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frozen [14]
3 years ago
13

Find the taylor polynomial t3(x) for the function f centered at the number

Mathematics
1 answer:
olga nikolaevna [1]3 years ago
5 0
e^{-4x}=\displaystyle\sum_{n=0}^\infty\frac{(-4x)^n}{n!}=1+(-4x)+\dfrac{(-4x)^2}2+\dfrac{(-4x)^3}6+\cdots
e^{-4x}=1-4x+8x^2-\dfrac{32x^3}3+\cdots

\sin2x=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^k(2x)^{2k+1}}{(2k+1)!}=(2x)-\dfrac{(2x)^3}6+\cdots
\sin2x=2x-\dfrac{4x^3}3+\cdots

e^{-4x}\sin2x=\left(1-4x+8x^2-\dfrac{32x^3}3+\cdots\right)\left(2x-\dfrac{4x^3}3+\cdots\right)
e^{-4x}\sin2x=2x-8x^2+\dfrac{44x^3}3+\cdots

\implies T_3(x)=2x-8x^2+\dfrac{44x^3}3
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rewona [7]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: f(0) = -3

\qquad \tt \rightarrow \: f(2) = 1

\qquad \tt \rightarrow \: f(4) = 1

____________________________________

\large \tt Solution  \: :

F(x) represents the value of y on curve for given value of x

\textsf{\large First -}

\qquad \tt \rightarrow \: f(0) =( 0) {}^{2}  - 3

[ since 0 < 2 ]

\qquad \tt \rightarrow \: f(0) = 0 {}^{}  - 3

\qquad \tt \rightarrow \: f(0) =   - 3

\textsf{\large Second}

\qquad \tt \rightarrow \: f(2) =   (2) {}^{2}  - 3

[ since 2 = 2 ]

\qquad \tt \rightarrow \: f(2) =   4{}^{}  - 3

\qquad \tt \rightarrow \: f(2) =   1

\textsf{\large Third -}

\qquad \tt \rightarrow \: f(4) =    - 4 + 5

[ since 4 > 2 ]

\qquad \tt \rightarrow \: f(4) =    1

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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2 years ago
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loris [4]

Answer:

By these examples you are able to see that the square of an odd number is always 1 more than a multiple of 4.

Step-by-step explanation:

For examples,

Let's consider squares of 3, 11, 25, 37 and 131.

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