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How do you find the taylor series of #1 1x^2#?

Answer 1

Emma Aldridge

The Taylor series basically hands us back #11x^2# ...

The Taylor series for #f(x)# about #0# is:

#f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+...#

For #f(x) = 11x^2#, we have

#f'(x) = 22x, f''(x) = 22, f'''(x) = 0,...#

#f(0) = 0#, #f'(0) = 0#, #f''(0) = 22#, #f'''(0) = 0#, ...

and the Taylor series is:

#f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+...#

#=0+0/(1!)x+22/(2!)x^2+0/(3!)x^3+...#

#=11x^2#

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Answer 2

Emilia Aguilar

To find the Taylor series of ( \frac{1}{1-x^2} ), we first note that this function can be expressed as a geometric series:

[ \frac{1}{1 - x^2} = 1 + x^2 + (x^2)^2 + (x^2)^3 + \ldots ]

Now, we can use the formula for the sum of an infinite geometric series:

[ S = \frac{a}{1 - r} ]

Where ( a ) is the first term and ( r ) is the common ratio. In this case, ( a = 1 ) and ( r = x^2 ).

So, the Taylor series expansion of ( \frac{1}{1-x^2} ) is:

[ 1 + x^2 + x^4 + x^6 + \ldots ]

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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.