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#1 2007-01-31 00:16:52

enita
Guest

series

Hi everyone!

I'm abit stuck on the question below. Can anyone help?

1. Evaluate √(1.05) correct to 5 significant firgures.

hint: use a series for (1+x)^(1/2)

Thank you

Kind Regards,

Enita

#2 2007-01-31 02:28:54

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

Re: series

We have the binomial expansion:

Now if we set x = 0.05 the we will have an expansion for

Substitute this value into the series to give:

No point in going further as we are beyond five significant figures accuracy so add these to give:

to 5 s.f.

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#3 2007-01-31 02:34:51

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: series

Glad that we only need to evaluate Taylor Polynomials to the 2nd derivative.
Taylor Polynomials to the 2nd derivative:
f(x[sub]0[/sub]+h)=f(x[sub]0[/sub])+f'(x[sub]0[/sub])h+(1/2)f"(x[sub]0[/sub])h²

In this case, f(x)=x)^(1/2)
x[sub]0[/sub]=1, h=0.05, so you can work out f'(1) and f"(1) alone.

Indeed, there is a special formula of (1+x)^(1/2), but I forgot. You may look it up your calculus book.

Last edited by George,Y (2007-01-31 02:35:37)


X'(y-Xβ)=0

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#4 2007-01-31 04:37:59

enita
Guest

Re: series

Thank you for you quick response!

how did you know to set x to equal 0.05 ?

Kind regards

Enita

#5 2007-01-31 04:56:04

gnitsuk
Member
Registered: 2006-02-09
Posts: 121

Re: series

We need to find:

And I have a series which gives me the value of:

where x can be any number whose modulus is less than 1.

Well, I notice that

which looks identical to my formula if only the 'x' in that formula were replaced by 0.05. And it's okay to use the formula because the modulus of 0.05 < 1.

Mitch.

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#6 2007-01-31 05:18:48

enita
Guest

Re: series

Thank you!

Nicely explained.

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