Little Red Riding Hood's grandma has moved out of her cabin in the woods and has taken an apartment in the city. Little Red Riding Hood with her basket of goodies took the bus downtown to visit her. Before she left, Little Red Riding Hood's mother gave her specific instructions.

"Get off the bus at the intersection of Church and Chapel streets. Then continue to walk past the intersection in the same direction as the bus was going. Grandma's apartment is in the fourth building past the intersection. And be sure you don't talk to strangers, dear."

Little Red Riding Hood got off the bus at the proper place and crossed the street. But to her dismay, the buildings were all built right up against each other. She couldn't tell where one ended and the next one began. Again and again she tried to count the buildings, but in the end her efforts produced only tears of frustration.

The wolf, who was standing on the street-corner, saw everything. "Dear little girl," he said, "what could possibly be so troubling that you should cry like that?"

"Go away," she replied. "I'm not supposed to talk to strangers."

The wolf was undaunted. "So you would prefer to stand there crying when you could ask me for assistance? All I desire is to put an end to your tears. Now tell me please, sweet child, what is the matter?"

Little Red Riding Hood could see that asking the wolf for help was much more likely to get her to Grandma's than crying. So she told him the whole story.

"The solution is oh so simple, tender lass," he said. "You see, the width of each building is exactly seventeen of your steps. You know how to multiply, don't you? You need only multiply those seventeen steps per building times the four buildings you want to count, and you will be able to find your Grandma's house. Just start at the corner, walk down the street counting your steps, and when you have counted to sixty-eight, you will be there."

The wolf stayed with her for her first few steps, then disappeared into an alley-way. Little Red Riding Hood carefully counted off the sixty-eight steps and was pleased to discover that it brought her quite close to a doorway. In she went, climbed the stairs, and knocked on the door. "Enter deary," came a raspy voice from within. And she did.

"Grandma!" she exclaimed. "What awful plastic surgery you've had..."

I'll let you provide your own ending to the story. Besides the lesson about not talking to strangers, what other lesson is there for us in this tale?

Suppose that the intersection of Church and Chapel streets is
the point, ` y = 1`` y = 1.0004 ^{17} `

Well it seems the wolf was also a calculus instructor who lost his job for having his freshman students for dinner. So here is the wolf's line of thinking:

If, then everybody knows that y(x) = x^{17}. The number, y(1) = 1, is very close to x = 1.0004, which is means that Grandma's building is very close to Church and Chapel. But how close? I know that x = 1. And it's easy to see from that that y'(x) = 17x^{16}. So whenever I'm nearby Church and Chapel, for every y'(1) = 170.0001thatxincreases, I should seey(x)increase by approximately17times that.

In this way the wolf was able to approximate the location of Grandma's
building as ` y = 1.0068`` y = 1.006821804``68` steps, she was only about one fifth
of a step from Grandma's doorway.

The wolf's calculation is just another application of something that we
have been using frequently in previous sections. That is, if `f(x)`
is a continuous function and its derivative, `f'(x)`, is also
continuous, you can approximate `f(x+h)` using

f(x + h) » f(x) + hf'(x) eq. 8.4-1where the

**Step 1:** `9.5` is close to `9`, which is a perfect square.
And indeed you know that the square root of `9` is `3`

**Step 2:** Find the derivative of the square root function.
We've done that one before.

_ f(x) = Öx 1 f'(x) =~~2Öx~~

**Step 3: **Find the value of the derivative at the `x` where you know the square root.
That would be at ` x = 9`

1 1 f'(9) =~~=~~~~2Ö9 6~~

**Step 4: **What's the difference between the number you know the square
root of and the number you are interested in taking the square root of? That
value is `h`.

h = 9.5 - 9 = 0.5When you are not looking at this page you might have to ask yourself, "Which way do I take the difference?" Always set it up so that if you add

**Step 5: **Apply the approximation forumla.

___ _ 1 1 Ö9.5 » Ö9 + hf'(9) = 3 + 0.5 ×Using a calculator, I find that squaring the approximation yields~~= 3~~~~6 12~~

In some textbooks they will use the symbol, `Dx`, instead of `h`.
For the problem we just did, it would look like

Dx = 9.5 - 9 = 0.5 f(x + Dx) » f(x) + Dxf'(x)Don't let that throw you. It's the same thing just using a different name.

In the movie, *Infinity*, Richard Feynman was depicted challenging a Chinese
merchant to an arithmetic contest in which the merchant was permitted to use his
abacus. The merchant beat him on simple addition and multiplication problems.
The problem where Feynman bested the merchant was taking the cube root of
`1729`. The merchant came up with `12`, but Feynman insisted
that they carry it out to some decimal places, and he used the method described
above to do just that.

**Step 1: ** `1729` is close to `1728`, which is `12 ^{3}`
(you know this because you recall from junior high that there are

**Step 2: ** Find the derivative of the cube root function:

f(x) = x^{1/3}1 f'(x) =~~x~~^{-2/3}3

**Step 3: **Find the value of the derivative at the `x` where you know the
cube root.

1^{ }1 1^{ }1 f'(1728) =~~1728~~^{-2/3}=~~×~~^{ }=~~3~~_{ }3 12^{2}432

**Step 4: **What's the difference between the number you know the cube
root of and the number you are interested in taking the cube root of? That
value is `h`.

h = 1729 - 1728 = 1

**Step 5: **Apply the approximation forumla.

How Feynman was able to divide out the fraction into the decimal,^{ }1 1729^{1/3}» 12 + hf'(1728) = 12_{ }432

**1)** Use the method we have discussed here to approximate
`258 ^{1/4}`

**2)** Use the method we have discussed here to approximate

__ Ö80

Transcendental functions are ones like sine, cosine, exponential,
log, and all the others that cannot be described in terms of simple
powers and roots of `x`. Even though these functions are
qualitatively different from the roots we approximated so far,
you can still apply the same approximation method to them as
well.
For example, lets apply the method to finding ` e ^{1.1}`.

**Step 1:** For what `x` that is near `1.1` do we know
the value of `e ^{x}`? Well we know that

**Step 2:** What is the derivative of `e ^{x}`? That's an
easy one.

f(x) = e^{x}f'(x) = e^{x}

**Step 3:** What is the value of the derivative at the `x`
where we know the function? Again this is an easy one.

f'(1) = e^{1}= e » 2.72

**Step 4:** What is the difference between the `x` where you
know the function's value and the `x` where you'd like to approximate
the function's value?

h = 1.1 - 1 = 0.1

**Step 5:** Apply the approximation formula.

eThe actual value of^{1.1}= e^{(1 + h)}» e + hf'(1) » 2.72 + 0.272 = 2.992

Approximate `cos(0.75)`.

**Step 1:** What `x` that is near `0.75` do
I know the cosine of? A little investigation yields that `0.75`
is very close to

pand that~~= 0.7853981... 4~~

_ Ö2 cos(p/4) =(if you don't remember this, see the trig id table)~~= 0.7071057... 2~~

**Step 2:** What is the derivative of ` cos(x)`

f(x) = cos(x)then

f'(x) = -sin(x)

**Step 3:** What is the value of the the derivative of
` cos(x) ` at the

_ Ö2 -sin(p/4) = -~~= -0.7071057... 2~~

**Step 4:** What is the difference between the `x`
that you know the cosine of and the one you want to know the cosine
of? That will be your `h`.

h = 0.75 - 0.7853981 = -0.0353981

**Step 5:** Apply the approximation formula.

_ Ö2 cos(0.75) = cos(p/4 + h) »The actual cosine of~~+ hf'(p/4) = 0.732137 2~~

All the approximations we have done so far have come out close.
But the method begs some questions.
Like, "How close will my approximation be?" and "How close does
my `x` have to be to the number I want to take this function
of for the method to work reasonably?" These two questions are not
silly details. They probe deeply into what derivatives
and continuous functions are all about.

For comparison, let's try the method on a very easy function.

f(x) = 3x + 2which is the function of a straight line with slope of

Since you know the method by now and since this is an easy function, we'll take all the steps at once.

f(0) = 2 f'(x) = 3 h = 1 - 0 = 1 f(1) = f(0 + h) » 2 + hf'(0) = 2 + (1 × 3) = 5But indeed,

So what is special about straight-line functions that makes
them so susceptible to our method? You could put together some
words about how the function is linear and so is the method,
and such an explanation would be correct. But another much
more concise way to put it is simply, "The method works
perfectly on straight-line functions because a straight-line
function *always has a second derivative of zero*."

The second derivative is the key to figuring out how
good an approximation the method yields each time.
If the derivative of `f(x)` were constant
(which is another way of saying the second derivative is zero),
then the method would be infallible. But for most functions
we are interested in applying the method to the derivative
is not constant.

Let's call the error that the approximation formula is off by
by symbol, `x` (that's the
Greek letter, xi),
which is the traditional math symbol for an error term.
Suppose we know what `f(x)` is at ` x = c``f(c+h)`.
Then if our approximation is off by some error amount, `x`,
you can represent that with the equation:

f(c + h) = f(c) + hf'(c) + x eq. 8.4-2You can bound the error term,

where^{ }1 |x| £~~|h~~^{2}f"(a)| eq. 8.4-3_{ }2

How we get equation 8.4-3 is **optional material.** You can read it by

**Worked example:** Develop a worst-case error term for when you approximate
the square root of `9.5` using the method introduced in this section
(

**Step 1: Find the second derivative.** We know that

_ f(x) = Öx = xSo applying the power rule to^{1/2}1 1 f'(x) =~~=~~~~x~~^{-1/2}2Öx 2

1 f"(x) = -~~x~~^{-3/2}4

**Step 2: Where does the second derivative have its maximum magnitude?**
This step is sometimes difficult, but not in this case. Observe that as `x`
increases, the magnitude (that is the absolute value) of `f"(x)` *always
decreases*. So the magnitude of `f"(x)` will be greatest at the lower
end of the interval. In this case the interval runs from ` x = 9 `` x = 9.5``f"(x)` at ` x = 9`

1 1 f"(9) = -~~9~~^{-3/2}= -~~4 108~~

**Step 3: Take the absolute value of half that times the square of the length of the interval.**
This is the application of the inequality in ` x = 9 `` x = 9.5``h`, of the
interval is ` h = 0.5 = 1/2`

1 1 1 1 1 |x| £And indeed using a calculator I find that the actual error is~~|h~~^{2}f"(9)| =~~×~~~~×~~~~=~~~~2 2 4 108 864~~

0.973 1 |x| = 0.001126331 =~~£~~~~864 864~~

No tale about Little Red Riding Hood would be complete without the wolf jumping up out of Grandma's bed and shouting, "All the better to eat you with!" So we pick up the story from there.

Of course Little Red Riding Hood was frightened beyond imagining. But the quick-minded little girl grabbed the phone and ran into the bathroom. Once she had the door locked with the wolf bellowing on the other side, she dialed 911.

"Please help me," she cried. "I'm locked in the bathroom and the wolf is going to eat me."

"Can you see any way to escape?" asked the 911 operator.

"I can see through the window," she replied, still panting with fear. "There's a fire escape out there."

"Now stay calm, little girl," said the operator. "Go out the window and down the fire escape. Then go to the Three Little Pigs' apartment. You'll be safe there. We've got an animal control officer on the way."

"But where do the Three Little Pigs live?" she whimpered.

"They're in the fifteenth building from the corner," was the answer.

Little Red Riding Hood regained her composure. "I know what to do," she said. "It's seventeen steps for each each building. That's 255 steps from the corner, right?"

"Well, not exactly," the operator replied. "The buildings get smaller as you go down the street. After four buildings you start counting only sixteen steps per building."

"I get it," said Little Red Riding Hood. "And after eight buildings I start counting only fifteen steps per building, then at the twelfth building count only fourteen steps per building, right?"

"Well not quite," the operator explained. "You see the rate at which they get smaller increases as you go down the block as well. So in the next four buildings the number of steps per building decreases by two instead of one. That means at the eighth building start counting fifteen steps per building, but after the tenth building start counting only fourteen steps per building. By the twelfth building you are down to only thirteen steps per building."

"Now I get it," said Little Red Riding Hood. "With each four buildings the rate at which the steps per building decreases -- that rate increases by one step, right?"

"Well, almost," droned the operator. "You see the rate of that rate changes as well. And the rate of the rate of the rate isn't constant either..."

Poor Little Red Riding Hood. This was making her head spin. And the wolf was starting to huff and puff just outside the door. She dropped the phone and ran down the fire escape.

With that story in mind, let's play the *Name That Function* game again.
I'm thinking of a cubic polynomial. At ` x = 0 ``1`. Also at ` x = 0``-2`,
the value of is second derivative
is `3`, and the value of its third derivative is `-5`.

f(0)This description of^{ }= 1 f'(0)^{ }= -2 f"(0)^{ }= 3 f^{(3)}(0) = -5 table 8.4-1

I've already given away that the function I'm thinking of is a cubic polynomial. So it must be in the form of

f(x) = BxYou only have to find^{3}+ Cx^{2}+ Dx + E eq. 8.4-8

f(0) = 1 = (0)B + (0)C + (0)D + E = E eq. 8.4-9aSo clearly you have

f'(x) = 3BxIt's just as clear from this that^{2}+ 2Cx + D eq. 8.4-9b f'(0) = -2 = (0)B + (0)C + D = D eq. 8.4-9c

f"(x) = 6Bx + 2C eq. 8.4-9d f"(0) = 3 = (0)B + 2C = 2C eq. 8.4-9eFrom this you can quickly see that

fAnd from that you should readily see that^{(3)}(x) = 6B eq. 8.4-9f f^{(3)}(0) = -5 = 6B eq. 8.4-9g

5 3 f(x) = -But suppose I had said that it was a 4th degree polynomial with the same values at~~x~~^{3}+~~x~~^{2}- 2x + 1 eq. 8.4-10 6 2

f(0)Before you were looking for the coefficients of^{ }= 1 f'(0)^{ }= -2 f"(0)^{ }= 3 f^{(3)}(0) = -5 f^{(4)}(0) = 7 table 8.4-2

f(x) = AxIf you go through the same procedure as we did for the cubic, you will find that the coefficients,^{4}+ Bx^{3}+ Cx^{2}+ Dx + E eq. 8.4-11

f'(x) = 4AxSince^{3}+ 3Bx^{2}+ 2Cx + D eq. 8.4-12a f"(x) = 12Ax^{2}+ 6Bx + 2C eq. 8.4-12b f^{(3)}(x) = 24Ax + 6B eq. 8.4-12c f^{(4)}(x) = 24A eq. 8.4-12d

7 5 3 f(x) =The point here is that adding the condition that the fourth derivative of the polynomial had to be equal to a particular value at~~x~~^{4}-~~x~~^{3}+~~x~~^{2}- 2x + 1 eq. 8.4-13 24 6 2

At this moment you might be trying to generalize this last point. And the generalization
is true. For any polynomial function, *imposing the condition that the *`n`*th
derivative is equal to
a particular value at *` x = 0 `*
effects only the *`n`*th degree term of the polynomial
and no other term.* And is there an easy way to arrive at the `n`th term of
the polynomial given such a condition? Yes. It will be self-evident as soon as you can see that

Recall thatThenth derivative ofx^{n}is equal to the constant,n!.

f(x) = xthen^{n}

fCan you see why? The power rule tells us that^{(n)}(x) = n!

fClearly the product of all those terms ahead of the^{(n)}(x) = n(n-1)(n-2)(n-3) × ... × 3 × 2 × 1 × x^{0}

If the `n`th derivative of `x ^{n}` is the

Now suppose I told you, for example, that the sixth derivative of a polynomial was equal
to `13` at ` x = 0`` A _{5}x^{5}`

13 = 6!ASo the condition that the sixth derivative of a polynomial is equal to_{6}13 A_{6}=~~6!~~

If you impose the condition that thenth derivative of a polynomial is equal toCat, then the coefficient of the x = 0nth degree term of the polynomialmustbe equal toC/n!.

But what if the function you are trying to match is not a polynomial? Can we use what
we learned about derivatives at ` x = 0`` f(x) = e ^{x}`

Let's begin using the approximation method we started with to find `e ^{0.2}`.
We know that

eThe real^{0.2}» e^{0}+ he^{0}= 1.2 eq. 8.4-14

eLook carefully.^{x}» e^{0}+ xe^{0}= 1 + x eq. 8.4-15

If ` f(x) = e ^{x}`

f(0) = PIn other words, at_{1}(0) f'(0) = P_{1}'(0)

f"(0) ¹ P"But_{1}(0)

You should be able to see that^{ }1 e^{x}» P_{2}(x) = 1 + x +~~x~~^{2}eq. 8.4-16a_{ }2!

Using^{ }1 1 1 e^{x}» P_{n}(x) = 1 + x +~~x~~^{2}+~~x~~^{3}+ ... +~~x~~^{n}eq. 8.4-16b_{ }2! 3! n!

nremembering that^{ }1 e^{x}» P_{n}(x) = å~~x~~^{k}eq. 8.4-16c k=0 k!

So with all that in mind, would it surprise you to learn that

In other words, as you add more and more terms to this polynomial, it gets closer and closer to^{ }1 1^{ }1 e^{x}= lim P_{n}(x) = lim 1 + x +~~x~~^{2}+~~x~~^{3}+ ... +~~x~~^{n}_{ }n~~> ¥ n~~~~> ¥ 2! 3! n!~~

But the more traditional way to write the limit on the right is to drop the limit sign and write^{ }n 1 e^{x}= lim P_{n}(x) = lim å~~x~~^{k}eq. 8.4-17a^{ }n~~> ¥ n~~~~> ¥ k=0 k!~~

¥When you see the^{ }1 e^{x}= å~~x~~^{k}eq. 8.4-17b k=0 k!

Equation 8.4-17b is an example of something called a *Maclaurin series*.
Click here to see
a brief biography of Maclaurin. If, at ` x = 0`` x = 0`*all* its derivatives (that is first derivative, second derivative, third derivative, and so on indefinitely)
are continuous. But not every `x` works for every Maclaurin series. We will cover that aspect
in a later section.

To create a Maclaurin series of a function, you must know the value of all its derivatives
at ` x = 0``e ^{x}`, is especially well suited
for this since it is its own derivative. So it is easy to know what any numbered derivative of

* The Maclaurin Formula:* In general, if you have a continuous function,

f(0)for^{ }= A_{0}f'(0)^{ }= A_{1}f"(0)^{ }= A_{2}f^{(3)}(0) = A_{3}. . .

or in ¥ |

In some cases this formula will not work for *all* `x`, but it will work for some
range of `x` near zero.

**Problem:** Find the Maclaurin series for ` f(x) = sin(x)`

**Step 1: Make a table of the first few derivatives of **`f(x)`.
This is often the hardest part of problems like this, but in this
case `sin(x)` is a very orderly function with repsect to
its derivatives.

f(x)^{ }= sin(x) f'(x)^{ }= cos(x) f"(x)^{ }= -sin(x) f^{(3)}(x) = -cos(x) f^{(4)}(x) = sin(x)

**Step 2: Try to discern a pattern as the derivatives go higher.**
Since the fourth derivative brings `sin(x)` back to itself, we can expect
that the pattern of the first four derivatives will repeat indefinitely for still
the higher derivatives.

**Step 3: Put ** `x = 0` ** into the table and evaluate.**

f(0)Here is where you really have to identify the pattern that the^{ }= sin(0) = 0 = A_{0}f'(0)^{ }= cos(0) = 1 = A_{1}f"(0)^{ }= -sin(0) = 0 = A_{2}f^{(3)}(0) = -cos(0) = -1 = A_{3}f^{(4)}(0) = sin(0) = 0 = A_{4}

**Step 4: Put it all into the Maclaurin formula.**
Since the even-power terms will be all zero, we put up only the odd-power terms.

It might seem at first as if this one would be hard to put into^{ }1 1 1 1 sin(x) = lim x -~~x~~^{3}+~~x~~^{5}-~~x~~^{7}+ ... ±~~x~~^{2n+1}n~~> ¥ 3! 5! 7! (2n + 1)! eq. 8.4-19a~~

¥ (-1)Remember in trig identity section we mentioned that^{k}sin(x) = å~~x~~^{2k+1}eq. 8.4-19b k=0 (2k + 1)!

Likewise with even functions. For *any* even function
(that is where ` f(-x) = f(x) ``n`th derivative
(if it exists) will be nonzero at ` x = 0`*only* when `n` is
an even number. Also for *any* even function, when you take the Maclaurin series, *only*
the terms with even-numberd exponents will be nonzero.

The graph shows the first six Maclaurin approximations to `sin(x)`.
That is in green is the first degree polynomial approximation, in brown
is the third degree polynomial approximation (has two nonzero terms),
and so on out to the black trace, which has six nonzero terms.
The actual trace of `sin(x)` is shown in blue, but for most of
the graph the traces of the Maclaurin approximation polynomials are
so close that they obscure the blue trace. You can see how the higher
degree approximations stay close to `sin(x)` over a larger
range of `x`. If you added more and more higher degree terms,
you would be able to get the Maclaurin approximation to stay near
`sin(x)` over as long an interval as you like.

Since `cos(x)` is the derivative of `sin(x)`, what would
you expect the Maclaurin series for `cos(x)` to be? Once you know
the Maclaurin series for `sin(x)`, you don't even have to apply
the Maclaurin formula to find it for `cos(x)`. All you need to do
is take the derivative, term by term, of the Maclaurin series for
`sin(x)`. That is, take the `n`th power term, reduce that
exponent by `1`, and multiply the coefficient by `n`. In
` n = 2k+1`

¥ (-1)And remember that^{k}(2k + 1)^{ }¥ (-1)^{k}cos(x) = å~~x~~^{2k+1-1}= å~~x~~^{2k}eq. 8.4-19c k=0 (2k + 1)! k=0 (2k)!^{ }1 1 1 1 cos(x) = lim 1 -~~x~~^{2}+~~x~~^{4}-~~x~~^{6}+ ... ±~~x~~^{2n}n~~> ¥ 2! 4! 6! (2n)!~~

Find the Maclaurin series (or at least the first four nonzero terms of it) for

1 f(x) =~~x + 1~~

**Step 1: Find the derivatives of the function.**
Go ahead and find the first three derivatives, `f'(x)`, `f"(x)`, and
`f ^{(3)}(x)`. Use the

**Step 2: Look for the pattern in the derivatives.**
Can you see what's happening as you continue to take higher derivatives
of this function? With each successive derivative you multiply
by the next negative integer and take the next higher power in
the denominator. Based upon that try to come up with a general
expression for `f ^{(n)}(x)`. When you think you
have it,

**Step 3: Evaluate the function and derivatives at**
`x = 0``x` into each
of the original function and its derivative expressions and see what you get.
Those numbers will become your coefficients,
`A _{0}` (from the original function),

**Step 4: Put it all into the ****Maclaurin formula**.
Simply gather up the `A _{n}`'s that you came up with in the
preceding step and plug them in. Then make any simplifications that seem
evident. When you are done,

Think carefully about the formula:

1If~~= lim 1 - x + x~~^{2}- x^{3}+ x^{4}- ... ± x^{n}x + 1 n~~> ¥~~

But let's try this Maclaurin series on a value that it can do, say,
` x = 1/2`` 1 + 1/2 = 3/2``2/3`. When I add up the first ten terms of
this Maclaurin series with ` x = 0.5`

1Pretty close, eh? Try it yourself with, say,~~» 0.6660151625 1 + 0.5~~

1 f(x) =Maclaurin is not going to get it for you. Why? Because Maclaurin requires that you find the function's value at~~eq. 8.4-20 x~~

In the previous paragraphs, though, we did discover that we can find a perfectly serviceable Maclaurin series for

1 f(x) =Suppose you substituted~~eq. 8.4-21 x + 1~~

1 f(x) = f(u + 1) =But the right-hand side of equation 8.4-22 is just the right-hand side of equation 8.4-20 by a different name. And we know how to do a Maclaurin series for that:~~eq. 8.4-22 u + 1~~

1 ¥So what happens when you substitute back~~= å (-1)~~^{k}u^{k}eq. 8.4-23a u + 1 k=0

1 ¥This is called a~~= å (-1)~~^{k}(x - 1)^{k}eq. 8.4-23b x k=0

Finding a Taylor series is, in reality, the same thing as finding a Maclaurin series, only
you've shifted the `x`-axis by some amount (in this case we shifted it by `1`).
In general the Taylor formula for finding the series taken around the point,
` x = a``f(x)` is

or in ¥ |

But once you know the Maclaurin formula, the recipe for the Taylor formula is easy.
To find the Taylor series around the point, ` x = a`

- Let
.`u = x - a` - Apply the Maclaurin formula to
to get a series in the variable,`f(u + a)``u`. - Substitute back
for`x - a``u`into the resulting Maclaurin series.

If learning to find a Maclaurin series were learning to tie your shoes when you
are at home, then finding a Taylor series is no more different than
tying your shoes at school. The two are really the same thing. Indeed a Maclaurin
series is nothing but a Taylor series taken around ` x = 0`

Find the first four nonzero terms of the Tayor series for the following functions (and observe which point you are asked to take the series around).

_around3)f(x) = Öx

earound^{x}+ e^{-x}4)f(x) =~~2~~

around5)f(x) = tan(x)

around6)f(x) = arctan(x)

1 f'(x) =View solution.~~1 + x~~^{2}

Suppose `c` is any real number and `n` is any
positive integer. Let

f(x) = (x + c)^{n}

- Work out the Maclaurin series for this function.
- Show that the Maclaurin series for this function terminates
by itself after
`n+1`terms (that is all the terms after the first`n+1`of them are zero). - Show that the Maclaurin series that you end up with is identical
to what you would end up with if you expanded
using the`(x + c)`^{n} .*binomial formula*

Move on to Power Tools for Taking Derivatives of Products

email me at *hahn@netsrq.com*