Taylors Remainder Confusion

1. Sep 20, 2010

jegues

1. The problem statement, all variables and given/known data
Use Taylor's remainder formula to show that the Taylor series for f(x) is about the point indicated converges to f(x) for all x.

$$f(x) = e^{5x}$$ about $$x=0$$

2. Relevant equations

3. The attempt at a solution

Since,

$$f^{n}(x) = 5^{n}e^{5x}$$,

Taylor's remainder formula for $$e^{5x}$$ and c = 0 gives

$$e^{5x} = 1 + 5x + \frac{5^{2}}{2!}x^{2} + \frac{5^{3}}{3!}x^{3} + ... + \frac{5^{n}}{n!}x^{n} + R_{n}$$

Where $$R_{n} = \frac{5^{n+1}e^{5z_{n}}}{(n+1)!}x^{n+1}$$

There are two cases we must consider:

Case 1: If x < 0 then,

****NOTE: The following line is the line I am confused about,****

$$x < z_{n} < 0,$$ and $$|R_{n}| < 5^{n+1}\frac{|x|^{n+1}}{(n+1)!}$$

How do they go about making this conclusion? Do they work from the this in the following manner?

$$x < z_{n} < 0$$

$$5x < 5z_{n} < 0$$

$$e^{5x} < e^{5z_{n}} < e^{0}$$

$$e^{5x}\frac{|x|^{n+1}}{(n+1)!} < e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} < e^{0}\frac{|x|^{n+1}}{(n+1)!}$$

And then take the limit as n goes to infinity on the outside and use squeeze theorem to prove the rest?

Where are they getting the second part of this line:

Also, they are using squeeze theorem around $$lim_{n \rightarrow \infty} |R_{n}|$$, how come they are only showing one function around $$|R_{n}|$$. (i.e. $$5^{n+1}\frac{|x|^{n+1}}{(n+1)!}$$) Where is the 0 term?

Once I understand this I will continue and finish off the problem, this is the only line that confuses me.

Last edited: Sep 20, 2010
2. Sep 20, 2010

Dick

It's because
$$0 \leq e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq e^{0}\frac{|x|^{n+1}}{(n+1)!}$$
The maximum value of e^(5*z_n) for z_n in the interval [x,0] is at z_n=0. Where e^(5*z_n)=e^0=1. That's the largest value R_n could possibly have.

3. Sep 20, 2010

jegues

Where did they get the less than or equal to signs? I don't understand where they're coming from.

Also, how do you know that,

$$e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}$$ is indeed larger or equal to 0?

Is it because the minimum value for e^(5*z_n) for z_n in the interval [x,0] is at z_n = x. Where e^(5*z_n) = e^(5x) which is always less than zero because x < 0.

Last edited: Sep 20, 2010
4. Sep 20, 2010

Dick

I put the less than or equal signs in. That part isn't very important. The point is just that (as you said) z_n is between x and 0. So 0<=exp(5*z_n)<=1. So |R_n| is bounded below by zero and above by the expression replacing exp(5*z_n) with 1. And yes, again. The value of the upper limit goes to 0 as n->infinity.

5. Sep 20, 2010

Dick

You want a bound for |R_n|. An absolute value is always >=0. And z_n is in [x,0]. And exp(z_n) is never negative. Are you trying to make this confusing?

6. Sep 20, 2010

jegues

No I'm not trying to make this confusing, I'm trying to understand it. Didn't you get things wrong when you were learning?

My concern is that I've been doing problems and I always have trouble proving that,

$$lim_{n \rightarrow \infty} R_{n} = 0$$

In order to do this I need to make use of the squeeze theorem, and in order for me to use squeeze theorem I have to be able to assign the correct boundaries around $$R_{n}$$ for a given range of x. (i.e. x < 0)

I'm trying to understand this so I can consistently show that either $$lim_{n \rightarrow \infty} R_{n} = 0$$ or not.

7. Sep 20, 2010

Dick

Sure, I got things wrong when learning. But this seems to be getting more confusing. If you can show |R_n|->0, then you have automatically shown R_n->0. You don't need a 'correct lower bound' for |R_n|. 0 will work just fine. Worry about the upper bound. Show that goes to 0.

8. Sep 21, 2010

jegues

Okay I'm going to give this another shot.

Case 1: x < 0

So, $$0 > z_{n} > x$$

Since,

$$|R_{n}| = 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}$$

We can see that in the range $$0 > z_{n} > x$$ the largest value we can obtain for $$e^{5z_{n}}$$ will be obtained when we set $$z_{n} = 0$$.

We can see that,

$$0 \leq |R_{n}| \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!}$$

$$0 \leq 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!}$$

Dividing all terms by $$5^{n+1}$$ ***NOTE: This step I'm not too sure about. I need to get the $$5^{n+1}$$ terms out so when I take the limit of,

$$5^{n+1} \frac{|x|^{n+1}}{(n+1)!}$$ as $$n \rightarrow \infty$$

I will get 0. With the $$5^{n+1}$$ term there I'll get $$\infty \cdot 0$$

EDIT: I am able to simply state that,

$$lim_{n \rightarrow \infty}5^{n+1} \frac{|x|^{n+1}}{(n+1)!} = 0$$ ???

How is this justified? (This is what the given solution states)

Plowing on...

$$0 \leq e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} \leq \frac{|x|^{n+1}}{(n+1)!}$$

Applying squeeze theorem,

EDIT: Now that I think about it, since I removed the, $$5^{n+1}$$ term it isn't really, $$|R_{n}|$$ anymore because that term's not there, right? So how do we deal with this?

$$lim_{n \rightarrow \infty} 0 \leq lim_{n \rightarrow \infty} |R_{n}| \leq lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} }$$

We can see that,

$$lim_{n \rightarrow \infty} 0 = lim_{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1)!} } = 0$$

Therefore we can conclude that,

$$lim_{n \rightarrow \infty} |R_{n}| = 0$$

So,

$$lim_{n \rightarrow \infty} R_{n} = 0$$ for all x < 0.

How does this look? Again, there's that one annoying $$5^{n+1}$$ I'm not sure how to get rid of in there, but other than that everything seems to make sense. After I figure that part out I'll move on to case 2: x > 0.

Last edited: Sep 21, 2010
9. Sep 21, 2010

Dick

DON'T divide by 5^(n+1). You've got
$$0 \leq |R_{n}| \leq 5^{n+1} \frac{|x|^{n+1}}{(n+1)!}$$
NOW do the squeeze by showing lim n->infinity |5*x|^(n+1)/(n+1)! is zero. If you can show the limit of |x|^(n+1)/(n+1)! zero you can surely show the limit of |5*x|^(n+1)/(n+1)! is zero.

10. Sep 21, 2010

jegues

So my task it to show that,

$$lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0$$

Well we can do the following,

Let $$y = 5x$$.

Now surely it's trivial to see that,

$$lim_{n \rightarrow \infty}\frac{|y|^{n+1}}{(n+1)!} = 0$$

In fact, now that I think about it, it was trivial to see that simply,

$$lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0$$

Is true, no?

I want to make sure what I'm doing is justified and I'm not just assuming the limit as n goes to infinity is 0.

Is this correct, do I need to show anything more? (As far as this limit is concerned)

11. Sep 21, 2010

Dick

Unless you've already proved it in the course, it probably wouldn't hurt to say why you think lim n->infinity |y|^(n+1)/(n+1)!=0 for all y. You seem pretty confident. And it's certainly true. Why do you think so?

12. Sep 21, 2010

jegues

We did prove it in class. Our professor said we can simply use it as a tool at our disposale without justification.

So I should be okay stating,

$$lim_{n \rightarrow \infty}\frac{|5x|^{n+1}}{(n+1)!} = 0$$

for all x < 0.

If that's okay I'm going to move on to case 2: x > 0 in my next post.

13. Sep 21, 2010

Dick

Why wouldn't it be ok? If it's true for all y, then it's true for y=5x. Good idea to move on.

14. Sep 21, 2010

jegues

Case 2: x > 0

So,

$$0 < z_{n} < x$$

Since,

$$|R_{n}| = 5^{n+1}e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!}$$

We can see that in the range $$0 < z_{n} < x$$ the largest value we can obtain for $$e^{5z_{n}}$$ is when $$z_{n} = x$$.

So we can see that,

$$0 \leq |R_{n}| \leq 5^{n+1} e^{5x} \frac{|x|^{n+1}}{(n+1)!}$$

Applying squeeze theorem,

$$lim_{n \rightarrow \infty} 0 \leq lim_{n \rightarrow \infty} |R_{n}| \leq e^{5x} lim_{n \rightarrow \infty} \frac{|5x|^{n+1}}{(n+1)!} }$$

Since,

$$lim_{n \rightarrow \infty} 0 = e^{5x} lim_{n \rightarrow \infty} \frac{|5x|^{n+1}}{(n+1)!} } = 0$$

Therefore,

$$lim_{n \rightarrow \infty} |R_{n}| = 0$$

and it follows that,

$$lim_{n \rightarrow \infty} R_{n} = 0$$ $$\forall x$$

Since we've examined both cases.

How's this look?