# Help with Taylor Series problem

• NHLspl09
In summary, you evaluated the Maclaurin series for f(x) by plugging in x=0, 0.01, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, & 10 to get the first three terms.
NHLspl09
Now although this is a problem for my EE course, it is more of a calculus question so I figured I would receive the best answers by posting it in this section. I have just started on the problem but could use some input on my thoughts. So here we go (there are two parts):
(problem screenshot is attatched)

## Homework Statement

I'm given the function f(x) = e^ax.

a) Derive an expression for the first three nonzero terms of the Taylor Series for f(x) about x=0 (Maclaurin Series).

b) Complete the following table assuming that a=0.1. The function f(x) is the exact function, the function fts2(x) uses the first two nonzero terms of the Taylor Series, and the function fts3(x) uses the first three nonzero terms of the Taylor Series.

## The Attempt at a Solution

a) Now I know (correct me if wrong) f(x) = f(0) + f ' (0) + ((f ''(0)/2!) x^2) + ... But am not sure if that's the correct answer to the question? Or do I need to take this further?

b) There is then a table with x=0, 0.01, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, & 10. Do I plug these values into the first three nonzero terms found in a? For example, first f(0), then f(0) + f ' (0), then f(0) + f ' (0) + ((f ''(0)/2!) x^2)?

Any help/input would be greatly appreciated!

Thank you

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Last edited:

Thank you! I'll continue to work on the problem and post any progress, if any. Was more looking for if my beginning steps were correct so I could base my work off of that rather than learning later on that the whole problem is incorrect. Any input would be greatly appreciated!

NHLspl09 said:
...

## Homework Statement

I'm given the function f(x) = e^ax.

a) Derive an expression for the first three nonzero terms of the Taylor Series for f(x) about x=0 (Maclaurin Series).
...

## The Attempt at a Solution

a) Now I know (correct me if wrong) f(x) = f(0) + f ' (0) + ((f ''(0)/2!) x^2) + ... But am not sure if that's the correct answer to the question? Or do I need to take this further?
f(x) = f(0) + f ' (0) x + ((f ''(0)/2!) x^2) + ...

You left out the x in the second term, the one with the first derivative.

In part (a), you're supposed to explicitly evaluate those derivatives to get the first few terms of the series.

SammyS said:
f(x) = f(0) + f ' (0) x + ((f ''(0)/2!) x^2) + ...

You left out the x in the second term, the one with the first derivative.

Thank you! My mistake, must of missed that.

vela said:
In part (a), you're supposed to explicitly evaluate those derivatives to get the first few terms of the series.

So correct me if I'm wrong, what you're saying is that f(0) = e^ax, f ' (0) = the first derivative, and f '' (0) = the second derivative, and that is what I need to do in order to solve for part a?

Yup. Find the derivatives, evaluate them at x=0, and plug those numbers into the formula.

NHLspl09 said:
...

So correct me if I'm wrong, what you're saying is that f(0) = e^ax, f ' (0) = the first derivative, and f '' (0) = the second derivative, and that is what I need to do in order to solve for part a?
f(0) = ea(0) = 1 , etc.

f ' (0) = ?

SammyS said:
f(0) = ea(0) = 1 , etc.

f ' (0) = ?

f ' (0) = a
f '' (0) = a2??

It would help if you'd show your work if you're unsure about a calculation.

vela said:
It would help if you'd show your work if you're unsure about a calculation.

f(0) = ea(0) = e0 = 1
f ' (0) = a*ea(0) = a*1 = a
f '' (0) = a2*ea(0) = a2*1 = a2

Does this help? This is the derivatives and breaking down to first, second, and third nonzero terms

Looks good.

vela said:
Looks good.

So then I would simply evaluate f(x) = 1 + a + a2 for the first [f(x) = 1], second [f ' (x) = 1 + a], and third [f '' (x) = 1 + a + a2] non zero terms with the values x=0, 0.01, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, & 10?

No. I have no idea where you came up with those expressions. You seem to be pushing symbols around and ignoring basic algebra. This is largely a plug-and-chug problem.

You have a function f(x)=eax. If you want to evaluate f(anything), you plug anything into it to get ea(anything). If you want f'(anything), you differentiate eax and then plug anything into the result. That's what you did so far to find f(0)=1, f'(0)=a, and f''(0)=a2.

So why did you do this? It's because f(0), f'(0), and f''(0) appear in the expression for the Maclaurin series for f(x)
$$f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots$$
So now plug your previous results into that expression to get
$$1+ax+\frac{(ax)^2}{2!}+\cdots$$
If this infinite series converges (which it will for all x), it will converge to the value of f(x), so you can write
$$f(x) = e^{ax} = 1+ax+\frac{(ax)^2}{2!}+\cdots$$
In part (b), you're told
\begin{align*}
f_\mathrm{TS2}(x) &= 1+ax \\
f_\mathrm{TS3}(x) &= 1+ax+\frac{(ax)^2}{2!}
\end{align*}
You want to calculate f(x), fTS2(x), and fTS3(x) for various values of x with a=0.1.

vela said:
No. I have no idea where you came up with those expressions. You seem to be pushing symbols around and ignoring basic algebra. This is largely a plug-and-chug problem.

You have a function f(x)=eax. If you want to evaluate f(anything), you plug anything into it to get ea(anything). If you want f'(anything), you differentiate eax and then plug anything into the result. That's what you did so far to find f(0)=1, f'(0)=a, and f''(0)=a2.

So why did you do this? It's because f(0), f'(0), and f''(0) appear in the expression for the Maclaurin series for f(x)
$$f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots$$
So now plug your previous results into that expression to get
$$1+ax+\frac{(ax)^2}{2!}+\cdots$$
If this infinite series converges (which it will for all x), it will converge to the value of f(x), so you can write
$$f(x) = e^{ax} = 1+ax+\frac{(ax)^2}{2!}+\cdots$$
In part (b), you're told
\begin{align*}
f_\mathrm{TS2}(x) &= 1+ax \\
f_\mathrm{TS3}(x) &= 1+ax+\frac{(ax)^2}{2!}
\end{align*}
You want to calculate f(x), fTS2(x), and fTS3(x) for various values of x with a=0.1.

Thank you for the explanation, which helps me to realize I correctly did the other two taylor series problems I was assigned! I definitely understand what you're saying and just think I wasn't explaining my self very well at all with where I was going... I am going to stop working on homework for the day and pick up early in the morning before work. When finished I will scan and post my final solution for any input you may have. Thanks again!

Just completed the problem and I feel it's correct. Can you see any errors or something that looks out of order?

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You wrote f(0)=a and f(0)=a2, which is wrong. Also, the problem asked for x=5 and x=10 in the table.

vela said:
You wrote f(0)=a and f(0)=a2, which is wrong. Also, the problem asked for x=5 and x=10 in the table.

Do you mean I wrote them wrong in my hand work on the left?? I only did that because the example he gave us had that for his problem, if that's what you meant. And also, thanks for catching the five and ten! I based this excel template off my other two completed problems so I missed that thank you!

What I mean is that f(0) isn't equal to a or a2. "f(0)" means you plug 0 into the function f(x), which was defined in this problem to be eax. It doesn't mean "plug 0 into the last function I happened to be thinking of."

I (and the grader) can infer from what you wrote down that what you were probably thinking was correct, but what you wrote down doesn't accurately reflect what you were thinking. It's similar to having spelling and grammatical errors in an essay. Readers may be able to figure out what you meant, but they shouldn't have to put in the extra effort.

vela said:
What I mean is that f(0) isn't equal to a or a2. "f(0)" means you plug 0 into the function f(x), which was defined in this problem to be eax. It doesn't mean "plug 0 into the last function I happened to be thinking of."

I (and the grader) can infer from what you wrote down that what you were probably thinking was correct, but what you wrote down doesn't accurately reflect what you were thinking. It's similar to having spelling and grammatical errors in an essay. Readers may be able to figure out what you meant, but they shouldn't have to put in the extra effort.

Ahhh! I see what you mean now.. Again, a silly error I made, in my similar previous problems I finished I had it correct, but didn't in this problem. I corrected it to read f(0), f ' (0), and f '' (0). Thank you!

And thanks to you all who have helped throughout the thread! You all really helped me to not only understand the problem, but in helping me understand this problem I also was able correctly eliminate three of the problems on my summer assignment for this course!

## 1. What is a Taylor Series and how is it used?

A Taylor Series is a representation of a mathematical function as an infinite sum of terms. It is used to approximate a function at a specific point by using a polynomial of increasing degree. This allows for easier calculations and analysis of functions.

## 2. How is a Taylor Series problem solved?

To solve a Taylor Series problem, one must first determine the function that the series is representing. Then, the coefficients of the polynomial can be calculated using derivatives of the function evaluated at the specific point. Finally, the polynomial can be written out and simplified.

## 3. What are the applications of Taylor Series?

Taylor Series have many applications in mathematics, physics, and engineering. They are commonly used to approximate functions, solve differential equations, and analyze the behavior of systems. They are also used in numerical methods for solving problems in computer science and data analysis.

## 4. How accurate is a Taylor Series approximation?

The accuracy of a Taylor Series approximation depends on the degree of the polynomial used. The higher the degree, the more terms in the series and the more accurate the approximation will be. However, using too many terms can lead to round-off errors and decrease the accuracy.

## 5. Are there any limitations to using Taylor Series?

While Taylor Series are a powerful tool in mathematics, there are limitations to their use. They can only approximate functions near a specific point and may not accurately represent the behavior of a function over its entire domain. Additionally, they may not converge for all functions and may not provide a good approximation for functions with singularities or discontinuities.

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