How would you use the Maclaurin series for #e^-x# to calculate #e^0.1#?

Answer 1

# e^(0.1) ~~ 1.10517 (5dp) #

We are aware of that

#e^(-x) = 1 - x + x^2/(2!)-x^3/(3!)+cdots#
is an alternating series and for #absx < 1# this series is absolutely convergent, so the error handled by cutting off after the #n# term is smaller than the higher order term left.
If we need to calculate #e^(-0.1)# within a precision #delta# then we solve first
#delta le x^n/(n!)# or #delta le (0.1)^n/(n!)# for instance, if #delta = 0.00001# then #n=4# would suffice. After that, #e^(0.1) = 1/e^(-0.1)#
# :. e^(0.1) = e^(-(-0.1)) # # :. e^(0.1) = 1-(-0.1)+(-0.1)^2/(2!)-(-0.1)^3/(3!)+(-0.1)^4/(4!) +#(higher terms) # :. e^(0.1) ~~ 1+0.1+0.01/2+0.001/6+0.0001/24 # # :. e^(0.1) ~~ 1.1+0.005+0.00016667+0.000041667 # # :. e^(0.1) ~~ 1.10517083 ... # # :. e^(0.1) ~~ 1.10517 (5dp) #
Compare with the calculator answer #e^0.1=1.10517091 ...#
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Answer 2

To calculate ( e^{0.1} ) using the Maclaurin series for ( e^{-x} ), we can utilize the series expansion of ( e^{-x} ), which is given by:

[ e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n ]

To find ( e^{0.1} ), substitute ( x = -0.1 ) into the series expansion of ( e^{-x} ) and compute the sum of the series up to a certain number of terms, ensuring an acceptable level of accuracy based on the desired precision.

[ e^{-0.1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} (-0.1)^n ]

Calculate each term of the series for ( n = 0, 1, 2, \ldots ), and sum them to find an approximation for ( e^{-0.1} ). Then, take the reciprocal of this value to find ( e^{0.1} ).

Since ( e^x ) is the reciprocal of ( e^{-x} ), we'll compute ( e^{0.1} ) as:

[ e^{0.1} = \frac{1}{e^{-0.1}} ]

This will give us the desired result.

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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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