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What is the Taylor's series for 1+3e^x+1 at x=0​

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Answer:

To find the Taylor series of a function f(x) about a point a, we can use the following formula:

f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

where f'(a), f''(a), f'''(a), ... denote the first, second, third, ... derivatives of f evaluated at a.

In this case, we have:

f(x) = 1 + 3e^(x+1)

To find the Taylor series about x=0, we need to evaluate the function and its derivatives at x=0.

f(0) = 1 + 3e^(0+1) = 1 + 3e

f'(x) = 3e^(x+1)

f'(0) = 3e^(0+1) = 3e

f''(x) = 3e^(x+1)

f''(0) = 3e^(0+1) = 3e

f'''(x) = 3e^(x+1)

f'''(0) = 3e^(0+1) = 3e

and so on.

Substituting these values into the formula for the Taylor series, we get:

f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...

= (1 + 3e) + 3ex + 3ex^2/2! + 3ex^3/3! + ...

= 1 + (3 + 3e)x + (3/2)e x^2 + (1/2)e x^3 + ...


Therefore, the Taylor series for 1+3e^x+1 about x=0 is:

1 + (3 + 3e)x + (3/2)e x^2 + (1/2)e x^3 + ...

User Giuseppedeponte
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