Maclaurin polynomial for f.f. ! }\\[5pt] We return to discuss convergence later in this section. f'''(x)&=\sin x & f'''(0)&=0\\[5pt] Find a formula for the nth Maclaurin polynomial. 1 }+\cdots=\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x^{2 n+1}}{(2 n+1) ! x x Calculus Power Series Constructing a Maclaurin Series 1 Answer Andrea S. Jun 21, 2017 ex +e2x = n=0(1 +2n) xn n! }(118)^2=0.03125.\), Similarly, to estimate \(R_2(11)\), we use the fact that, Since \(f'''(x)=\dfrac{10}{27x^{8/3}}\), the maximum value of \(f'''\) on the interval \((8,11)\) is \(f'''(8)0.0014468\). Use Taylors theorem to find an estimate on \(R_n(1)\). Explain the meaning and significance of Taylors theorem with remainder. A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function. }{|x|^n}=\dfrac{|x|}{n+1}\), \(\displaystyle \lim_{n}\dfrac{|a_{n+1}|}{|a_n|}=\lim_{n}\dfrac{|x|}{n+1}=0\). + # f To show that the series converges to \(e^x\) for all \(x\), we use the fact that \(f^{(n)}(x)=e^x\) for all \(n0\) and \(e^x\) is an increasing function on \((,)\). Next, we need to check the endpoints. the derivatives evaluated at 0. Direct link to conleygibbs's post What is the Maclaurin ser, Posted 11 years ago. If the series Equation 6.4 is a representation for f at x = a, we certainly want the series to equal f(a) at x = a. We begin by looking at linear and quadratic approximations of f(x)=x3f(x)=x3 at x=8x=8 and determine how accurate these approximations are at estimating 113.113. Maclaurin series eliminate that choice and force us to choose a=0. is starting to emerge. It's kind of 2 plus ) It starts as : At the end of this video, Sal says that he will show how is the power series of e connected to the power series of cos(x) and sin(x). thing for the sine of x that we did last video. x And what makes Since this is a geometric series, it converges to \(\dfrac{1}{x}\) as long as \(|1x|<1.\) Therefore, the Taylor series found in Example does converge to \(f(x)=\dfrac{1}{x}\) on \((0,2).\), We now consider the more general question: if a Taylor series for a function \(f\) converges on some interval, how can we determine if it actually converges to \(f\)? \[p_5(x)=x\dfrac{x^3}{3!}+\dfrac{x^5}{5!} This formula helps in finding the approximate value of the function. 1999-2023, Rice University. Become a problem-solving champ using logic, not rules. f''(x)&=\dfrac{1}{x^2} & f''(1)&=1\\[5pt] 1. To determine the interval of convergence, we use the ratio test. The proof relies on supposing that e is rational and arriving at a contradiction. 1 0 }-\dfrac{x^{7}}{7 ! Let \(p_n\) be the \(n^{\text{th}}\)-degree Taylor polynomial of \(f\) at \(a\) and let, be the \(n^{\text{th}}\) remainder. ) \(|R_n(x)|\dfrac{e^b}{(n+1)!}|x|^{n+1}\). In maths, we come across several complicated functions that are difficult to solve. + + x^n/(2^n)n! x To find these Taylor polynomials, we need to evaluate ff and its first three derivatives at x=1.x=1. }+\ldots\end{array} \), \(\begin{array}{l}\frac{1}{1-x}\end{array} \), \(\begin{array}{l}\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+. We can see how to derive these expansions using the Maclaurin series formula in the following solved examples section. Determine the interval of convergence. x2 + f + 1(x1)21(x1)2 at a=0a=0 (Hint: Differentiate 11x.)11x. f''(x)&=\cos x & f''(0)&=1\\[5pt] infinite number of terms, it would look like # f^((n))(x)=e^x => f^((n))(0)=1 #, So the Maclaurin series is: We will find the derivatives of the given function f(x) = sin x. f(0) = f '' (0) = f '' (0) = f(4)(0) = . = 0, \(= x-\dfrac{x^{3}}{3 ! Here, we state an important result. Use Taylors theorem to bound the error. Using the same idea as in part a., the result is \(\displaystyle \lim_{n}R_n(x)=0\) for all \(x\), and therefore, the Maclaurin series for \(\sin x\) converges to \(\sin x\) for all real \(x\). ( = 1 0! x x ( 2 Since \(f''(x)=\dfrac{2}{9x^{5/3}}\), the largest value for \(|f''(x)|\) on that interval occurs at \(x=8\). }=0\), for any real number \(x\). In this project, we use the Maclaurin polynomials for \(e^x\) to prove that \(e\) is irrational. = Also, \(g\) is zero at \(t=a\) and \(t=x\) because, \[ \begin{align*} g(a) &=f(x)f(a)f(a)(xa)\dfrac{f''(a)}{2!}(xa)^2++\dfrac{f^{(n)}(a)}{n! We consider this question in more generality in a moment, but for this example, we can answer this question by writing. (x a)n Hence, the Maclaurin series formula is: N n=0 f (n)(0) n! For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by. We find that. \(\displaystyle \sum_{n=0}^\dfrac{|x|^n}{n! }(x-a)^{2}+\dfrac{f^{\prime \prime \prime}(a)}{3 ! Graphs of the function and its Maclaurin polynomials are shown in Figure \(\PageIndex{3}\). a function whose derivative is always equal to that function. &=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}++\dfrac{x^n}{n! things about the number e-- is that when you take the Use Taylors Theorem with Remainder to prove that the Maclaurin series for ff converges to ff on that interval. Also, f(0)=1, so we can conclude the Maclaurin Series expansion will be simply: Your Mobile number and Email id will not be published. I'll use the yellow here. Let's go to n = 4. f (0)(x) = f (x) = ex2 f '(x) = ex2 2x f ''(x) = ex2 2 + 2x ex2 2x = ex2(4x2 + 2) x 0 = e ( 0) 2 1 . assuming 0! = f = \nonumber \], \(\displaystyle \lim_{n}\dfrac{|x|^2}{(2n+3)(2n+2)}=0\), for all \(x\), we obtain the interval of convergence as \((,).\) To show that the Maclaurin series converges to \(\sin x\), look at \(R_n(x)\). ? ) ) Approximating e with a Maclaurin series (which is like a Taylor polynomial centered at x=0 with infinitely many terms). Using the first Taylor polynomial at \(x=8\), we can estimate, \[\sqrt[3]{11}p_1(11)=2+\dfrac{1}{12}(118)=2.25. We now show how to use this definition to find several Taylor polynomials for \(f(x)=\ln x\) at \(x=1\). }|xa|^{n+1} \nonumber \]. Indeed, the first 4 terms for the sine are. approximate e to the x using a Maclaurin series-- }+\cdots=\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x^{2 n}}{(2 n) ! }=1-\frac{x^{2}}{2 ! . the pattern here. How do you find the Maclaurin series of #f(x)=cos(x^2)# x \nonumber \], Solving this inequality for \(x\), we have that the fifth Maclaurin polynomial gives an estimate to within \(0.0001\) as long as \(|x|<0.907.\), Use the fourth Maclaurin polynomial for \(\cos x\) to approximate \(\cos\left(\dfrac{}{12}\right).\). n We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false. Thus, \(sn!R_n(1)=0\). Using the Maclurin series formula, f (x) = f (0)+f (0)x+ f (0) 2! Using the results from part 2, show that for each remainder, Use Taylors theorem to write down an explicit formula for, Use Taylors theorem to find an estimate on. Since the fourth derivative is \(\sin x\), the pattern repeats. x + \nonumber \], c. By Note, there exists a c in the interval \((8,11)\) such that the remainder when approximating \(\sqrt[3]{11}\) by the first Taylor polynomial satisfies, \[R_1(11)=\dfrac{f''(c)}{2!}(118)^2. The Taylor series for \(f\) at 0 is known as the Maclaurin series for \(f\). Find the Taylor series for f(x)=12xf(x)=12x at x=2x=2 and determine its interval of convergence. Each new topic we . Find the Maclaurin series for f(x)=cosx.f(x)=cosx. In the next example, we find the Maclaurin series for ex and sinxsinx and show that these series converge to the corresponding functions for all real numbers by proving that the remainders Rn(x)0Rn(x)0 for all real numbers x. n &\quad+\left[\dfrac{f^{(n)}(t)}{(n1)!}(xt)^{n1}\dfrac{f^{(n+1)}(t)}{n! 2 which can be written as summation x^n/(2^n)n! Integrate the approximation ex1+x+x22++x6720ex1+x+x22++x6720 evaluated at x2 to approximate 01ex2dx.01ex2dx. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo Step 1: Find Coefficients Let f ( x) = ex. &=\dfrac{|x|^2}{(2n+3)(2n+2)}\end{align*}. It requires a bit of manipulation to figure out, but the constant happens to be while the n! If you are redistributing all or part of this book in a print format, Then the series has the form, \[\sum_{n=0}^c_n(xa)^n=c_0+c_1(xa)+c_2(xa)^2+ \dots. Since \(g\) is a polynomial function (in \(t\)), it is a differentiable function. Want to cite, share, or modify this book? + . ex 1 x = 1 + x 2 + x2 6 + x3 24 + . And how did , Posted 11 years ago. point on that curve is the same as the value of f(n)(a)n!. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. ( 7 votes) Upvote ! Use Note to prove that the Maclaurin series for \(f\) converges to \(f\) on that interval. n Jan 13, 2023 OpenStax. This is equal to the reconcile the negatives in some interesting + x4 4! maclaurin e^{x} en. Learn the why behind math with our certified experts, \(e^{x}=1+\dfrac{x}{1 ! Since \(f(x)=e^x\),we know that \(f(x)=f(x)=f''(x)==f^{(n)}(x)=e^x\) for all positive integers \(n\). ? The proof relies on supposing that \(e\) is rational and arriving at a contradiction. f However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to f.f. A: In this question , we will solve the derivative of the given function at some value and rearrange . Since. ? ( + x^3/(3!) Maclaurin Series: Expansion, Formula & Examples with Solutions Triangle Inequalities Using Similar Polygons Conducting an Experiment Confidence Interval for Population Mean Confidence Interval for Population Proportion Confidence Interval for Slope of Regression Line Confidence Interval for the Difference of Two Means Confidence Intervals ) x 2 x, f Find formulas for the Maclaurin polynomials \(p_0,\,p_1,\,p_2\) and \(p_3\) for \(f(x)=\dfrac{1}{1+x}\). }+..\end{array} \), \(\begin{array}{l}\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7! Direct link to Stev's post I managed to find it late, Posted 10 years ago. 3 }}f\left( 0 \right) + \cdots \], Putting the values in the above series, we have If we happen to know that \(f^{(n+1)}(x)\) is bounded by some real number \(M\) on this interval \(I\), then, \[|R_n(x)|\dfrac{M}{(n+1)! So before, you probably + \frac{{{x^3}}}{{3!}} Similarly, at \(x=0,\), \(\displaystyle \sum_{n=0}^(1)^n(01)^n=\sum_{n=0}^(1)^{2n}=\sum_{n=0}^1\). Use these polynomials to estimate 6.6. x where f^(n)(0) is the n-th derivative of f(x) evaluated at 0, and 'n!' ( So we have to find f ? At x=2,x=2, we see that. Use the ratio test to show that the interval of convergence is (,).(,). + f''(0)x^2/(2!) If \(f\) has derivatives of all orders at \(x=a\), then the Taylor series for the function \(f\) at \(a\) is, \[\sum_{n=0}^\dfrac{f^{(n)}(a)}{n!}(xa)^n=f(a)+f(a)(xa)+\dfrac{f''(a)}{2!}(xa)^2++\dfrac{f^{(n)}(a)}{n! The \(n^{\text{th}}\) partial sum of the Taylor series for a function \(f\) at \(a\) is known as the \(n^{\text{th}}\)-degree Taylor polynomial. x Where, ( # f(x) = f(0) + f'(0)x/(0!) Practice, practice, practice. In the following exercises, compute the Taylor series of each function around x=1.x=1. x 1 }+\frac{x^{4}}{4 ! Maclaurin series formula helps inwriting a function as a series (or sum) of terms involving the derivatives of the function. }+\dfrac{x^{5}}{5 ! In order to apply the ratio test, consider, \[\begin{align*} \dfrac{|a_{n+1}|}{|a_n|}&=\dfrac{|x|^{2n+3}}{(2n+3)!}\dfrac{(2n+1)! Which is probably one of the most important mathematical power series ! This simplifies to e^x/2 = 1 + x/2 + x^2/(2^2)2! Write the Maclaurin polynomials \(p_0(x),p_1(x),p_2(x),p_3(x),p_4(x)\) for \(e^x\). \nonumber \], That is, \(f\) can be represented by the geometric series \(\displaystyle \sum_{n=0}^(1x)^n\). first mind blowing thing about the number e. It's just, you could keep As stated above, Maclaurin polynomials are Taylor polynomials centered at zero. [T] Plot exe4(x)exe4(x) where e4(x)=1+x+x22+x36+x424e4(x)=1+x+x22+x36+x424 on [0,2].[0,2]. Estimate the remainder for a Taylor series approximation of a given function. We now state Taylors theorem, which provides the formal relationship between a function \(f\) and its \(n^{\text{th}}\)-degree Taylor polynomial \(p_n(x)\). n Therefore. We now consider the more general question: if a Taylor series for a function ff converges on some interval, how can we determine if it actually converges to f?f? To prove that Rn(x)0,Rn(x)0, we typically use the bound. i.e., the Maclaurin series formula is obtained by substituting a = 0 in the above formula. + Consider the function of the form The Taylor series for ff at 0 is known as the Maclaurin series for f.f. 2 cos So that is cosine of x. To find the series of f (x) = e5x, there are a couple of approaches. plus 1 over 2 factorial, plus 1 over 3 factorial, And something pretty neat That is, the series converges for 0

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