# Find the error when approximating a function

Let $$f:(-\infty, 1) \to \mathbb{R}$$ be a function, $$f(x)=e^x+ln(1-x)$$.

Find $$n \in \mathbb{N}$$ such that the error when approximating $$e^{1.1}+\ln(1.1)$$ by their Taylor polynomial $$T_{n,f,0}(x)$$, with $$n < 0.0001$$.

I'm really confused and I have no idea where to start.

I know that $$f(x)=T_{n,f,0}(x)+R_{n}(x)$$ where $$R_{n}(x)$$ is the reminder of the Taylor polynomial.

$$e^{1.1}+\ln(1.1)=(1-\frac{x^3}{6}-\frac{5 x^4}{24}-\frac{23 x^5}{120}-...)-n$$ Where $$n$$ is the given error?

Sorry if this is so confusing but I'm really lost!

• Should "$t \in \mathbb{N}$" be "$n \in \mathbb{N}$"? – angryavian Nov 23 '18 at 21:29
• Yes! I got confused with another problem. I will fix it now. – Moria Nov 23 '18 at 21:30
• Don't just blindly replace all $t$s with $n$s; now $n < 0.0001$ does not make any sense. Presumably you want to find $n$ such that the error of the Taylor polynomial $T_{n,f,0}$ is $< 0.0001$. – angryavian Nov 23 '18 at 21:32
• It must be a typo, it was written like that in an old exam. But that's what I have to do and I literally have no idea where to start. – Moria Nov 23 '18 at 21:37

## 1 Answer

Your $$f(x)$$ does not match the function you are asked to form the Taylor series of because it has $$\ln (1-x)$$ and the $$e^x$$ term is $$0.1$$, not $$1.1$$ when $$x=0.1$$. Your Taylor series is not correct-it should have an $$x$$ on the left, and the Taylor series of $$e^{1+x}+\ln(1+x)$$ is not what you have shown-at $$x=0$$ it should be $$e$$, not $$1$$.

Once you get the right Taylor series, $$n$$ is the number of terms you need to keep to get the error at $$x=0.1$$ to be less than $$0.0001$$. You can look up the remainder term of the Taylor series, which will be $$0.1^{n+1}$$ times the $$n+1^{st}$$ derivative of the function. You can bound the $$n+1^{st}$$ derivative in the interval $$[0,0.1]$$ and evaluate the upper bound for the error this gives you.

• Thank you!! I'll have to ask my teacher about it. It was written like that in an old exam. Now it's clearer how to solve this problem. – Moria Nov 23 '18 at 22:04