# Taylor Series and Size of Functions

I am looking for help with this question

By looking at the Taylor series, decide which of the following functions $$\ln(1+y^2) \; \; \; \; \; \; \; \; \sin(y^2)\; \; \; \; \; \; \; \;1-\cos(y)$$ is largest and which is smallest for values of $y$ near $0$.

I've done the Taylor expansions about $0$. I've written them below with the first few terms of the expansions.

$$f(x)=\ln(1+x^2)$$ $$=x^2- \frac {x^4}{2}+\frac {x^6}{3}-\frac {x^8}{4}+\frac{x^{10}}{5}$$

$$--$$

$$f(x)=\sin(x^2)$$ $$=x^2- \frac {x^6}{6}+\frac {x^{10}}{120}-\frac {x^{14}}{5040}+\frac{x^{18}}{362880}$$

$$--$$

$$f(x)=1-\cos(x)$$ $$=\frac {x^2}{2}- \frac {x^4}{24}+\frac {x^6}{720}-\frac {x^8}{40320}+\frac{x^{10}}{3628800}$$

How do I find which function is largest/smallest for values of $x$ near $0$? Each of these functions tends toward the point of origin. I'm having troubles grasping the intuition of the question.

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One way would be to subtract one series from another and see whether the answer comes out positive or negative near $x=0$ - that may help you to understand how you could have arrived at the same answer just by looking. – Mark Bennet Mar 28 '12 at 18:20
Note that if $f$ is larger than $g$ near $0$, then also $f/x^2$ is larger than $g/x^2$ near $0$. – TMM Mar 28 '12 at 18:20
By the way, don't forget the "$\dots$"! These Taylor series are not finite. – TMM Mar 28 '12 at 18:28
You could always cheat and use a calculator to evaluate each of the functions at, say, $x=.001$ to see how they compare. – Gerry Myerson Mar 29 '12 at 6:18

They are all equall to zero for the exact argument 0, but if you are not exactly at the origin but only close, say at $\epsilon$, then $\epsilon^2>\epsilon^2/2$. Similarly if the first term is equal, then the next one decides which one is larger for small arguments.
When the powers are not identical, then (for the limit $x\rightarrow 0$) the lower power dominates the other one ($x^2 < n x^4$ for small enough $x$. the larger n is, the smaller x has to be for this to be true, but it is true nontheless when you are "near" 0).