Only one of these calculations is correct to evaluate the Maclaurin series forf(x) =1+xSelect one:1.Step 1:= 1+x+x +x+1+x1.=1+x' +x +x° +1+x?Step 2:4.9.Step 3:=x+x' +x°+x'+1+x?1.Step 1:+ .3=1-x1-x+x-xStep 2:1= 1 -x² + x* – x6 +1+x²Step 3:= x -x'+ x° - x' + ...31+x²2.Step 1:1= 1+x +x² +x° +1-x

Given function is fx=x1+x2 We have to find the maclurins series of fx=x1+x2 The maclurins series of…

Answered: Only one of these calculations is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Only one of these calculations is correct to evaluate the Maclaurin series for
f(x) =
1+x
Select one:
1.
Step 1:
= 1+x+x +x+
1+x
1.
=1+x' +x +x° +
1+x?
Step 2:
4.
9.
Step 3:
=x+x' +x°+x'+
1+x?
1
.
Step 1:
+ .
3
=
1-x
1-x+x-x
Step 2:
1
= 1 -x² + x* – x6 +
1+x²
Step 3:
= x -x'+ x° - x' + ..
.3
1+x²
2.
Step 1:
1
= 1+x +x² +x° +
1-x

Expert Solution

Step 1

Given function is $f (x) = \frac{x}{1 + x^{2}}$

We have to find the maclurins series of $f (x) = \frac{x}{1 + x^{2}}$

The maclurins series of the function $f (x) = \frac{1}{1 + x}$ is given by

$\frac{1}{1 + x} = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n}$

Hence, $\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + x^{4} - x^{5} + .......$

Now, substitute $x = x^{2}$ in $\frac{1}{1 + x} = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{n}$

Hence,

$\begin{matrix} \frac{1}{1 + x^{2}} =_{n = 1}^{\infty} {(- 1)}^{n} {(x^{2})}^{n} \\ =_{n = 1}^{\infty} {(- 1)}^{n} x^{2 n} \end{matrix}$

Therefore, $\frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - x^{6} + x^{8} - x^{10} + .........$

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Only one of these calculations is correct to evaluate the Maclaurin series for f(x)=- 1+x2 Select one: Step 1: = 1+x+x² + 1+x 1 Step 2: = 1+x? + 1+x? %3D Step 3: x7 + = X 1+x? 1 Step 1: = 1-x+x2 - x° +.. 1-X 1 Step 2: = 1-x2 +x -x° + 1+x2 Step 3: = x -x' +x5 -x + 1+x2 Step 1: 1 = 1+x +x2 1- X