How do I obtain constant terms in binomial expansion?
Let us try to view the general term of a binomial expansion in a slightly different way.Let us consider an example where we need to find the constant term in the expansion of [math](x - \frac{2}{x^2})^9[/math]General term for the above binomial is: [math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-\frac{2}{x^2})^{r}[/math][math]T_{r+1} = \ ^{9}C_{r}(x)^{9- r}(-2)^{r}(x)^{-2r}[/math][math]T_{r+1} = \ ^{9}C_{r}(-2)^{r}(x)^{9-3r}[/math]Now for a term in the expansion to be constant, the power of x should be 0.So, 9 - 3r = 0r =3Therefore, the 4th term in the expansion of [math](x - \frac{2}{x^2})^9[/math] is the constant term.[math]T_{3+1} = \ ^{9}C_{3}(-2)^{3}(x)^{9-3*3}[/math][math]T_{4} = \ 84 (-8) \ = \ - 672 [/math]Method: Step 1: Find the general term in the expansion of the binomial.Step 2: Collect all the powers of x terms and make it one entityStep 3: Set the power of x equal to 0 and find the value of rStep 4: Substitute back the value of r in the general term to get the constant term.Note: If r is fractional, then there is no constant term in the expansion.I hope it helps!
Find the first five non-zero terms of Maclaurin series (Taylor series centered at x=0) for the function below?
Since e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ..., xe^x = x(1 + x + x^2/2! + x^3/3! + x^4/4! + ...) ........= x + x^2 + x^3/2! + x^4/3! + x^5/4! + ... I hope this helps!
Find the first five non-zero terms of the power series representation centered at 0 for the function?
Start with the geometric series 1/(1 - t) = Σ(n = 0 to ∞) t^n, convergent for |t| < 1. Replace t with -x^2: 1/(1 + x^2) = Σ(n = 0 to ∞) (-1)^n x^(2n), convergent for |-x^2| = |x|^2 < 1 ==> |x| < 1. Integrate both sides from 0 to t: arctan t = Σ(n = 0 to ∞) (-1)^n t^(2n+1)/(2n+1), convergent (save endpoint(s)) for |t| < 1. Let t = x/8: arctan(x/8) = Σ(n = 0 to ∞) (-1)^n x^(2n+1)/(8^(2n+1) * (2n+1)), convergent (save endpoint(s)) for |x/8| < 1 ==> |x| < 8. So, R = 8. I hope this helps!
The function f(x) = sin(4x) has a Maclaurin series. Find the first 4 nonzero terms in the series?
4x - (4x)^3/3! + (4x)^5/5! -(4x)^7/7
Evaluate the indefinite integral as a power series?
if g(t) = ln ( 1 - t ) then g ' = - 1 / [ 1 - t ] = - Σ t^n , | t | < 1....integrate this series to get ln ( 1 - t)....divide the new series , term by term , with t.....multiply that answer by [ 1 / 4 ]......last series is for | t | < 1 , but t ╪ 0
What is the binomial expansion for (1+x)^-1?
Hello,Binomial expansion for this is followed the same way as expanding larger power expansionsSo (1+x)^-1 = 1 + (-1)x + (-1)(-1–1)/2! x^2 + (-1)(-1–1)(-1–2)/3! x^3 +…….So simplifying we get1 -x + x^2 - x^3 +….This followes a simple formula given by(1 + x)^n = 1 + nx + n(n-1)/2!. x^2 + n(n-1)(n-2)/3!. x^3 +…..Bear in mind this will result in an infinite series and will converge for values belonging to |x| < 1
How do I find the power series representation of [math]\frac{(3 + 2x)}{(1 - x)^2}[/math]?
Let’s see… First, note that Long Division yields that:\begin{align} \frac{3+2x}{(1-x)^2} & = \frac{-2(1-x)+5}{(1-x)^2} \\ & = \frac{-2}{1-x} + \frac{5}{(1-x)^2} \end{align}Once there, the power series for the first expression [math]\displaystyle \frac{-2}{1-x}[/math] can be obtained using the infinite geometric series formula:\begin{align} \frac{1}{1-x} & = x^0 + x^1 + \dots \qquad (-1
How to evaluate the indefinite integral as a power series?
Start with the geometric series 1/(1 - t) = 1 + t + t^2 + t^3 + t^4 + ... Let t = -x^2: 1/(1 + x^2) = 1 - x^2 + x^4 - x^6 + x^8 - ... Intgerate both sides from 0 to t: arctan t = t - t^3/3 + t^5/5 - t^7/7 + t^9/9 - ... Let t = x^2: arctan(x^2) = x^2 - x^6/3 + x^10/5 - x^14/7 + x^18/9 - ... Integrate both sides one more time: ∫ arctan(x^2) dx = (x^3/3 - x^7/21 + x^11/55 - x^15/105 + x^19/171 - ...) + C. I hope this helps!
How to evaluate the indefinite integral as a power series?
Start with the infinite geometric series: 1/(1 - t) = ∑ t^n (from n=0 to ∞). If we integrating both sides of this with respect to to, we get: -ln(1 - t) = ∑ t^(n + 1) (from n=0 to ∞) ==> ln(1 - t) = -∑ t^(n + 1) (from n=0 to ∞). Dividing both sides by 5t yields that the integrand has power series: ln(1 - t)/(5t) = (-1/5)∑ t^n (from t=0 to ∞). --- Integrating the above term-by-term yields: ∫ ln(1 - t)/(5t) dt = ∫ [(-1/5)∑ t^n (from t=0 to ∞)] dt = C - (1/5)∑ t^(n + 1)/(n + 1) (from t=0 to ∞). So, the first five non-zero terms are (ignoring the constant of integration): (1/5)∑ t^(n + 1)/(n + 1) (from t=0 to 4) = (-1/5)(t^1/1 + t^2/2 + t^3/3 + t^4/4 + t^5/5) = (-1/5)t - (1/10)t^2 - (1/15)t^3 - (1/20)t^4 - (1/25)t^5. (B) Since the infinite geometric series that we started out with has a radius of convergence of 1 about t = 0, we see that this series also have a radius of convergence of 1 about t = 0; however, even though the infinite geometric series doesn't converge for t = -1, this series does. I hope this helps!