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