Bounds of the remainder of a Taylor series

In summary: By both sides of ##a=1##, do you...mean check on both sides of a=1 or check on both sides of x'=1?Checking on both sides of a=1.
  • #1
songoku
2,266
319
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:

1626763591792.png
##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:
Physics news on Phys.org
  • #2
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:
  • #3
FactChecker said:
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
 
  • #4
songoku said:
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##.
 
  • Like
Likes songoku
  • #5
vela said:
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
 
  • #6
songoku said:
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.
 
  • #7
songoku said:
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?
 
  • Like
Likes songoku and FactChecker
  • #8
vela said:
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
 
  • #9
songoku said:
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.
 
  • Like
Likes songoku
  • #10
FactChecker said:
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}##
1626843190821.png


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|##
1626843789889.png


But sorry I don't understand how to find the interval where ##|x-1| \le |x'-1|##
 
  • #11
songoku said:
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|].
 
  • Like
Likes songoku
  • #12
FactChecker said:
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
 
  • #13
songoku said:
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.
 
  • #14
FactChecker said:
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:
g


and also

1626878217806.png
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.

FactChecker said:
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
 
  • #15
FactChecker said:
You need to check on both sides of a=1, not just the side that x' is on.
Are you sure about this? The Lagrange form of the remainder is
$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ 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.

songoku said:
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|##.
 
  • Like
Likes songoku
  • #16
vela said:
Are you sure about this? The Lagrange form of the remainder is
$$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},$$ 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)
 
  • #17
vela said:
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
 
Last edited:
  • #18
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.
 
  • #19
songoku said:
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)##.
 
  • Like
Likes songoku
  • #20
vela said:
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
 
  • #21
songoku said:
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).
 
Last edited:
  • Like
Likes songoku and FactChecker
  • #22
haruspex said:
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
 
  • #23
songoku said:
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?
 
  • Like
Likes songoku
  • #24
haruspex said:
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
 
  • #25
songoku said:
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
You should find a separate bound for the two cases, depending on where x is. And remember that the original problem was to find the bounds, not just the upper bound. So you may need to calculate a maximum and a minimum. (They might all have the same magnitude, but they might not in all example problems.)
 
  • Like
Likes songoku
  • #26
FactChecker said:
You should find a separate bound for the two cases, depending on where x is. And remember that the original problem was to find the bounds, not just the upper bound. So you may need to calculate a maximum and a minimum. (They might all have the same magnitude, but they might not in all example problems.)
For ##[x,1)## the upper bound will be ##\frac{24}{(1+x)^5}## and lower bound is ##\frac 3 4##

For ##(1,x]## the upper bound will be ##\frac{3}{4}## and lower bound will be ##\frac{24}{(1+x)^5}##

Is this correct? Thanks
 
  • #27
Looks ok to me.
 
  • Like
Likes songoku
  • #28
Thank you very much for the help and explanation FactChecker, vela, Mark44, haruspex
 

1. What is the purpose of calculating the bounds of the remainder of a Taylor series?

The bounds of the remainder of a Taylor series help us understand the accuracy of approximating a function with a polynomial. It tells us how close the approximation is to the actual value of the function at a specific point.

2. How do you calculate the bounds of the remainder of a Taylor series?

The bounds of the remainder of a Taylor series can be calculated using the Lagrange form of the remainder, which involves the use of the nth derivative of the function and the value of the (n+1)th derivative at a specific point.

3. Can the bounds of the remainder of a Taylor series be negative?

No, the bounds of the remainder of a Taylor series cannot be negative. It represents the absolute value of the difference between the actual value of the function and the approximation, so it is always a positive value.

4. How does the degree of the polynomial affect the bounds of the remainder of a Taylor series?

The degree of the polynomial used in the Taylor series approximation affects the bounds of the remainder. As the degree increases, the bounds of the remainder become smaller, indicating a more accurate approximation of the function.

5. Can the bounds of the remainder of a Taylor series be used to determine the convergence of the series?

Yes, the bounds of the remainder of a Taylor series can be used to determine the convergence of the series. If the bounds of the remainder approach 0 as the number of terms in the series increases, then the series is said to converge.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
780
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
770
  • Calculus and Beyond Homework Help
Replies
4
Views
733
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
128
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
450
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top