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#### losheungwai

##### Newbie level 5
expand the function about the point a through third order terms:
f(x) = e^(-2x)
a = 5

"third order terms" does it mean differatiate f(x) three times?

#### CrabMan

##### Member level 1
From what i think you are asking then yes third order refers to differential of function w.r.t x three times.

#### Element_115

Do you mean:
Use the Taylor Expansion equation for e^x (to the third term) and solve for x=5 ?

If so just look in a math book or google it.

But if it's "a=5" and not "x=5" I might be off the mark on this one.

Cheers

#### zorro

Hi,

my interpretation of the following statement:

expand the function about the point "a" through third order terms:
f(x) = e^(-2x)
a = 5

is to obtain an expression like this:

f(x) ≈ ∑(from n=0 to 3) Cn (x-5)^n

i.e. a truncated Taylor expansion whose Cn coeffiicients have to be found.
Regards

Z

#### saruman1983

##### Member level 2
Taylor expansion:

f(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2+f'''(x-a)^3+...

From what i understand, you should neglect all (x-a)^i terms with i>3. So, in this case:

f(x)=e^(-2x)
f'(x)=-1/2*e^(-2x)
f''(x)=1/4*e^(-2x)
f'''(x)=-1/8*e^(-2x)

and f(x)≈e^(-10)-1/2*e^(-10)(x-5)+1/4*e^(-10)(x-5)^2-1/8*e^(-10)(x-5)^3

### losheungwai

Points: 2

#### zorro

Warning: In saruman's post, factorials are missing and derivatives are wrong.

### losheungwai

Points: 2

#### saruman1983

##### Member level 2
Indeed. I'm sorry, just answered it hastefully

Taylor expansion:

f(x)=f(a)/0!+f'(a)/1!*(x-a)+f''(a)/2!*(x-a)^2+f'''(a)/3!*(x-a)^3+...

...and so on

I apologize for any errors

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