The Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] is :
[tex]1 + 3e^x+ x^2 = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]
What do you mean by Taylor's series ?
The Taylor's series is a way to represent a function as a power series, which is a sum of terms involving the variable raised to increasing powers. The series is centered around a specific point, called the center of the series. The Taylor's series approximates the function within a certain interval around the center point.
The general formula for the Taylor's series of a function f(x) centered at [tex]x = a[/tex] is:
[tex]f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...[/tex]
where [tex]f'(a), f''(a), f'''(a),[/tex] etc. are the derivatives of f(x) evaluated at [tex]x = a[/tex].
Finding the Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] :
We need to find the derivatives of the function at [tex]x=0[/tex]. We have:
[tex]f(x) = 1 + 3e^x + x^2[/tex]
[tex]f(0) = 1 + 3e^0 + 0^2 = 4[/tex]
[tex]f'(x) = 3e^x+ 2x[/tex]
[tex]f'(0) = 3e^0 + 2(0) = 3[/tex]
[tex]f''(x) = 3e^x + 2[/tex]
[tex]f''(0) = 3e^0 + 2 = 5[/tex]
[tex]f'''(x) = 3e^x[/tex]
[tex]f'''(0) = 3e^0 = 3[/tex]
Substituting these values into the general formula for the Taylor's series, we get:
[tex]f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...[/tex]
[tex]f(x) = 4 + 3x + 5x^2/2 + 3x^3/6 + ...[/tex]
Simplifying, we get:
[tex]f(x) = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]
Therefore, the Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] is :
[tex]1 + 3e^x+ x^2 = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]
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