emcrazy14
Urgent help needed:
Expand the following up to the term in x^3:



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emcrazy14
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\[\frac{ 1+x }{ \sqrt{1x} }\]

Jibrilpresbury
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idk

anonymous
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expand in a power series i take it

anonymous
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about zero i hope
first term is \(f(0)=1\)

anonymous
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first derivative is
\[f'(x)=\frac{3x}{2\sqrt{1x}^3}\] so second term is
\[f'(0)x=\frac{3}{2}x\]

anonymous
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yikes, now we need the derivative of that beast

anonymous
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do i have to do that too?

emcrazy14
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Yes, please?

anonymous
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grrr

anonymous
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lets cheat

anonymous
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looks like it is
\[f''(x)=\frac{7x}{4\sqrt{1x}^5}\]

anonymous
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making the next term
\[\frac{f''(0)}{2}x^2=\frac{7}{8}x^2\]

anonymous
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one more?

emcrazy14
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So I find the derivative of that too?

anonymous
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\[f'^{(3)}(x)=\frac{333x}{8\sqrt{1x}^7}\]

anonymous
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i think

anonymous
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if so then \(\frac{f^{(3)}(0)}{3!}x^3=\frac{33}{16}x^3\)

anonymous
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hmm let me check if this is right

anonymous
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no messed up \(3!=6\) doh

anonymous
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third term is \(\frac{11}{16}x^3\)

anonymous
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k lets check the whole dam thing and see if it is right


emcrazy14
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Okay, thank you so much! :)

anonymous
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yw