Finding a bound on the difference of two functions What is this asking? I am going through a textbook and don't understand the problem 
Find a bound on the difference of $\sin x$ and  $x-\frac{x^3}{6}+\frac{x^5}{120}$ for $x \in [0,1]$
 A: Hint.  The problem means, find a constant $M$ such that
$$\left|\sin x-\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)\right|\le M$$
for all $x\in[0,1]$.  You should recognise the second expression as the degree $5$ Taylor polynomial for $\sin x$ in powers of $x$.  We can write an equality with a remainder term,
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5(x)\ .$$
The difference is $R_5(x)$, and I'm sure your textbook will have theorems for finding bounds on this quantity - look up something like "Taylor series with remainder" in the index.
Read your text carefully, see what you can find, and try to finish the problem.
Comments


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*A "sneaky trick": your expression can also be viewed as the degree $6$ Taylor polynomial,
$$\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+0x^6+R_6(x)\ ,$$
and this may give you a better bound than the above.

*Some books use a slightly different notation: they might have $R_6$ where I have written $R_5$ and $R_7$ where I have written $R_6$.  Check your textbook carefully.
