Why Does the Maclaurin Series for 1/(1+2x^2) Have Only Four Terms Up to x^6?

[~Sam~]

Homework Statement

Find the Maclaurin polynomial for P(6)X for 1/(1+2x²) about x=0

Homework Equations

I'm pretty sure I need to use the Maclaurin polynomial for 1/(1-x)= 1+x+x²+x³+...+xⁿ +O(xⁿ⁺¹)

The Attempt at a Solution

I simply rewrote as 1/(1-(-2x²)) then proceeded to sub and expanded. The answer was not what I expected and not what I got. Apparently it's suppose to be 1-2x²+4x⁴-8x⁶. Can anyone expain to me why is there only four terms?

 

[Hootenanny]

Apparently it's suppose to be 1-2x²+4x⁴-8x⁶. Can anyone expain to me why is there only four terms?

Why would you expect it to be more than four terms? Don't forget that you are expanding in powers of x².

Perhaps it would be more obvious if we first take the expansion,

$$\frac{1}{1-\left(-\theta\right)} \sim 1 - \theta + \theta^2 - \theta^3 + \theta^4$$

Now simply make the substitution $\theta = 2x^2$,

$$\frac{1}{1-\left(-\theta\right)} \sim 1 - 2x^2 + 4x^4 - 8x^6 + 16x^8$$

Note that the fact that you are expanding in powers of x², means that the powers of each term in the Maclaurin series doubles.

 

[~Sam~]

Why would you expect it to be more than four terms? Don't forget that you are expanding in powers of x².

Perhaps it would be more obvious if we first take the expansion,

$$\frac{1}{1-\left(-\theta\right)} \sim 1 - \theta + \theta^2 - \theta^3 + \theta^4$$

Now simply make the substitution $\theta = 2x^2$,

$$\frac{1}{1-\left(-\theta\right)} \sim 1 - 2x^2 + 4x^4 - 8x^6 + 16x^8$$

Note that the fact that you are expanding in powers of x², means that the powers of each term in the Maclaurin series doubles.

I understand that..but it ask for of order 6..P6(x) yet the answer given is 1-2x²+4x²-8x⁶...and not 1-2x²+4x²-8x⁶+16x⁸-32x¹⁰. Is the answer given in the book just wrong?

 

[Hootenanny]

I understand that..but it ask for of order 6..P6(x)

What does order six mean?

 

[~Sam~]

What does order six mean?

The notation P_n means the n-th order Taylor polynomial for f about a. We have the general taylor formula f(a)+ f'(a)(x-a) + f''(a)/2! ((x-a)¹) +...+fⁿ (a)/n! ((x-a)ⁿ) Which basically means P₆ would be taking the formula to the 6th derivative, hence order 6.

 

[Hootenanny]

The notation P_n means the n-th order Taylor polynomial for f about a. We have the general taylor formula f(a)+ f'(a)(x-a) + f''(a)/2! ((x-a)¹) +...+fⁿ (a)/n! ((x-a)ⁿ) Which basically means P₆ would be taking the formula to the 6th derivative, hence order 6.

No. P₆ means a ploynomial of order six, that is a polynomial where the greatest power is six.