Applying Taylor Series - Math

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

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Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Tap to reveal answer

Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

← Didn't Know|Knew It →

Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Tap to reveal answer

Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

← Didn't Know|Knew It →

Question

Give the term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Tap to reveal answer

Answer

The term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!}$ = $\frac{f ' ' (0)}{2}$ .

We can find by differentiating twice in succession:

$f (x) = $7^{x}$$

$f ' (x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )= \ln 7 \cdot $7^{x}$$

$f ' '(x) = $\frac{\mathrm{d}$ }{\mathrm{d} x} \left (\ln 7 \cdot $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( $7^{x}$ \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot $7^{x}$$

$f ' '(x)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{x}$$

$f ' '(0)=\left ( \ln 7 \right $)^{2}$ \cdot $7^{0}$ = \left ( \ln 7 \right $)^{2}$ \cdot 1 = \left ( \ln 7 \right $)^{2}$$

The coefficient we want is

$$\frac{f ' ' (0)}{2}$ =\frac{\left ( \ln 7 \right $)^{2}$$}{2}$ ,

so the corresponding term is $\frac{\left( \ln 7 \right $)^{2}$$}{2} $x^{2}$ .

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