# Bounds of the remainder of a Taylor series

Homework Statement:
Given that ##f(x)=\frac{1}{1+x}##. Find bound of remainder ##R_3 (x)## assuming ##|x-1| \leq2##
Relevant Equations:
Taylor series

Taylor Inequality
I have found the Taylor series up to 4th derivative:
$$f(x)=\frac{1}{2}-\frac{1}{4}(x-1)+\frac{1}{8}(x-1)^2-\frac{1}{16}(x-1)^3+\frac{1}{32}(x-1)^4$$

Using Taylor Inequality: ##a=1, d=2## and ##f^{4} (x)=\frac{24}{(1+x)^5}##

I need to find M that satisfies ##|f^4 (x)| \leq M##

From ##|x-1| \leq 2 \rightarrow 0 \leq x+1 \leq 4## , then ##0 \leq (x+1)^5 \leq 4^5##

Putting the boundary of ##(1+x)^5## to ##f^{4} (x)##, there is no upper boundary of ##f^{4} (x)## which does not make sense. And actually ##x+1## can not be zero because ##f^{4} (x)## would be undefined so my approach is certainly wrong.

How to find M? Thanks

Last edited:

FactChecker
Gold Member
For any given ##x : |x+1|\le 2## you can get a bound that is finite.
CORRECTION: Should be ##x : |x-1|\le 2##

Last edited:
For any given ##x : |x+1|\le 2## you can get a bound that is finite.
I am sorry, but how to get ##|x+1|\le 2##?

Thanks

vela
Staff Emeritus
Homework Helper
Putting the boundary of ##(1+x)^5## to ##f^{4} (x)##, there is no upper boundary of ##f^{4} (x)## which does not make sense. And actually ##x+1## can not be zero because ##f^{4} (x)## would be undefined so my approach is certainly wrong.
It actually does make sense that you can't find a bound since the function ##\frac 1{1+x}## blows up as ##x \to -1##.

• songoku
It actually does make sense that you can't find a bound since the function ##\frac 1{1+x}## blows up as ##x \to -1##.
So the answer the question is: no bound for ##R_3 (x)##?

Thanks

FactChecker
Gold Member
I am sorry, but how to get ##|x+1|\le 2##?

Thanks
Sorry. I should have said ##|x-1|\le 2##, which is given in the problem statement. I will edit my post.

vela
Staff Emeritus
Homework Helper
So the answer the question is: no bound for ##R_3 (x)##?
There's no bound on the interval ##-1 \le x \le 3##, but is that the interval relevant for calculating the remainder?

• songoku and FactChecker
There's no bound on the interval ##-1 \le x \le 3##, but is that the interval relevant for calculating the remainder?
Sorry, that's the only interval I can get. I also have checked the interval of convergence, which is ##-1 < x < 3## and still get no upper bound for ##R_3 (x)##. I can't find any other interval related to the question

FactChecker
Gold Member
Sorry, that's the only interval I can get. I also have checked the interval of convergence, which is ##-1 < x < 3## and still get no upper bound for ##R_3 (x)##. I can't find any other interval related to the question
For any particular value ## x'## in (-1,3) we do not care about the entire interval [-1,3]. We only care about ##|x-1|\le |x'-1|##. The function derivatives are bounded in that.

• songoku
For any particular value ## x'## in (-1,3) we do not care about the entire interval [-1,3]. We only care about ##|x-1|\le |x'-1|##. The function derivatives are bounded in that.
I draw graph of ##\frac{24}{(1+x)^5}## It is monotonically decreasing in interval ##-1<x<3## and the range will be ##(\frac{3}{128}, \infty )##

I also draw the graph of ##|x-1|## But sorry I don't understand how to find the interval where ##|x-1| \le |x'-1|##

FactChecker
Gold Member
But sorry I don't understand how to find the interval where ##|x-1| \le |x'-1|##
When trying to bound the error of the partial series at a point ##x'##, it is only necessary to look at the interval of points ##x## that are at least as close to the 1 as ##x'## is. That is for x in the interval [1-|x'-1|, 1+|x'-1|].

• songoku
When trying to bound the error of the partial series at a point ##x'##, it is only necessary to look at the interval of points ##x## that are at least as close to the 1 as ##x'## is. That is for x in the interval [1-|x'-1|, 1+|x'-1|].
So I need to find ##M## where ##|f^{3} (x')| \le M## and ##x' \in [a,x]##

##a \le x' \le x \rightarrow 1 \le x' \le 3## so the value of M is ##\frac{3}{4}##

Is this correct? Thanks

FactChecker
Gold Member
So I need to find ##M## where ##|f^{3} (x')| \le M## and ##x' \in [a,x]##

##a \le x' \le x \rightarrow 1 \le x' \le 3## so the value of M is ##\frac{3}{4}##

Is this correct? Thanks
You need to check on both sides of a=1, not just the side that x' is on. I think that we are using different notations, but the idea seems right.

I think that we are using different notations
I also think so. Honestly I am still not really understand the notation ##|x-1| \le |x'-1|##. I only use Taylor inequality I posted in #1 but after you introduced ##x'##, I searched through the web and found this link:

https://brilliant.org/wiki/taylor-series-error-bounds/

This is part of what is in there: and also At first, I thought ##x## in ##f^{n+1} (x)## is exactly the same as ##x## in ##|x-a| \le d## but in that link, it is written as ##f^{n+1} (z)## instead where ##a<z<x## so I thought ##a## will always be the lower boundary of ##z##, that's why I only checked the value bigger than 1.

You need to check on both sides of a=1, not just the side that x' is on.
By both sides of ##a=1##, do you mean for 2 intervals, something like ##p<x'<1## and ##1<x'<3## where ##p## is a value that I need to find first?

Thanks

vela
Staff Emeritus
Homework Helper
You need to check on both sides of a=1, not just the side that x' is on.
$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ where ##c## is between ##a## and ##x##. It seems as long as you find an upper bound ##M## for ##f^{(n+1)}(c)## on ##(a,x)## (or ##(x,a)## if ##x<a##), you should be fine.

I also think so. Honestly I am still not really understand the notation ##|x-1| \le |x'-1|##.
Take ##x'## as fixed. You're finding an upper bound for ##R_n(x)## in the interval ##|x-a|<d## with ##d = |x'-a|##.

• songoku
FactChecker
Gold Member
$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ where ##c## is between ##a## and ##x##. It seems as long as you find an upper bound ##M## for ##f^{(n+1)}(c)## on ##(a,x)## (or ##(x,a)## if ##x<a##), you should be fine.

Take ##x'## as fixed. You're finding an upper bound for ##R_n(x)## in the interval ##|x-a|<d## with ##d = |x'-a|##.
You are proposing that there is a stronger theorem about the bound on the error. I am not aware of it.
CORRECTION: Apparently there is a stronger theorem. It is referenced in @songoku 's post #14)

Take ##x'## as fixed. You're finding an upper bound for ##R_n(x)## in the interval ##|x-a|<d## with ##d = |x'-a|##.
I will try to use that:

$$d=|x'-a|$$
$$2=|x'-1|$$
$$x'=3 ~\text{or} ~x'=-1$$

Since ##x' \in (a,x)## or ##x' \in (x,a)##, there will be 2 intervals, which are ##-1 < x' < 1## or ##1 < x' < 3##. No upper bound for the remainder in interval ##-1 <x'<1## so the one I should use is ##1<x'<3##

Is this correct? Thanks

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Mark44
Mentor
I've deleted a couple of my posts and replies to them, as I really misread the graph in post #10. As a result, my responses didn't add anything to the thread.

vela
Staff Emeritus
Homework Helper
Is this correct? Thanks
No.

Say you want to find an upper bound for ##R_3(1.5)##, i.e., ##x'=1.5##. The remainder is equal to
$$R_3(1.5) = \frac {f''''(c)}{4!} (1.5-1)^4$$ for some ##c## in the interval ##(1,1.5)##. If you find an upper bound ##M## for ##f''''(x)## on the interval ##(1,1.5)##, then you can say
$$R_3(1.5) \le \frac {M}{4!} (1.5-1)^4.$$ Note that you don't really care that the series converges between ##-1 < x \le 3## to find an upper bound for the remainder ##R_3(1.5)##.

• songoku
No.

Say you want to find an upper bound for ##R_3(1.5)##, i.e., ##x'=1.5##. The remainder is equal to
$$R_3(1.5) = \frac {f''''(c)}{4!} (1.5-1)^4$$ for some ##c## in the interval ##(1,1.5)##. If you find an upper bound ##M## for ##f''''(x)## on the interval ##(1,1.5)##, then you can say
$$R_3(1.5) \le \frac {M}{4!} (1.5-1)^4.$$ Note that you don't really care that the series converges between ##-1 < x \le 3## to find an upper bound for the remainder ##R_3(1.5)##.
Let me try again:

I want to find an upper bound for ##R_3(x)##. The remainder is equal to
$$R_3(x) = \frac {f''''(c)}{4!} (x-1)^4$$ for some ##c## in the interval ##(1,x)## or ##(x,1)##. There will be 2 intervals since I do not know whether ##x>1## or ##x<1##

From the question: ##|x-1| \le 2## so ##-1 \le x \le 3##

Then how to proceed to find the relevant interval for the remainder? Thanks

haruspex
Homework Helper
Gold Member
2020 Award
how to proceed to find the relevant interval for the remainder?
If I understand the question, it can be any interval containing x and 1. So for x<1 you can choose [x,1).

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• songoku and FactChecker
If I understand the question, it can be any interval containing x and 1. So for x<1 you can choose [x,1).
So for ##x<1## the interval is ##[x,1)## and for ##x>1## the interval is ##(1,x]##.

Then how to find the upper bound? Thanks

haruspex
Homework Helper
Gold Member
2020 Award
So for ##x<1## the interval is ##[x,1)## and for ##x>1## the interval is ##(1,x]##.

Then how to find the upper bound? Thanks
Is there a local extremum within the interval? What about values at the ends of the interval?

• songoku
Is there a local extremum within the interval? What about values at the ends of the interval?
Since ##\frac{24}{(1+x)^5}## is monotonically decreasing in interval ##-1<x \le 3##, there is no local extremum so the upper bound will the located at the left end-point of the interval.

I think I only need to find upper bound either on ##[x,1)## or ##(1,x]## so the one I can use is ##(1,x]##. The upper bound will be ##\frac 3 4##

Is this correct? Thanks

FactChecker
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