Maclaurin Series

[ed84c]

Question: Find a Maclaurin approximation to calculate

 

[math]

 

\frac{1}{\sqrt{e}}

 

[/math]

 

so I go for the M.A of

[math]

{e}^{-0.5x}

[/math]

 

which I have worked out to be

 

[math]

 

1-0.5x+\frac{x^2}{2!}-\frac{0.5x^3}{3!}+\frac{x^4}{4!}-\frac{0.5x^5}{5!}

 

[/math]

 

But If I now dow f(1) it does not approximate

 

[math]

{e}^{-0.5x}

[/math]

 

(although it does around f(0))

 

Any help?

[Klaynos]

I've not worked it through, but how far from 0 does it give an approximate value? 1 might just be too far away

 

Then you'd either have to use more terms, or a Taylor series around 1...

 

also have you droped an x on the original term[math]\frac{1}{\sqrt{e}}[/math]?

[ed84c]

at 0.3 its 0.2 away.

 

And no I didnt drop the x; that was the question. Maybe Im doing it the wrong way?

 

This is only the first exersize in the chapter which does not go into taylor series'.

[Klaynos]

are you sure e is exponential and it's not asking for an expansion of f(e)=f(x)=[math]\frac{1}{\sqrt{x}}[/math]

[ed84c]

OK i did it like that but it came out around 10?

[Country Boy]

Question: Find a Maclaurin approximation to calculate

 

[math]

 

\frac{1}{\sqrt{e}}

 

[/math]

 

so I go for the M.A of

[math]

{e}^{-0.5x}

[/math]

 

which I have worked out to be

 

[math]

 

1-0.5x+\frac{x^2}{2!}-\frac{0.5x^3}{3!}+\frac{x^4}{4!}-\frac{0.5x^5}{5!}

 

[/math]

HOW did you get that? You seem to just have 0.5 times the MacLaurin coefficient of [math]e^x[/math] when n is odd, 1 times if n is even. If f(x)= [math]e^{-x/2}[/math] then f(0)= 1 as you have. f'x)= [math]-1/2 e^{-x/2}[/math] so f'(0)= -1/2 also as you have. f"(x)= [math]1/4 e^{-x/2}[/math] so f"(0)= 1/4 and the MacLaurin coefficient is (1/4)/2= 1/8, not at all what you have! In general the nth derivative of [math]e^{-x/2}[/math] is [math](-1)^nx^n/2^n[/math] and so the nth coefficient of the MacLaurin series is [math](-1)^n/(2^n n!)[/math].

 

You could also do that just by replacing x in the MacLaurin series for [math]e^x[/math] by -x/2: [math]x^n[/math] becomes [math](-1)^n x^n/2^n[/math].

 

In particular, taking x= 1, you should get [math]1/\sqrt{e}[/math] is approximately 1- 1/2+ 1/8- 1/48+ 1/384+ 1/3840= 0.6044, a pretty good approximation to [math]1/\sqrt{e}[/math] which is about 0.6053

 

But If I now dow f(1) it does not approximate

 

[math]

{e}^{-0.5x}

[/math]

 

(although it does around f(0))

 

Any help?