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To find the Maclaurin series for the given function [tex]\( f(x) = 3 e^{x^2} \)[/tex], we need to expand this function around [tex]\( x = 0 \)[/tex]. Let us first understand that the Maclaurin series for any function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = f(0) + f'(0)\frac{x}{1!} + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} + \cdots \][/tex]
In this case, [tex]\( f(x) = 3 e^{x^2} \)[/tex], we will calculate the necessary derivatives and evaluate them at [tex]\( x = 0 \)[/tex].
Step 1: Calculate [tex]\( f(0) \)[/tex]
[tex]\[ f(0) = 3 e^{0^2} = 3 e^0 = 3 \][/tex]
Step 2: Calculate the first derivative [tex]\( f'(x) \)[/tex] and find [tex]\( f'(0) \)[/tex]
Let’s find the first derivative:
[tex]\[ f(x) = 3 e^{x^2} \][/tex]
[tex]\[ f'(x) = 3 \left( e^{x^2} \right)' \][/tex]
Using the chain rule for [tex]\( e^{x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} e^{x^2} = e^{x^2} \cdot \frac{d}{dx} (x^2) = 2x e^{x^2} \][/tex]
Thus:
[tex]\[ f'(x) = 3 \cdot 2x e^{x^2} = 6x e^{x^2} \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 6 \cdot 0 \cdot e^{0^2} = 0 \][/tex]
Step 3: Calculate the second derivative [tex]\( f''(x) \)[/tex] and find [tex]\( f''(0) \)[/tex]
Let's find the second derivative:
[tex]\[ f''(x) = \frac{d}{dx} (6x e^{x^2}) \][/tex]
Apply the product rule here:
[tex]\[ f''(x) = 6 \left( x \frac{d}{dx} (e^{x^2}) + e^{x^2} \frac{d}{dx} (x) \right) \][/tex]
[tex]\[ f''(x) = 6 \left( x \cdot 2x e^{x^2} + e^{x^2} \right) \][/tex]
[tex]\[ f''(x) = 6 \left( 2x^2 e^{x^2} + e^{x^2} \right) \][/tex]
[tex]\[ f''(x) = 6 \left( 2x^2 e^{x^2} + e^{x^2} \right) = 6 e^{x^2} (2x^2 + 1) \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 6 e^{0^2} (2(0)^2 + 1) = 6 \cdot 1 \cdot 1 = 6 \][/tex]
Step 4: Calculate the third derivative [tex]\( f'''(x) \)[/tex] and find [tex]\( f'''(0) \)[/tex]
Let's find the third derivative:
[tex]\[ f'''(x) = \frac{d}{dx} (6 e^{x^2} (2x^2 + 1)) \][/tex]
Apply the product rule:
[tex]\[ f'''(x) = 6 \left[ \frac{d}{dx} (e^{x^2}) (2x^2 + 1) + e^{x^2} \frac{d}{dx} (2x^2 + 1) \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 2x e^{x^2} (2x^2 + 1) + e^{x^2} (4x) \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 2x (2x^2 + 1) e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ (4x^3 + 2x) e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 4x^3 e^{x^2} + 2x e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 4x^3 e^{x^2} + 6x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 e^{x^2} (4x^3 + 6x) \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f'''(0) = 6 \cdot e^{0^2} (4(0)^3 + 6 \cdot 0) = 0 \][/tex]
Step 5: Calculate the fourth derivative [tex]\( f^{(4)}(x) \)[/tex] and find [tex]\( f^{(4)}(0) \)[/tex]
Continuing, we need the fourth derivative:
[tex]\[ f^{(4)}(x) = \frac{d}{dx} \left[ 6 e^{x^2} (4x^3 + 6x) \right] \][/tex]
Again, use the product rule:
[tex]\[ f^{(4)}(x) = 6 \left[ \frac{d}{dx} (e^{x^2}) (4x^3 + 6x) + e^{x^2} \frac{d}{dx} (4x^3 + 6x) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 2x e^{x^2} (4x^3 + 6x) + e^{x^2} (12x^2 + 6) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 2x (4x^3 + 6x) e^{x^2} + e^{x^2} (12x^2 + 6) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ (8x^4 + 12x^2) e^{x^2} + 12x^2 e^{x^2} + 6 e^{x^2} \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 8x^4 e^{x^2} + 24x^2 e^{x^2} + 6 e^{x^2} \right] \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f^{(4)}(0) = 6 (8(0)^4 + 24(0)^2 + 6)e^{0^2} = 36 \][/tex]
So, summarizing the coefficients:
- [tex]\( f(0) = 3 \)[/tex]
- [tex]\( f'(0) = 0 \)[/tex]
- [tex]\( f''(0) = 6 \)[/tex]
- [tex]\( f'''(0) = 0 \)[/tex]
- [tex]\( f^{(4)}(0) = 36 \)[/tex]
Thus, the first four non-zero terms of the Maclaurin series for [tex]\( f(x) = 3 e^{x^2} \)[/tex] are:
[tex]\[ 3 + \frac{6}{2!} x^2 + \frac{36}{4!} x^4 \][/tex]
Simplifying:
[tex]\[ 3 + 3 x^2 + \frac{3}{2} x^4 \][/tex]
Which results in:
[tex]\[ f(x) = 3 + 3x^2 + \frac{3}{2}x^4 \][/tex]
So, the first four non-zero terms of the Maclaurin series for [tex]\( f(x) = 3 e^{x^2} \)[/tex] are:
[tex]\[ 3, 3x^2, \frac{3}{2}x^4 \][/tex]
[tex]\[ f(x) = f(0) + f'(0)\frac{x}{1!} + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} + \cdots \][/tex]
In this case, [tex]\( f(x) = 3 e^{x^2} \)[/tex], we will calculate the necessary derivatives and evaluate them at [tex]\( x = 0 \)[/tex].
Step 1: Calculate [tex]\( f(0) \)[/tex]
[tex]\[ f(0) = 3 e^{0^2} = 3 e^0 = 3 \][/tex]
Step 2: Calculate the first derivative [tex]\( f'(x) \)[/tex] and find [tex]\( f'(0) \)[/tex]
Let’s find the first derivative:
[tex]\[ f(x) = 3 e^{x^2} \][/tex]
[tex]\[ f'(x) = 3 \left( e^{x^2} \right)' \][/tex]
Using the chain rule for [tex]\( e^{x^2} \)[/tex]:
[tex]\[ \frac{d}{dx} e^{x^2} = e^{x^2} \cdot \frac{d}{dx} (x^2) = 2x e^{x^2} \][/tex]
Thus:
[tex]\[ f'(x) = 3 \cdot 2x e^{x^2} = 6x e^{x^2} \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = 6 \cdot 0 \cdot e^{0^2} = 0 \][/tex]
Step 3: Calculate the second derivative [tex]\( f''(x) \)[/tex] and find [tex]\( f''(0) \)[/tex]
Let's find the second derivative:
[tex]\[ f''(x) = \frac{d}{dx} (6x e^{x^2}) \][/tex]
Apply the product rule here:
[tex]\[ f''(x) = 6 \left( x \frac{d}{dx} (e^{x^2}) + e^{x^2} \frac{d}{dx} (x) \right) \][/tex]
[tex]\[ f''(x) = 6 \left( x \cdot 2x e^{x^2} + e^{x^2} \right) \][/tex]
[tex]\[ f''(x) = 6 \left( 2x^2 e^{x^2} + e^{x^2} \right) \][/tex]
[tex]\[ f''(x) = 6 \left( 2x^2 e^{x^2} + e^{x^2} \right) = 6 e^{x^2} (2x^2 + 1) \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 6 e^{0^2} (2(0)^2 + 1) = 6 \cdot 1 \cdot 1 = 6 \][/tex]
Step 4: Calculate the third derivative [tex]\( f'''(x) \)[/tex] and find [tex]\( f'''(0) \)[/tex]
Let's find the third derivative:
[tex]\[ f'''(x) = \frac{d}{dx} (6 e^{x^2} (2x^2 + 1)) \][/tex]
Apply the product rule:
[tex]\[ f'''(x) = 6 \left[ \frac{d}{dx} (e^{x^2}) (2x^2 + 1) + e^{x^2} \frac{d}{dx} (2x^2 + 1) \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 2x e^{x^2} (2x^2 + 1) + e^{x^2} (4x) \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 2x (2x^2 + 1) e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ (4x^3 + 2x) e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 4x^3 e^{x^2} + 2x e^{x^2} + 4x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 \left[ 4x^3 e^{x^2} + 6x e^{x^2} \right] \][/tex]
[tex]\[ f'''(x) = 6 e^{x^2} (4x^3 + 6x) \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f'''(0) = 6 \cdot e^{0^2} (4(0)^3 + 6 \cdot 0) = 0 \][/tex]
Step 5: Calculate the fourth derivative [tex]\( f^{(4)}(x) \)[/tex] and find [tex]\( f^{(4)}(0) \)[/tex]
Continuing, we need the fourth derivative:
[tex]\[ f^{(4)}(x) = \frac{d}{dx} \left[ 6 e^{x^2} (4x^3 + 6x) \right] \][/tex]
Again, use the product rule:
[tex]\[ f^{(4)}(x) = 6 \left[ \frac{d}{dx} (e^{x^2}) (4x^3 + 6x) + e^{x^2} \frac{d}{dx} (4x^3 + 6x) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 2x e^{x^2} (4x^3 + 6x) + e^{x^2} (12x^2 + 6) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 2x (4x^3 + 6x) e^{x^2} + e^{x^2} (12x^2 + 6) \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ (8x^4 + 12x^2) e^{x^2} + 12x^2 e^{x^2} + 6 e^{x^2} \right] \][/tex]
[tex]\[ f^{(4)}(x) = 6 \left[ 8x^4 e^{x^2} + 24x^2 e^{x^2} + 6 e^{x^2} \right] \][/tex]
Evaluating at [tex]\( x = 0 \)[/tex]:
[tex]\[ f^{(4)}(0) = 6 (8(0)^4 + 24(0)^2 + 6)e^{0^2} = 36 \][/tex]
So, summarizing the coefficients:
- [tex]\( f(0) = 3 \)[/tex]
- [tex]\( f'(0) = 0 \)[/tex]
- [tex]\( f''(0) = 6 \)[/tex]
- [tex]\( f'''(0) = 0 \)[/tex]
- [tex]\( f^{(4)}(0) = 36 \)[/tex]
Thus, the first four non-zero terms of the Maclaurin series for [tex]\( f(x) = 3 e^{x^2} \)[/tex] are:
[tex]\[ 3 + \frac{6}{2!} x^2 + \frac{36}{4!} x^4 \][/tex]
Simplifying:
[tex]\[ 3 + 3 x^2 + \frac{3}{2} x^4 \][/tex]
Which results in:
[tex]\[ f(x) = 3 + 3x^2 + \frac{3}{2}x^4 \][/tex]
So, the first four non-zero terms of the Maclaurin series for [tex]\( f(x) = 3 e^{x^2} \)[/tex] are:
[tex]\[ 3, 3x^2, \frac{3}{2}x^4 \][/tex]
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