How do you find the taylor series series for # (4-x)^(1/2)#?

Answer 1
There's a few different ways this can be done. Probably the most straight-forward approach is to let #f(x)=(4-x)^(1/2)# and write the Taylor series about #x=0# as #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots#
Since #f(x)=(4-x)^(1/2)#, #f(0)=4^{1/2}=2#, #f'(x)=\frac{1}{2}(4-x)^{-1/2}*(-1)#, #f'(0)=-\frac{1}{2}*4^{-1/2}=-\frac{1}{4}#,
#f''(x)=\frac{1}{4}(4-x)^{-3/2}*(-1)#, #f''(0)=-\frac{1}{4}*4^{-3/2}=-\frac{1}{32}#, etc...
Eventually this leads to the Taylor series #2-\frac{1}{4}x-\frac{1}{64}x^2-\frac{1}{512}x^{3}-\cdots#, which converges to #f(x)=(4-x)^{1/2}# for #|x|<4#.
Another approach is to use a bit of slick algebra along with the generalized binomial theorem (discovered by Newton) that #(1+x)^{p}=1+px+\frac{p(p-1}}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots# for #|x|<1#.
We have: #(4-x)^{1/2}=4^{1/2}(1-x/4)^{1/2}=2(1+(-x/4))^{1/2}#. Now use the generalized binomial theorem with #-x/4# in place of #x# and #p=1/2# to get, for #|-x/4|<1 \Leftrightarrow |x|<4#
#(4-x)^{1/2}#
#=2(1+(-x/4))^{1/2}#
#=2(1+(1/2)(-x/4)+\frac{1/2*-1/2}{2!}(-x/4)^2+\frac{1/2*-1/2*-3/2}{3!}(-x/4)^3+\cdots)#
#=2-x/4-\frac{1}{64}x^2-\frac{1}{512}x^{3}-\cdots#
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Answer 2

To find the Taylor series for ( (4-x)^{1/2} ), follow these steps:

  1. Determine the function and the point about which you want to expand the Taylor series. In this case, the function is ( (4-x)^{1/2} ) and the point is ( x = 0 ).
  2. Find the derivatives of the function up to the desired order.
  3. Evaluate the derivatives at the point of expansion (( x = 0 )).
  4. Write out the Taylor series using the formula:

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

where ( f(a) ) is the value of the function at the point of expansion and the derivatives are evaluated at that point.

For ( (4-x)^{1/2} ):

  • ( f(0) = (4-0)^{1/2} = 2 )
  • ( f'(x) = -\frac{1}{2}(4-x)^{-1/2} )
  • ( f''(x) = \frac{1}{4}(4-x)^{-3/2} )
  • ( f'''(x) = -\frac{3}{8}(4-x)^{-5/2} )

Now evaluate these derivatives at ( x = 0 ):

  • ( f'(0) = -\frac{1}{2}(4-0)^{-1/2} = -\frac{1}{4} )
  • ( f''(0) = \frac{1}{4}(4-0)^{-3/2} = \frac{1}{8} )
  • ( f'''(0) = -\frac{3}{8}(4-0)^{-5/2} = -\frac{3}{32} )

Plug these values into the Taylor series formula:

[ (4-x)^{1/2} = 2 - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{3}{32}x^3 + \cdots ]

This is the Taylor series expansion for ( (4-x)^{1/2} ) centered at ( x = 0 ).

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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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