1. ## Maclaurin Series

Hi, can someone help me with this question?

2. Hello tottijohn
Originally Posted by tottijohn
Hi, can someone help me with this question?
I presume that you know that the expansion of $\frac{1}{1-x}$ is
$1+x+x^2 +x^3 + ...$
So:
$\frac{x^2}{1-2x} = x^2\Big(1+(2x) + (2x)^2 + (3x)^3+ ...\Big)$
Can you take it from here?

Based on your guidance and reading up more on Taylor's Theorem (I initially thought this qn is on Maclaurin Integral Test), here is my final take on the question:

x^2[1+2x+(2x)^2+(2x)^3+(2x)^4+(2x)^5+...]
=x^2+2x^3+4x^4+8x^5+16x^6+32x^7+...

Thus, the coefficient of x^5 is 8. Am i correct?

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Also, if the function given is x^2/(1-3x^3), is the expansion:
x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...] ???

Lastly, I cant seem to get the maths code for the equation correct, so pardon me for type the maths equation as above. Keep getting a syntax error.

4. Hello tottijohn
Originally Posted by tottijohn

Based on your guidance and reading up more on Taylor's Theorem (I initially thought this qn is on Maclaurin Integral Test), here is my final take on the question:

x^2[1+2x+(2x)^2+(2x)^3+(2x)^4+(2x)^5+...]
=x^2+2x^3+4x^4+8x^5+16x^6+32x^7+...

Thus, the coefficient of x^5 is 8. Am i correct?
Yes, you are. (Sorry, I just noticed a $3$ crept into my answer. Of course, I meant $(2x)^3$ not $(3x)^3$.
Also, if the function given is x^2/(1-3x^3), is the expansion:
x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...] ???
Again, correct!
Lastly, I cant seem to get the maths code for the equation correct, so pardon me for type the maths equation as above. Keep getting a syntax error.
The code for the expansion is OK:
$$x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...]$$
displays as
$x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...]$
To write the fraction in LaTeX, write:
$$\frac{x^2}{1-3x^3}$$
which displays as:
$\frac{x^2}{1-3x^3}$