How do you find MacLaurin's Formula for #f(x)=sinhx# and use it to approximate #f(1/2)# within 0.01?

Answer 1

#sinh(1/2)~~0.52#

We know the definition for #sinh(x)#: #sinh(x)=(e^x-e^-x)/2#
Since we know the Maclaurin series for #e^x#, we can use it to construct one for #sinh(x)#. #e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/2+x^3/(3!)...#
We can find the series for #e^-x# by replacing #x# with #-x#: #e^-x=sum_(n=0)^oo(-x)^n/(n!)=sum_(n=0)^oo(-1)^n/(n!)x^n=1-x+x^2/2-x^3/(3!)...#
We can subtract these two from each other to find the numerator of the #sinh# definition:
#color(white)(-e^-x.)e^x=color(white)(....)1+x+x^2/2+x^3/(3!)+x^4/(4!)+x^5/(5!)...# #color(white)(e^x)-e^-x=-1+x-x^2/2+x^3/(3!)-x^4/(4!)+x^5/(5!)...# #e^x-e^-x=color(white)(lllllllll)2xcolor(white)(lllllllll)+(2x^3)/(3!)color(white)(lllllll)+(2x^5)/(5!)...#
We can see that all the even terms cancel and all the odd terms double. We can represent this pattern like so: #e^x-e^-x=sum_(n=0)^oo 2/((2n+1)!)x^(2n+1)#
To complete the #sinh(x)# series, we just need to divide this by #2#: #(e^x-e^-x)/2=sinh(x)=sum_(n=0)^oo cancel2/(cancel2(2n+1)!)x^(2n+1)=#
#=sum_(n=0)^oo x^(2n+1)/((2n+1)!)=x+x^3/(3!)+x^5/(5!)...#
Now we want to calculate #f(1 \/ 2)# with an accuracy of at least #0.01#. We know this general form of the Lagrange error bound for an nth degree taylor polynomial about #x=c#: #|R_n(x)|<=|M/((n+1)!)(x-c)^(n+1)|# where #M# is an upper bound on the nth derivative on the interval from #c# to #x#.
In our case, the expansion is a Maclaurin series, so #c=0# and #x=1 \/ 2#: #|R_n(x)|<=|M/((n+1)!)(1/2)^(n+1)|#
The higher order derivatives of #sinh(x)# will either be #sinh(x)# or #cosh(x)#. If we consider the definitions for them, we see that #cosh(x)# will always be larger than #sinh(x)#, so we should work out the #M#-bound for #cosh(x)#
The hyperbolic cosine function is always increasing, so the largest value on the interval will be at #1 \/ 2#: #sinh(1/2)=(e^(1/2)+e^(-1/2))/2=(sqrte+1/sqrte)/2=sqrte/2+1/(2sqrte)=M#
Now we plug this into the Lagrange error bound: #|R_n(x)|<=(sqrte/2+1/(2sqrte))/((n+1)!)(1/2)^(n+1)#
We want #|R_n(x)|# to be smaller than #0.01#, so we try some #n# values until we get to that point (the lesser amount of terms in the polynomial, the better). We find that #n=3# is the first value that will give us an error bound smaller than #0.01#, so we need to use a 3rd degree taylor polynomial. #sinh(1/2)~~sum_(n=0)^3(1/2)^(2n+1)/((2n+1)!)=336169/645120~~0.52#
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Answer 2

To find MacLaurin's formula for ( f(x) = \sinh x ), we first need to compute the derivatives of ( f(x) ) at ( x = 0 ).

  1. The first derivative of ( f(x) = \sinh x ) is ( f'(x) = \cosh x ), evaluated at ( x = 0 ), ( f'(0) = \cosh 0 = 1 ).
  2. The second derivative of ( f(x) = \sinh x ) is ( f''(x) = \sinh x ), evaluated at ( x = 0 ), ( f''(0) = \sinh 0 = 0 ).
  3. The third derivative of ( f(x) = \sinh x ) is ( f'''(x) = \cosh x ), evaluated at ( x = 0 ), ( f'''(0) = \cosh 0 = 1 ).
  4. The fourth derivative of ( f(x) = \sinh x ) is ( f''''(x) = \sinh x ), evaluated at ( x = 0 ), ( f''''(0) = \sinh 0 = 0 ).

Since the even derivatives are ( 0 ) and the odd derivatives are ( 1 ) at ( x = 0 ), the Maclaurin series for ( f(x) = \sinh x ) is:

[ \sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \ldots ]

To approximate ( f\left(\frac{1}{2}\right) ) within ( 0.01 ), we use the first few terms of the Maclaurin series:

[ f\left(\frac{1}{2}\right) \approx \frac{1}{2} + \frac{\left(\frac{1}{2}\right)^3}{3!} = \frac{1}{2} + \frac{1}{48} ]

[ = \frac{24}{48} + \frac{1}{48} = \frac{25}{48} ]

Therefore, ( f\left(\frac{1}{2}\right) ) is approximately ( \frac{25}{48} ), which is within ( 0.01 ) of the actual value.

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Answer 3

MacLaurin's formula for ( f(x) = \sinh(x) ) is:

[ \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \frac{x^9}{9!} + \ldots ]

To approximate ( \sinh\left(\frac{1}{2}\right) ) within ( 0.01 ), we'll use the first few terms of the Maclaurin series:

[ \sinh\left(\frac{1}{2}\right) \approx \frac{1}{2} + \frac{\left(\frac{1}{2}\right)^3}{3!} + \frac{\left(\frac{1}{2}\right)^5}{5!} ]

[ \sinh\left(\frac{1}{2}\right) \approx \frac{1}{2} + \frac{1}{48} + \frac{1}{3840} ]

[ \sinh\left(\frac{1}{2}\right) \approx 0.5 + 0.0208333333 + 0.0002604167 ]

[ \sinh\left(\frac{1}{2}\right) \approx 0.52009375 ]

This approximation is within (0.01) of the actual value of ( \sinh\left(\frac{1}{2}\right)).

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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.

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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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