# Taylor series expansion of tangent

1. Sep 5, 2008

### kreil

1. The problem statement, all variables and given/known data
find the first four nonzero terms in the power series expansion of tan(x) about a=0

2. Relevant equations
$$\Sigma_{n=0}^{\infty} \frac{f^n (a)}{n!}(x-a)^n$$

3. The attempt at a solution

Well the series has a zero term at each even n (0,2,4 etc)

for n=1 I got x, for n=3 I got $\frac{x^3}{3}$

i ran into a problem while trying to compute n=5, since I got $\frac{x^5}{10}$ rather than $\frac{2x^5}{15}$ which is the answer. I assume that I must have differentiated wrong, so if someone could check the below derivatives I would appreciate it!

$$tan(x)$$(n=0)

$$sec^2(x)$$(n=1)

$$2tan(x)sec^2(x)$$(n=2)

$$2tan^2(x)sec^2(x)+2sec^4(x)$$(n=3)

$$4tan^3(x)sec^2(x)+4tan(x)sec^4(x)+8sec^4(x)tan(x)$$(n=4)

$$8tan^4(x)sec^2(x)+12tan^2(x)sec^4(x)+16tan^2(x)sec^4(x)+4sec^6(x)+32tan^2xsec^4(x)+8sec^6(x)$$(n=5)

As you can see in the n=5 expression plugging in x=0 reduces it to equaling 12; if it were instead equal to 16 I would have the correct expression.

Also, this method is very tedious so if anyone knows of an easier way please let me know!

2. Sep 5, 2008

### HallsofIvy

Staff Emeritus
This is incorrect. Differentiating the "sec^2(x)" part of the first term of the previous derivative gives 2sec(x)(sec(x) tan(x)) so the multiplying by 2tan(x) gives 4tan^2(x)sec^2(x). This should be 4 tan^2(x)sec^2(x)+ 2sec^4(x).

3. Sep 5, 2008

### alphachapmtl

4tan^2 sec^2 + 2sec^4 (n=3)

4. Sep 6, 2008

Thanks!

5. Sep 6, 2008

### alphachapmtl

tan (n=0)
sec^2 (n=1)
2 tan sec^2 (n=2)
4 tan^2 sec^2 + 2 sec^4 (n=3)
16 tan sec^4 + 8 tan^3 sec^2 (n=4)
16 sec^6 + 88 sec^4 tan^2 + 16 sec^2 tan^4 (n=5)
272 sec^6 tan + 416 sec^4 tan^3 + 32 sec^2 tan^5 (n=6)
272 sec^8 + 2880 sec^6 tan^2 + 1824 sec^4 tan^4 + 64 sec^2 tan^6 (n=7)
7936 sec^8 tan + 24576 sec^6 tan^3 + 7680 sec^4 tan^5 + 128 sec^2 tan^7 (n=8)

See attachment,
Ref: Mathematica by Example, (By Martha L. Abell, James P.Braselton), page 188.

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