How do you find the Taylor's formula for #f(x)=x^p# for x-1?
where
We have to distinguish two cases:
...
so the Taylor series has a finite number of terms:
In this case the derivatives of all orders are:
so the Taylor series is:
Note that if we use the definition of generalized binomial coefficient, we have:
then we can unify the notation as:
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To find Taylor's formula for the function ( f(x) = x^p ) centered at ( x = 1 ), we use Taylor's formula:
[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots ]
- Compute ( f(a) = f(1) = 1^p = 1 ).
- Compute ( f'(x) = px^{p-1} ) and evaluate at ( x = 1 ): ( f'(1) = p ).
- Compute ( f''(x) = p(p-1)x^{p-2} ) and evaluate at ( x = 1 ): ( f''(1) = p(p-1) ).
- Compute ( f'''(x) = p(p-1)(p-2)x^{p-3} ) and evaluate at ( x = 1 ): ( f'''(1) = p(p-1)(p-2) ), and so on.
The Taylor series for ( f(x) = x^p ) centered at ( x = 1 ) is:
[ f(x) = 1 + p(x - 1) + \frac{p(p-1)}{2!}(x - 1)^2 + \frac{p(p-1)(p-2)}{3!}(x - 1)^3 + \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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