Applying Taylor Series - Math

Card 0 of 4

Question

Give the $x^{2}$ term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Answer

The $x^{2}$ 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 $x^{2}$ term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Answer

The $x^{2}$ 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 $\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ .

Compare your answer with the correct one above

Question

Give the $x^{2}$ term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Answer

The $x^{2}$ 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 $\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ .

Compare your answer with the correct one above

Question

Give the $x^{2}$ term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ .

Answer

The $x^{2}$ 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 $\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ .

Compare your answer with the correct one above

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