e^3x/2 expanding to 4 terms - point me in the right direction?

Question

I think it has to do with binomial theorem - need pointing in the right direction. Cheers

Answer 1

I will assume that you mean e^(3x/2)

Recall the formula for Taylor's expansion about x = 0:
f(x) = f(0) + [f'(0)/1!]x + [f''(0)/2!]x^2 + [f'''(0)/3!]x^3 + ...
The n-th term looks like: [(nth derivative of f evaluated at 0)/n!]x^n

So if we let f(x) = e^(3x/2).
Then:
f'(x) = (3/2) e^(3x/2)
f''(x) = (9/4) e^(3x/2)
f'''(x) = (27/8) e^(3x/2)

Plugging in 0 for x, we have:
f(0) = 1
f'(0) = 3/2
f''(0) = 9/4
f'''(0) = 27/8

So the first four terms for the expansion is:
f(x) = 1 + (3/2) x + [(9/4)/2] x^2 + [(27/8)/3!] x^3
= 1 + (3/2) x + (9/8) x^2 + (27/48) x^3

I hope that is what you are looking for!

Answer 2

e^3x/2= e^3/2 * e^x
e^x = 1 + x + x^2/2! +x^3/3!+ ..+x^n/n!+ .
So e^3x/2 =e^(3/2)[1 + x + x^2/2! +x^3/3!+ ..+x^n/n!+ ]

Answer 3

e^x = 1 + x/1! + x^2/2! + x^3/3!
All you need to do is to replace 'x' in the general series with '3x/2' and simplify

Answer 4

It has nothing to do with the binomial theorem but it is to do with McClarins expansion.

f(x) = f(0)+xf'(0)+(x/2!).f"(0)+..............

Answer 5

exp[x]=1+x+x^2/2+x^3/6...wherever you see x, substitute 3x/2

exp[3x/2]=1+(3x/2)+(3x/2)^2/2+...

Answer 6

e^(3x/2)

let t = 3x/2

expand e^t (4 series terms) and replace t by 3x/2.