How do I find the sum of a Taylor series $\infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ ?

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How do I find the sum of a Taylor series $\infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ ?

How do I find the value of $f (x) = \infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $f(x)=sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ . The interval of convergence is (0,4)

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Answer 1 · 2018-04-10T10:20:56

See below.

Explanation:

For $| y | < 1$ $abs y < 1$ we have

$\infty \sum k = 0 {(- 1)}^{k} y^{k} = \frac{1}{1 + y}$ $sum_(k=0)^oo (-1)^ky^k = 1/(1+y)$

considering now

$y = \frac{x - 2}{2}$ $y = (x-2)/2$

we have

$\infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}} = \frac{1}{2} (\frac{1}{1 + \frac{x - 2}{2}})$ $sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1)) = 1/2(1/(1+(x-2)/2))$

for $\frac{| x - 2 |}{2} < 1$ $abs (x-2)/2 < 1$

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How do I find the sum of a Taylor series $\infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ ?

How do I find the value of $f (x) = \infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $f(x)=sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ . The interval of convergence is (0,4)

1 Answer

Explanation:

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How do I find the sum of a Taylor series ∞∑n=0(−1)n(x−2)n2n+1sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))?

How do I find the value of f(x)=∞∑n=0(−1)n(x−2)n2n+1f(x)=sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1)). The interval of convergence is (0,4)

1 Answer

Explanation:

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How do I find the sum of a Taylor series $\infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ ?

How do I find the value of $f (x) = \infty \sum n = 0 \frac{{(- 1)}^{n} {(x - 2)}^{n}}{2^{n + 1}}$ $f(x)=sum_(n=0)^oo((-1)^n(x-2)^n)/(2^(n+1))$ . The interval of convergence is (0,4)