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erik [133]
3 years ago
6

PLZ HELP!! need this asap!

Mathematics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

x = 1 and x = 3

Step-by-step explanation:

Compare the outputs of the functions for the same input. If they're the same, then f(x) = g(x).

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Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$           $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

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Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$         (with error 0.0001)

 

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

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