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Nadya [2.5K]
3 years ago
6

Derive the first three (non-zero) terms of Taylor's series expansion for the function

Mathematics
1 answer:
mestny [16]3 years ago
7 0

Answer:

We want to find the first 3 terms of the Taylor's series expansion for f(x) = sin(x) around x = 0.

Remember that a Taylor's series expansion of a function f(x) around the point x₀ is given by:

f(x) = f(x_0) + \frac{1}{2!}f'(x_0)*(x - x_0) + \frac{1}{3!}*f''(x_0)*(x - x0)^2 + ...

Where in the formula we have the first 3 terms of the expansion (but there are a lot more).

So, if:

f(x) = sin(x)

x₀ = 0

The terms are:

f(x_0) = sin(0) = 0

\frac{1}{2!}*f(x_0)'*(x - x_0) = \frac{1}{2} cos(0)*(x - 0) = x/2

\frac{1}{3!}*f(x_0)''*(x - x_0)^2 = \frac{1}{6}*-sin(0)*(x - 0)^2 = 0

\frac{1}{4!}*f(x_0)'''*(x - x_0)^3 = \frac{1}{24}*-cos(0)*(x- 0)^3 = -\frac{x^3}{24}

We already can see that the next term is zero (because when we derive the cos part, we will get a sin() that is zero when evaluated in x = 0), then the next non zero term is:

\frac{1}{6!}*f(x_0)''''*(x - x_0)^5 = \frac{1}{2*3*4*5*6} *(x - 0)^5 = \frac{x^5}{720}

Then we can write:

sin(x) = \frac{x}{2} - \frac{x^3}{24} + \frac{x^5}{720}

Evaluating this in x = 0.2, we get:

sin(0.2) = \frac{0.2}{2} - \frac{0.2^3}{24} + \frac{0.2^5}{720} = 0.099667

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(note that in English, if it is not worth X dollars, means if it is not worth AT LEAST X dollars")
contrapositive of 
<span>
"If an item is not worth five dimes, then it is not worth two quarters.”
is the negation of the converse, which become
"If an item is worth two quarters, then it is worth (at least) five dimes."


The converse of the previous statement is therefore
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as before.

Note that the converse does NOT logically mean the same as the original statement.

</span>
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3 years ago
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