# Why do I not understand what seems to be basic Calculus?

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• mcastillo356
In summary, the conversation discusses Taylor's Formula and its use in finding the error in a Taylor approximation. It also introduces Taylor's Theorem and provides a proof for it using mathematical induction. The conversation then provides an example of using Taylor's Theorem to approximate the square root of 26 and finding the corresponding error bound. The use of absolute values in the calculations is necessary to ensure accurate results.
mcastillo356
Gold Member
TL;DR Summary
Is there something about basic analysis I am missing?
Hi PF, stucked with this proof:

Taylor's Formula

The following theorem provides a formula for the error in a Taylor approximation ##f(x)\approx{P_{n} (x)}## similar to that provided for linear approximation (...)

Taylor's Theorem

If the ##(n+1)##st-order derivative, ##f^{(n+1)} (t)##, exists for all ##t## in an interval containing ##a## and ##x##, and if ##P_{n} (x)## is the ##n##th-order Taylor polynomial for ##f## about ##a##, that is,

##P_{n} (x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^n##.

then the error ##E_{n}(x)=f(x)-P_{n}(x)## in the approximation ##f(x)\approx{P_{n}(x)}## is given by

##E_{n}(x)=\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##

where ##s## is some number between ##a## and ##x##. The resulting
formula

##f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^{n}+\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}##, for some ##s## between ##a## and ##x##,

is called Taylor's formula with Lagrange remainder; the Lagrange remainder term is the explicit formula given above for ##E_{n}(x)##.

(Note that the error term (Lagrange remainder) in Taylor's formula looks just like the next term in the Taylor polynomial would look if we continued the Taylor polynomial to include one more term (of degree ##n+1##), except that the derivative ##f^{(n+1)}## is not evaluated at ##a## but rather at some (generaly unknown) point ##s## between ##a## and ##x##. This makes it easy to remember Taylor's formula.)

PROOF Observe that the case ##n=0## of Taylor's formula, namely,

##f(x)=P_{0}(x)+E_{0}(x)=f(a)+\dfrac{f'(s)}{1!}(x-a)##

is just the Mean-Value Theorem

##\dfrac{f(x)-f(a)}{x-a}=f'(s)## for some ##s## between ##a## and ##x##.

Also note that the case ##n=1## is just the error formula for linearization given in the previous theorem

We will complete the proof for higher ##n## using mathematical induction.(...). Suppose, therefore, that we have proved the case ##n=k-1##, where ##k\geq{2}## is an integer. Thus, we are assuming that if ##f## is any function whose ##k##th derivative exists on an interval containing ##a## and ##x##, then

##E_{k-1}(x)=\dfrac{f^{(k)}(s)}{k!}(x-a)^k##

where ##s## is some number between ##a## and ##x##. Let us consider the next higher case ##n=k##. As in the proof of Theorem 11 (previous), we assume ##x>a## (the case ##x<a## is similar) and apply the Generalized Mean-Value Theorem to the functions ##E_{k}(t)## and ##(t-a)^{k+1}## on ##[a,x]##. Since ##E_{k}(a)=0##, we obtain a number ##u## in ##(a,x)## such that

##\dfrac{E_{k}(x)}{(x-a)^{k+1}}=\dfrac{E_{k}(x)-E_{k}(a)}{(x-a)^{k+1}-(a-a)^{k+1}}=\dfrac{E'_{k}(u)}{(k+1)(u-a)^{k}}##

Now
(Troublesome step):
##E'_{k}(u)=\dfrac{d}{dt}\left(f(t)-f(a)-f'(a)(t-a)-\dfrac{f''(a)}{2!}(t-a)^2-\ldots-\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}\right)\Bigg | _{t=u}##

##=f'(u)-f'(a)-f''(a)(u-a)-\ldots-\dfrac{f^{(k)}(a)}{(k-1)!}(u-a)^{k-1}##

This last expression is just ##E_{k-1}(u)## for the function ##f'## instead of ##f##. By the induction assumption it is equal to

##\dfrac{(f')^{(k)}(s)}{k!}(u-a)^{k}=\dfrac{f^{(k+1)(s)}}{k!}(u-a)^{k}##

for some ##s## between ##a## and ##u##. Therefore,

##E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}##

We have shown that the case ##n=k## of Taylor's Theorem is true if the case ##n=k-1## is true, and the inductive proof is complete

Question: why doesn't differentiate properly ##\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}## at what I find the difficult, troublesome step?; why is not ##\dfrac{f^{(k+1)}(a)}{k!}(t-a)^{k}##?

Attempt: it has to be a lack of induction proof knowledge; or basic calculus understanding.

PS: I'm going to post with no preview.

Edited to improve LaTeX: Big | for differentiation

Last edited:
noahN
mcastillo356 said:
Summary:: Is there something about basic analysis I am missing?

hen the error En(x)=f(x)−Pn(x) in the approximation f(x)≈Pn(x) is given by

En(x)=f(n+1)(s)(n+1)!(x−a)n+1

where s is some number between a and x. The resulting

En(x) is a summation
$$E_n(x)=\sum_{k=n+1}^\infty \frac{f^{(k)}(a)(x-a)^k}{k!}$$

So s =s(x) should satisfy
$$f^{(n)}(s(x))=f^{(n)}(a)+\frac{(n+1)!}{(x-a)^n }\sum_{k=n+2}^\infty \frac{f^{(k)}(a)(x-a)^k}{k!}$$
where I do not find a benefit to introduce s(x).

Last edited:
Lnewqban and mcastillo356
mcastillo356 said:
Question: why doesn't differentiate properly ##\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}## at what I find the difficult, troublesome step?; why is not ##\dfrac{f^{(k+1)}(a)}{k!}(t-a)^{k}##?
Why should ##f^{(k)}(a)## become ##f^{(k+1)}(a)##? ##f^{(k)}(a)## is just a number (a constant), only the ##(t-a)^{k}## part changes under differentiation.

benorin, Lnewqban, Infrared and 2 others
Dragon27 said:
is just a number (a constant)

Love

anuttarasammyak said:
En(x) is a summation
$$E_n(x)=\sum_{k=n+1}^\infty \frac{f^{(k)}(a)(x-a)^k}{k!}$$

This is not necessarily the case. The function ##f## is not assumed to be infinitely differentiable, much less analytic.

Hi, PF, my textbook gives this example to show the meaning or truth of Taylor's Theorem:

EXAMPLE 4 Use the 2nd-order Taylor polynomial for ##\sqrt{x}## about ##x=25## found in Example 1(a) to approximate ##\sqrt{26}##. Estimate the size of the error, and specify an interval that you can be sure contains ##\sqrt{26}##.

SOLUTION In Example 1(a) we calculated ##f''(x)=-(1/4)x^{-3/2}## and obtained the Taylor polynomial

##P_{2}(x)=5+\dfrac{1}{10}(x-25)-\dfrac{1}{1000}(x-25)^2##

The required approximation is

##\sqrt{26}=f(26)=5+\dfrac{1}{10}(x-25)-\dfrac{1}{1000}(x-25)^{2}=5,099##

Now ##f'''(x)=(3/8)x^{-5/2}##. For ##25<s<26##, we have

##|f'''(s)|\leq{\dfrac{3}{8}\dfrac{1}{25^{5/2}}=\dfrac{3}{8\times{3,125}}=\dfrac{3}{25.000}}##

Thus, the error in the approximation satisfies

##|E_{2}(26)|\leq{\dfrac{3}{25.000\times{6}}(26-25)^{3}}=\dfrac{1}{50.000}=0,000\;02##

Therefore, ##\sqrt{26}## lies in the interval ##(5,098\;98,\quad{5,099\;02})##

Question: It's ok, but, how do I explain that I obviously needed absolute values for ##f'''(s)## and ##E_{2}(26)##?; previously I've readen Taylor's Theorem, and it doesn't mention. How can I match this?

Thanks!

PS: Hope LaTeX is right

mcastillo356 said:
Question: It's ok, but, how do I explain that I obviously needed absolute values for ##f'''(s)## and ##E_{2}(26)##?; previously I've readen Taylor's Theorem, and it doesn't mention. How can I match this?
Because you want the magnitude of the error; i.e., as a nonnegative number, and you don't care if the result is under or over the actual value.
mcastillo356 said:
PS: Hope LaTeX is right
Looks good.

Mark44 said:
Because you want the magnitude of the error; i.e., as a nonnegative number, and you don't care if the result is under or over the actual value.
I actually think is the same to take absolute value or express it as a negative number. Just notice that ##E(x)=f(x)-L(x)=-(L(x)-f(x))##. It's a gap, no matter the sign. Taking the absolute value is more intuitive, but of the same significance and meaning as if we would compute it without absolute values, just a meaningless issue.

Sorry PF, I must explain last post. Basically I've simplified.

##E_{n}(x)=f(x)-P_{n}(x)##

I've used the fact, the following trick: ##E_{1}(x)=f(x)-P_{1}(x)##; now, ##P_{1}(x)=L(x)##

Hi, PF, one last hurdle about the proof, the last step, the last expression

"
This last expression is just ##E_{k-1}(u)## for the function ##f'## instead of ##f##. By the induction assumption is equal to

##\dfrac{(f')^{(k)}(s)}{k!}(u-a)^{k}=\dfrac{f^{(k+1)}(s)}{k!}(u-a)^{k}##

for some ##s## between ##a## and ##u##. Therefore,

##E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}##
"
##E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}## is given by induction, isn't it?. I was obsessed with the Mean Value Theorem, and didn't make progress (thought this last step ought to come from MVT)

The last expression for ##E_{k}(x)## comes from putting the expression before that (which is what we calculated ##E'_{k}(u)## should be) into the formula for ##\dfrac{E_{k}(x)}{(x-a)^{k+1}}## above (which contains the ##E'_{k}(u)## we wanted).

mcastillo356
Understood, I think.
Thanks @Dragon27 for the last tip, thanks though PF!

## 1. Why is Calculus considered difficult?

Calculus is a branch of mathematics that deals with the study of change and rates of change. It involves complex concepts and techniques, making it challenging for many people to understand. Additionally, the abstract nature of calculus can also make it difficult to grasp.

## 2. Why do I struggle to understand basic Calculus concepts?

Understanding calculus requires a strong foundation in algebra and trigonometry. If you struggle with these subjects, you may find it difficult to understand basic calculus concepts. It is important to review and master these prerequisite topics before diving into calculus.

## 3. How can I improve my understanding of Calculus?

Practice, practice, practice! Calculus is a subject that requires a lot of practice to fully understand. Make sure to work through plenty of problems and seek help from a tutor or teacher if needed. Additionally, try to visualize and understand the concepts rather than just memorizing formulas.

## 4. Why is it important to understand Calculus?

Calculus is a fundamental tool in many fields, including engineering, physics, economics, and statistics. It allows us to model and solve real-world problems involving change and rates of change. Additionally, understanding calculus can help develop critical thinking and problem-solving skills.

## 5. How long does it take to understand Calculus?

The time it takes to understand calculus varies from person to person. It depends on your prior knowledge and understanding of prerequisite topics, as well as your dedication to studying and practicing. With consistent effort and practice, most people can grasp the basics of calculus within a few months.

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