How do you find the Maclaurin Series for # (sin(x/11))^2#?
Apply the identity of trigonometry:
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To find the Maclaurin series for ( (\sin(x/11))^2 ), we can use the Maclaurin series for ( \sin(x) ).
The Maclaurin series for ( \sin(x) ) is:
[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ]
To find the Maclaurin series for ( (\sin(x/11))^2 ), substitute ( x/11 ) into the series for ( \sin(x) ), then square the result.
[ (\sin(x/11))^2 = \left( \frac{x}{11} - \frac{(x/11)^3}{3!} + \frac{(x/11)^5}{5!} - \frac{(x/11)^7}{7!} + \cdots \right)^2 ]
[ = \frac{x^2}{11^2} - 2\frac{x^4}{11^4} + \frac{5x^6}{11^6} - \frac{14x^8}{11^8} + \cdots ]
Thus, the Maclaurin series for ( (\sin(x/11))^2 ) is:
[ \frac{x^2}{121} - \frac{2x^4}{1331} + \frac{5x^6}{161051} - \frac{14x^8}{1771561} + \cdots ]
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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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