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Mathematics

1 answer:

xeze [42]3 years ago

6 0

Answer:

Step-by-step explanation:

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Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that <u>The Maclaurin series is given by:</u>

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

8 0

4 years ago

How many answers do you have to answer to message people on brainly

tiny-mole [99]

You don’t have to answer people to message

3 0

3 years ago

Read 2 more answers

In triangle xyz, z2 = x2 + y2. triangle xyz has sides x, qy, z opposite to the corresponding vertices x, y, z which equation is

Mrac [35]

${\text{In triangle XYZ}},{\text{ square of one side is equal to the sum of squares of other two sides, }} \hfill \\ {\text{,i}}{\text{.e}}{\text{., if triangle XYZ has sides of measure }}x\,{\text{unit, }}y\,{\text{unit and }}z\,{\text{unit}} \hfill \\ {\text{opposite to the corresponding vertices }}X,{\text{ }}Y\,{\text{and }}Z\,{\text{respectively, then}} \hfill \\ {z^2} = {x^2} + {y^2} \hfill \\ {\text{Since, the given triangle satisfying the condition of Pythagoras theorem, }} \hfill \\{\text{thus, the given triangle would be a right angled triangle and thereby the measure}} \hfill \\{\text{of one angle of triangle will be a right angle say at vertex Z}}{\text{.}} \hfill \\\Rightarrow m\left( {\angle Z} \right) = 90^\circ \hfill \\\Rightarrow m\left( {\angle XZY} \right) = 90^\circ \hfill \\{\text{Thus,the measure of angle XZY is equal to }}90{\text{ }}degrees. \hfill \\$