How do you find the Taylor polynomial of degree 4 for #f(x) = cosh(x)# about #x = 0 # by using the given Taylor polynomial for #e^x#?

Question

Aurora Alvarado · Accepted Answer

#cosh x = 1+x^2/2+x^4/24+o(x^4)#

The MacLaurin series for #e^x# is:

#e^x = sum_(n=0)^oo x^n/(n!)#

Truncating the series at #n=4# we obtain the Taylor polynomial of degree 4:

#e^x = 1 + x +x^2/2 +x^3/6+x^4/24 +o(x^4)#

and substituting #-x# we have:

#e^(-x) = 1 - x +x^2/2 -x^3/6+x^4/24 +o(x^4)#

Using the exponential form of #cosh x# then we have:

#cosh x = (e^x+e^(-x))/2 = 1/2( 1 + x +x^2/2 +x^3/6+x^4/24 + 1 - x +x^2/2 -x^3/6+x^4/24)+o(x^4)#

and we can see that the terms of even order cancel each other, so that:

#cosh x = 1+x^2/2+x^4/24+o(x^4)#

Isabella Ackerman · Answer

undefined To find the Taylor polynomial of degree 4 for $ f(x) = \cosh(x) $ about $ x = 0 $ using the given Taylor polynomial for $ e^x $, you can use the fact that $ \cosh(x) $ is related to $ e^x $.

Since $ \cosh(x) = \frac{e^x + e^{-x}}{2} $, you can use the Taylor polynomial for $ e^x $ to find the Taylor polynomial for $ \cosh(x) $.

The Taylor polynomial for $ e^x $ about $ x = 0 $ up to degree 4 is:

$$ P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} $$

Now, substitute this polynomial into the expression for $ \cosh(x) $ and simplify to obtain the Taylor polynomial for $ \cosh(x) $ about $ x = 0 $ up to degree 4.

$$ \cosh(x) = \frac{e^x + e^{-x}}{2} = \frac{P_4(x) + P_4(-x)}{2} $$

$$ = \frac{1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24}}{2} $$

$$ = 1 + \frac{x^2}{2} + \frac{x^4}{24} $$

Thus, the Taylor polynomial of degree 4 for $ \cosh(x) $ about $ x = 0 $ is:

$$ P_4(x) = 1 + \frac{x^2}{2} + \frac{x^4}{24} $$

Charles Anctil · Answer

undefined To find the Taylor polynomial of degree 4 for $ f(x) = \cosh(x) $ about $ x = 0 $ using the Taylor polynomial for $ e^x $, we can use the fact that $ \cosh(x) $ can be expressed in terms of exponential functions.

1. Recall that $ \cosh(x) = \frac{e^x + e^{-x}}{2} $.

2. Expand $ e^x $ using its Taylor polynomial about $ x = 0 $. Let's denote this Taylor polynomial as $ P_4(x) $:
$$ P_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} $$

3. Replace $ e^x $ in the expression for $ \cosh(x) $ with its Taylor polynomial:
$$ \cosh(x) = \frac{P_4(x) + P_4(-x)}{2} $$

4. Simplify the expression to obtain the Taylor polynomial for $ \cosh(x) $ about $ x = 0 $. This will be the Taylor polynomial of degree 4.

5. The resulting polynomial will represent the Taylor polynomial of degree 4 for $ \cosh(x) $ about $ x = 0 $ using the given Taylor polynomial for $ e^x $.