# Taylor expansionWatch

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#1
Can anyone help me with calculating this fourth order Taylor expansion about x=0? I have been working on it for about half an hour and have consulted books but seem to be getting ridiculously large coefficients.

Q: Find a Taylor series expansion for order 4 about x = 0 for:

f(x) = (1 + x)^(37/2)

Thanks,

ATK
0
13 years ago
#2
(Original post by atkelly)
Q: Find a Taylor series expansion for order 4 about x = 0 for:

f(x) = (1 + x)^(37/2)
f(x) = (1 + x)^(37/2)
f'(x) = (37/2)(1 + x)^(37/2-1)
f''(x) = (37/2)(37/2-1)(1 + x)^(37/2-2)
f''(x) = (37/2)(37/2-1)(37/2-2)(1 + x)^(37/2-3)
f'''(x) = (37/2)(37/2-1)(37/2-2)(37/2-3)(1 + x)^(37/2-4)

f(x) ~ 1 + 37x/2 + 1295x2/8 + 14245x3/16 + 1324785x4/16

Not sure if that's right.
0
13 years ago
#3
(Original post by Gnarl)
f(x) = (1 + x)^(37/2)
f'(x) = (37/2)(1 + x)^(37/2-1)
f''(x) = (37/2)(37/2-1)(1 + x)^(37/2-2)
f''(x) = (37/2)(37/2-1)(37/2-2)(1 + x)^(37/2-3)
f'''(x) = (37/2)(37/2-1)(37/2-2)(37/2-3)(1 + x)^(37/2-4)

f(x) ~ 1 + 37x/2 + 1295x2/8 + 14245x3/16 + 1324785x4/16

Not sure if that's right.
Nearly right. You forgot to divide f''''(0) by 4!

1 + (37/2)x + (1295/8)x^2 + (14245/16)x^3 + (441595/128)x^4
0
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