How do you find the third degree Taylor polynomial for #f(x)= ln x#, centered at a=2?

Answer 1

#ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3#.

The general form of a Taylor expansion centered at #a# of an analytical function #f# is #f(x)=sum_{n=0}^oof^((n))(a)/(n!)(x-a)^n#. Here #f^((n))# is the nth derivative of #f#.
The third degree Taylor polynomial is a polynomial consisting of the first four (#n# ranging from #0# to #3#) terms of the full Taylor expansion.
Therefore this polynomial is #f(a)+f'(a)(x-a)+(f''(a))/2(x-a)^2+(f'''(a))/6(x-a)^3#.
#f(x)=ln(x)#, therefore #f'(x)=1/x#, #f''(x)=-1/x^2#, #f'''(x)=2/x^3#. So the third degree Taylor polynomial is: #ln(a)+1/a(x-a)-1/(2a^2)(x-a)^2+1/(3a^3)(x-a)^3#.
Now we have #a=2#, so we have the polynomial: #ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3#.
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Answer 2

To find the third-degree Taylor polynomial for ( f(x) = \ln(x) ) centered at ( a = 2 ), we first find the values of the function and its derivatives at the point ( x = 2 ):

  1. ( f(2) = \ln(2) )
  2. ( f'(x) = \frac{1}{x} ), so ( f'(2) = \frac{1}{2} )
  3. ( f''(x) = -\frac{1}{x^2} ), so ( f''(2) = -\frac{1}{4} )
  4. ( f'''(x) = \frac{2}{x^3} ), so ( f'''(2) = \frac{1}{4} )

The third-degree Taylor polynomial is given by:

[ P_3(x) = f(2) + f'(2)(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!}(x-2)^3 ]

Substitute the values:

[ P_3(x) = \ln(2) + \frac{1}{2}(x-2) - \frac{1}{8}(x-2)^2 + \frac{1}{24}(x-2)^3 ]

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Answer 3

To find the third-degree Taylor polynomial for ( f(x) = \ln(x) ) centered at ( a = 2 ), we first need to find the derivatives of ( f(x) ) up to the third derivative evaluated at ( x = 2 ).

  1. ( f(x) = \ln(x) )
  2. ( f'(x) = \frac{1}{x} )
  3. ( f''(x) = -\frac{1}{x^2} )
  4. ( f'''(x) = \frac{2}{x^3} )

Now, evaluate these derivatives at ( x = 2 ):

  1. ( f(2) = \ln(2) )
  2. ( f'(2) = \frac{1}{2} )
  3. ( f''(2) = -\frac{1}{4} )
  4. ( f'''(2) = \frac{1}{4} )

Now, use the Taylor series formula:

( T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 )

Substitute the values:

( T_3(x) = \ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + \frac{1}{24}(x - 2)^3 )

This is the third-degree Taylor polynomial for ( f(x) = \ln(x) ) centered at ( a = 2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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