Maclaurin series for calc 2?
The Maclaurin series for sin(x) is ∞ ∑ (-1)^n x^(2n+1) / (2n+1)! n=0 Just replace x by 4x² to get the Maclaurin series for sin(4x²): ∞ ∑ (-1)^n (4x²)^(2n+1) / (2n+1)! n=0 which simplifies to ∞ ∑ (-1)^n [4^(2n+1)][x^(4n+2)] / (2n+1)! n=0 The integral of sin(4x²) equals the integral of this sum. Assuming that interchanging summation and integration is legitimate here (which it is), the integral of sin(4x²) is given by ∞ ∑ ∫ {(-1)^n [4^(2n+1)][x^(4n+2)] / (2n+1)!} dx n=0 which equals ∞ ∑ (-1)^n [4^(2n+1)][x^(4n+3)] /[(4n + 3) (2n+1)!] n=0 For the definite integral from 0 to 0.62, the value is ∞ ∑ (-1)^n [4^(2n+1)][0.62^(4n+3)] /[(4n + 3) (2n+1)!] n=0 since the infinite sum, evaluated at 0, is 0. Using only the first two terms. 0.62 ∫ sin(4x²) dx ≈ 0 0.62 ∫ (4x² - (32x^6)/3) dx 0 = (4/3)x³ - (32/21) x^7 (evaluated at 0 and 0.62) ≈ 0.264
Maclaurin series of e^x X sinx?
You need to find the nth derivative of f(x) = e^x sin(x) at x=0. At first this may seem daunting, but just try computing things for a little while: f'(x) = e^x cos(x) + e^x sin(x) f''(x) = (-e^x sin(x) + e^x cos(x)) + (e^x cos(x) + e^x sin(x)) = 2 e^x cos(x) f'''(x) = 2(-e^x sin(x) + e^x cos(x)) f'^(4) (x) = 2(-e^x cos(x) - e^x sin(x) - e^x sin(x) + e^x cos(x)) = -4 e^x sin(x) ... You can show (through an inductive argument) that in general the pattern goes like this: f^(0) (x) = e^x [sin(x)] f^(1) (x) = e^x [cos(x) + sin(x)] f^(2) (x) = 2 e^x [cos(x)] f^(3) (x) = 2 e^x [cos(x) - sin(x)] f^(4) (x) = -4 e^x [sin(x)] f^(5) (x) = -4 e^x [cos(x) + sin(x)] f^(6) (x) = -8 e^x [cos(x)] f^(7) (x) = -8 e^x [cos(x) - sin(x)] ... where after two terms the constant gets doubled, after four terms the sign swaps, and the thing in []'s cycles from sin(x) to cos(x) + sin(x) to cos(x) to (cos(x) - sin(x). Evaluating this at 0, e^x is just 1, the sin(x) is just 0, and cos(x) is just 1, leaving the pattern f^(0) (0) = 1*0 = 0 f^(1) (0) = 1 f^(2) (0) = 2 f^(3) (0) = 2 f^(4) (0) = -4*0 - 0 f^(5) (0) = -4 f^(6) (0) = -8 f^(7) (0) = -8 ... The general pattern is then that every 4th term is 0, every two terms you double, and every four terms you add a negative. The Maclaurin series is then 0*x^0/0! + 1*x^1/1! + 2*x^2/2! + 2*x^3/3! + 0*x^4/4! + (-4)*x^5/5! + (-8)*x^6/6! + (-8)*x^7/7! + ... = x + x^2 + x^3/3 - x^5/30 - x^6/90 - x^7/630 + ... where the coefficients behave according to the rule I gave above. You can write this rule in exponential notation easily if you split it according to how far you are from a multiple of 4: f^(4k+0) (0) = 0 f^(4k+1) (0) = (-4)^k f^(4k+2) (0) = 2*(-4)^k f^(4k+3) (0) = 2*(-4)^k The derivation of this rule is much simpler using complex exponentials, but you might not be aware of them so I won't do that.
Maclaurin Series for sinx?
The problem states to use the Maclaurin series for sinx to compute sin(3°) and correct to five decimal places. I understand that the form will be ∑n=0^inf (f^n(0) / n!) * x^n I also understand that sin(3°) is 0.0523 But how do I use the series to determine this?
What's the Maclaurin series for Tan [sin(x)]?
What's the Maclaurin series for Tan [sin(x)]?I doubt if there is a simple short cut. The series for the tangent function uses Bernoulli numbers and is not as simple as the series for sine or cosine. If it were, you could substitute the sine series into each term. That would be messy even if the tangent series were simple.So it seems you have no option but to differentiate multiple times. The first derivative is [math]\sec^2(\sin(x))\cos(x) = 1[/math] when [math]x = 0[/math]. So the constant term is [math]\tan(\sin(0)) = 0[/math] and the term in [math]x[/math] is [math]x[/math]. The next derivative is [math]2\sec^2(\sin(x))\tan(\sin(x))\cos(x) - \sec^2(\sin(x))\sin(x) = 0[/math] when [math] x = 0[/math]. So the first three terms are [math]1 + x + 0x^2/2![/math]. You can carry on, but I doubt if there is a simple formula.
Expand e^sin x by McLaurin's series?
Using the series for exp and sine (in this order): e^(sin x) = 1 + sin x + (1/2!) (sin x)^2 + (1/3!) (sin x)^3 + (1/4!) (sin x)^4 + ... = 1 + (x - x^3/3! + ...) + (1/2)(x - x^3/3! + ...)^2 + (1/6)(x - x^3/3! + ...)^3 + (1/24)(x - x^3/3! + ...)^4 + ... = 1 + (x - x^3/3! + ...) + (1/2)(x^2 - 2x^4/3! + ...) + (1/6)(x^3 + ...) + (1/24)(x^4 + ...) + ... = 1 + x + (1/2)x^2 - (1/8) x^4 + ... I hope this helps!
Find the Maclaurin series for the function f(x)= (sinx)/x, x≠0; 1, x=0?
The Maclaurin series for sin(x) is sin(x) = x - x^3/3! + x^5/5! - x^7/7! +... so sin(x)/x = 1 - x^2/3! + x^4/5! - x^6/7! + ... if x≠0.
Maclaurin series ln(1 + sinx) - how do they get this answer?
ln(1 + sinx) = x - (1/2)x^2 + (1/6)x^3 - (1/12)x^4 I know I could get it by finding the derivatives, but shouldn't I be able to take the expansion of ln(1 + x), then plug in the expansion of sin(x) INTO the x in the previous expansion and get the same answer? ln(x) = sinx - (1/2)sinx^2 + (1/6)sinx^3 - (1/12)sinx^4 when I plug sinx = x - (x^3)/6 in, I am not getting the right answer. How is getting a plus (1/6)x^3 ever going to be possible when it will be a minus from plugging in the first sinx? Am I just messing up the arithmetic but have the right idea? This goes for all things, like ln(1 + e^x) for example... so confused about combining expansions
How do I prove the Maclaurin series for sin(x) and cos(x)?
We can prove the expansion of circular functions by using indeterminate coefficients and repeated differentiation. First let’s assign [math]\sin x[/math] the infinite sequence[math]\displaystyle\sin x=A+Bx+Cx^2+Dx^3+Ex^4+\cdots\tag*{}[/math]Differentiating gives the new sequence[math]\displaystyle\cos x=B+2Cx+3Dx^2+4Ex^3+5Fx^4+\cdots\tag*{}[/math]So[math]\displaystyle-\sin x=2C+6Dx+12Ex^2+20Fx^3+30Gx^4+\cdots\tag*{}[/math]Setting [math]x=0[/math] in the first two equations gives [math]A=0[/math] and [math]B=1[/math]. Comparing the coefficients of the two expansions for [math]\sin x[/math] gives[math]\displaystyle\begin{align*}A=-2C & \quad\implies\quad C=0\\B=-6D & \quad\implies\quad D=-\frac 16\\C=-12E & \quad\implies\quad E=0\\D=-20F & \quad\implies\quad F=\frac 1{120}\end{align*}\tag*{}[/math]And so on. Therefore[math]\displaystyle\sin x=\sum\limits_{m\geq0}\frac {(-1)^m x^{2m+1}}{(2m+1)!}\tag*{}[/math]A similar thing can be done for [math]\cos x[/math].
Assume that sin(x) equals its Maclaurin series for all x . Use the Maclaurin series for sin(5x^2) to evaluate?
substitute t = 5x^2 ....sin(t) first by definition, the correct mclaurin series expansion for the function f(t) = sin(t) is ∞ ∑ ( -1 )^n * [ t^(2n+1) / (2n+1)! ] ..............but how can we get this? as the following: n=0 first, i am going to define some terms such as n = the number of derivatives & number of factorials f^(n)(t) = is first original and the derived equations. f^(n)(0) = plugging 0 in the original equation and derived equations. an = is coefficient numbers of the variables. IT WILL BE AS TABLE: n ---- f^(n)(t) ------f^n(0)-------an 0------sin(t)---------0-----------(0)/... = 0/1 = 0 (0! is equal to one ) 1------cos(t)--------1-----------1/1! 2-----(-sin(t)--------0-----------0/2! = 0 3---(-cos(t))------(-1)--------(-1/3!) 4------sin(t)----------0------------ 0/4! = 0 5---(cos(t))-------1--------(1/5!) now sin(t) = 0 + 1*t^1 + 0 - [ 1*t^3/3! ] + 0 + [ 1*t^5/5! ] then it will be since its alternating, and its increment for both: the exponent of the variable and to the denominator (with factorials) are odd number, it will be as 2n+1 so the series is ∞ ∑ ( -1 )^n * [ t^(2n+1) / (2n+1)! ] = t - t^3/3! + t^5/5! - .... n = 0 let's substitute back the t = 5x^2 ∞ ∑ ( -1 )^n * [ (5x^2)^(2n+1) / (2n+1)! ] = (5x^2) - (5x^2)^3/3! + . . .. n = 0 ∞ ∑ ( -1 )^n * [ (5x)^(4n+2) / (2n+1)! ] = (5x^2) - 125x^6/3! + .. n = 0 5x^2 - (125/6) * x^6 + .. =====> lets start ingratiating (5/3) * x^3 - (125/42) * x^7 + ... ====> from 0.75 to 0 (5/3) * (0.75^3 - 0^3) - (125/42) * (0.75^7 - 0^7) (5/3) * (27/64) - (125/42) * (2187/16384) (70155/229376) ≈ 0.31 ======== please e-mail me if u have a question.
Which came first, the derivative of sin(x) or the Maclaurin Series of sin(x) (as it looks like one is used to prove the other)?
In the mid-Eighteenth Century, several texts gave arguments for the power series representations for sin(x), ln(1+x), e^x, and other transcendental functions. These results were viewed to be appropriate content for a pre-calculus curriculum, and were then used to argue that if x were infinitesimal, then sin(x) = x and cos(x) = 1, key facts that are then used to find the derivative of sine. Euler’s textbooks are perhaps the best known examples of this approach.By the beginning of the Nineteenth Century, standards of rigor were changing, and direct arguments for these power series expansions fell out of fashion, to be replaced by methods based on Brook Taylor’s work relating the higher derivatives to the coefficients of the power series representation.Certainly mathematicians were working with the power series representation for sin(x) long before the notion of “derivative” as a function was in use.