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3 years ago

78 15 694 13 459 14 ?

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:

12 45 17 98

Step-by-step explanation:

12 36 78

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Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

4 years ago

R=6<br> Find The Area Of A Circle

zheka24 [161]

Answer:

36π

Step-by-step explanation:

7 0

2 years ago

I need help asap

weqwewe [10]

The answer I believe is D. BC is congruent to DF

6 0

4 years ago

What is the answer ?

MrRissso [65]

The angle B will be equal to ∠B=53.13 degrees.

<h3>What is the triangle?</h3>

Triangle is a shape made of three sides in a two-dimensional plane. the sum of the three angles is 180 degrees.

It is given that:-

Three sides of the triangle are:- a= 3, b=4 and c =5

The angle B will be calculated by the following formula;-

$\angle B=Cos^{-1}\dfrac{a^2+c^2-b^2}{2ac}\\\\\\\angle B=Cos^{-1}\dfrac{3^2+5^2-4^2}{2\times 3\times 5}\\\\\\\angle B=Cos^{-1}(0.6)\\$

∠B=53.13 degrees

Hence the angle B will be equal to ∠B=53.13 degrees.

To know more about triangles follow

brainly.com/question/17335144

#SPJ1

6 0

2 years ago

Can a triangle have side lengths 4, 4, 7? <br> 1.no <br> 2. Yes <br> 3. not enough information

Marina CMI [18]

Yes! The lengths of each side must be less than the sum of the other two lengths. It looks like this:
4+4>7
7+4>4
4+7>4

8 0

3 years ago

78 15 694 13 459 14 ?​

78 15 694 13 459 14 ?