Taylor Expansion Homework: Find First 3 Terms for (1+2x)^-1/4 & sin(x^2)/x

[Gwilim]

Homework Statement

Find the first three non-zero terms of the Taylor expansion about zero in each of the following cases:
a) (1+2x)^-1/4
b) sin(x^2)/x

2. The attempt at a solution
I don't even know where to begin. I've only just read about what Talyors expansion is, and I don't know what's meant by 'about zero'. If someone could show me some answers and the steps they took to arrive at them though, I'm sure I would learn a lot from it.

 

[HallsofIvy]

A power series is of the form
$$\sum_{n=0}^{\infty} a_n (x- a)^n$$
and the sum is "about a". In particular, a sum "about 0" is of the form
[tex]\sum_{n=0}^{\infty} a_n x^n[/itex]<br /> <br /> The <b>Taylor's series</b> of f(x), about a, is given by <br /> $$\sum_{n= 0}^{\infty}\frac{1}{n!} \frac{d^n f(a)}{dx^n} (x- a)^n$$<br /> so your problem is really just finding the first three non-zero derivatives at x= 0.<br /> <br /> The Taylor's series "about 0" is also called the "McLaurin series".[/tex]

 

[Gwilim]

Thanks! Mind checking my answers and 'tidying them up' to fit the question?
a) f(x) = (1+2x)^-1/4
by the chain rule
f'(x) = -1/2(1+2x)^-5/4
f''(x) = 5/4(1+2x)^-9/4
f(0) = 1
f'(0) = -1/2
f''(0) = 5/4
b)
f(x) = sin(x^2)/x
by the chain rule and the product rule
f'(x) = 2cos(x^2) - f(x)
f''(x) = -4xsin(x^2) - f'(x)
f(0) = ... wait I've been doing this wrong haven't I? f(0) doesn't exist. lol

 

[Gwilim]

While I'm waiting for a reply, more practice differentiating (I'm very rusty), these questions from a different years past paper:

Homework Statement

Find the first three non-zero terms of the Taylor expansion about zero in each of the following cases:
a) (1+x^2)^1/2
b) sin(2x)/x

2. The attempt at a solution

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

b)
f(x) = sin(2x)/x
f'(x) = 2cos(2x)/x - sin(2x)/x^2
f''(x) = -4sin(2x)/x - 4cos(2x)/x^2 - sin(2x)/x^3

 

[Defennder]

You're right for the first part. Just substitute the terms into the formulae given and you're done. Turns out that f(0) isn't defined for b), and neither does f'(0) and f''(0).

 

[Gwilim]

so what answer would I give for question b) in the exam? Would they still expect me to differentiate the expression twice for any marks though, even though a cursory glance shows that f(0) is not defined? Additionally, in a) of the second set of questions I posted, if I have differentiated correctly f'(0)=0 and f''(0)=0, so these are not non-zero terms in the Taylor expansion. Am I correct in assuming that further derivatives will also give 0 when x=0, and if so what answer would I give for this question?

 

[HallsofIvy]

Thanks! Mind checking my answers and 'tidying them up' to fit the question?

Please note that I have edited my first response (so I can pretend never to have made such a silly mistake!). I have added the "1/n!" term.

a) f(x) = (1+2x)^-1/4
by the chain rule
f'(x) = -1/2(1+2x)^-5/4
f''(x) = 5/4(1+2x)^-9/4
f(0) = 1
f'(0) = -1/2
f''(0) = 5/4

So the first 3 terms of the "Taylor's series about 0" are 1, (-1/2) x, and (5/4)(1/2!)x^2= (5/8)x^2. The "quadratic Taylor's polynomial about 0" is 1- (1/2)x+ (5/8)x^2.

b)
f(x) = sin(x^2)/x
by the chain rule and the product rule
f'(x) = 2cos(x^2) - f(x)

If you use the product rule then you need to write the function as f(x)= sin(x^2)x^(-1) and f'(x)= (2x)(cos(x^2)(x^-1)- x^(-2)sin(x^2)= 2cos(x^2)- sin(x^2)/x^2.
Of course, you could also use the quotient rule: f'(x)= [(2x)cos(x^2)(x)- (1)sin(x^2)]/x^2= 2cos(x^2)- sin(x^2)/x^2 again.

f''(x) = -4xsin(x^2) - f'(x)
f(0) = ... wait I've been doing this wrong haven't I? f(0) doesn't exist. lol

No, f(0) doesn't exist but its limit does. It has a "removable discontinuity" at x= 0 since sin(x^2)/x = (x)(sin(x^2)/x^2)-> (0)(1)= 0. The function defined by f(x)= sin(x^2)/x if x is not 0, 0 if x= 0. For that function, f(0)= 0, f'(0)= 1. You will need to recalculate f" since your f' is wrong.

 

[Gwilim]

Thanks again for the help, and for pointing out the error in differentiating b).

No, f(0) doesn't exist but its limit does. It has a "removable discontinuity" at x= 0 since sin(x^2)/x = (x)(sin(x^2)/x^2)-> (0)(1)= 0. The function defined by f(x)= sin(x^2)/x if x is not 0, 0 if x= 0. For that function, f(0)= 0, f'(0)= 1. You will need to recalculate f" since your f' is wrong.

Let me see if I understand this. f(x) is very small for very small values of x, or as x tends to 0, f(x) tends to 0. So we take f(0)= 0 for the purposes of providing an answer.

Making corrections:
f'(x) = 2cos(x^2) - sin(x^2)/x^2
f''(x) = -4sin(x^2) + 2sin(x^2)/x^3 - 2cos(x^2)/x
f'(0) = 1 (assuming sin(x^2)/x^2 can be said to be equal to 1 when x = 0)
f''(0) = oh no... the sin term thankfully goes to 0, but the other two terms tend to infinity? and are of opposite signs? I'm just going to guess this is equal to 0.

since f(0) = 0 and f''(0) probably equals 0, I have to differentiate again to get the second term? And assuming the third derivative also tends to zero I'll have to differentiate again to get the third term. That seems like a lot of work for the amount of marks the question is worth.