Can anyone please check my work? I am particularly unsure about number 1, simply because I am not really sure what that 2 at the end of the equation is... I read it at e^x^2.

Thank you!

Step-by-step explanation:

1) I believe you are correct about the exponent.  f(x) = e^(-x²).

A Maclaurin series is a Taylor series about x=0:

f(x) = f(0) + f'(0)/1! x + f''(0)/2! x² + f'''(0)/3! x³ + ...

First, we find f''(x).

f(x) = e^(-x²)

f'(x) = -2x e^(-x²)

f''(x) = 4x² e^(x²) − 2 e^(x²)

Evaluate at x = 0:

f''(0) = -2

Find the coefficient:

f''(0) / 2! = -2/2! = -1

2) ∑ (xᵏ / k!) = eˣ

∑ 1 / (4ᵏ / k!)

= ∑ (¼ᵏ / k!)

= e^¼

= ∜e

3) The x⁴ term in the Maclaurin series will have a coefficient of f⁽⁴⁾(0) / 4!.  From the pattern, that coefficient is 36.  Therefore:

36 = f⁽⁴⁾(0) / 4!

f⁽⁴⁾(0) = 864

4) f(x) = x cos(x²)

cos(t) = ∑ (-1)ⁿ t²ⁿ / (2n)!

cos(x²) = ∑ (-1)ⁿ x⁴ⁿ / (2n)!

x cos(x²) = ∑ (-1)ⁿ x⁴ⁿ⁺¹ / (2n)!

When n = 0, the first term of the Maclaurin series is x.

When n = 1, the next non-zero term of the Maclaurin series is -x⁵/2.

So the 5th order Maclaurin polynomial is f(x) = x − x⁵/2.

5) e^x = ∑ (xⁿ / n!)

e^(x²) = ∑ (x²ⁿ / n!)

x² e^(x²) = ∑ (x²ⁿ⁺² / n!)

fahrenheit

step-by-step explanation:

fahrenheit is more precise than celsius. the ambient temperature on most of the inhabited world ranges from -20 degrees fahrenheit to 110 degrees fahrenheit—a 130-degree range. on the celsius scale, that range is from -28.8 degrees to 43.3 degrees—a 72.1-degree range. this means that you can get a more exact measurement of the air temperature using fahrenheit because it uses almost twice the scale.

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