# Taylor Series Help

1. Nov 17, 2013

### xtrubambinoxpr

1. The problem statement, all variables and given/known data

Determine the real number to which the series $\sum^{∞}_{k=1}$ $(2-e)^k/2^k(k!)$

2. Relevant equations

I know that e^x = the series of x^k / k

3. The attempt at a solution

I would assume to sub in 2-e for x, but then that takes away the x.

2. Nov 17, 2013

### Dick

It's almost that easy, but your "I would assume" is wrong for a couple of reasons. But I have no idea what you are thinking. Can you explain why getting rid of the x is bad?

3. Nov 17, 2013

### xtrubambinoxpr

Well because i need to have a x in the equation for it to maintain a taylor polynomial wouldnt i? plus when I plug it in and try to solve, it doesn't come out to any specific numbers. like if it was just 2-e i could go with the flow and get it, but the 2^k is confusing me badly

4. Nov 17, 2013

### Dick

No, you don't need to find the taylor series, you are given it. You want to figure out the value of the series sum. Can you simplify (2-e)^k/2^k using rules of exponents?

5. Nov 17, 2013

### xtrubambinoxpr

$\frac{(2-e)^k}{2^k}*\frac{1}{(k!)}$ = ($\frac{2-e}{2}$)^k * $\frac{1}{k!}$

is this correct??

6. Nov 17, 2013

### Dick

Yes! Now the other issue is that the series you are given starts at k=1. The taylor series for e^x starts at k=0. You have to account for the difference.

7. Nov 17, 2013

### xtrubambinoxpr

So make all the k on the equations k=k-1 to make it equal to the original series at k=0.
That would make perfect sense. But where is the 1/2 coming from in the equation?

8. Nov 17, 2013

### Dick

I'm not sure I get your question, the 1/2^k is what you were given. That's where it's coming form. The original series (with your simplification) you are given is ((2-e)/2)/1!+((2-e)/2)^2/2!+... The taylor series for e^((2-e)/2) is 1+((2-e)/2)/1!+((2-e)/2)^2/2!+...

9. Nov 17, 2013

### xtrubambinoxpr

Omg wow..
So that means (2-e)/2 is x for e^x
Right?! That makes perfect sense!!
But how do I find the sum from that? Using limits? And ratio test?

10. Nov 17, 2013

### Dick

You don't have to do anything else. You have a series for e^((2-e)/2). That's what it sums to. Nothing else needed. But if you write down the series for e^((2-e)/2) that differs from the series you were given in one important respect related to the k=1. Reread my last post.

11. Nov 17, 2013

### xtrubambinoxpr

Gotcha. So the series converges to the real number e^[(2-e)/2]?

But the k=1 vs k=0 part so something just with a change that's minor because they both equal the same *** provided*** that they are set up correctly to the original e^x term. I understand what you mean.

12. Nov 17, 2013

### Dick

No, you don't understand what I mean. The taylor series for e^((2-e)/2) starts with 1. The series you were given doesn't start with 1. Otherwise they are identical. Just saying k=k-1 doesn't fix that. They are still different.

13. Nov 17, 2013

### xtrubambinoxpr

Ok so how do I fix it and what does it converge to? Outta my whole calc class this is the only topic I can't grasp so I'm sorry if I sound like an idiot :/

14. Nov 17, 2013

### Dick

I wouldn't say an idiot, but you are missing something pretty easy. Write out the first few terms of the series you are given. Then write out the first few terms of the taylor series for e^((2-e)/2). Isn't it kind of clear that they aren't the same? One of them is 1 larger than the other. Which one?

15. Nov 17, 2013

### xtrubambinoxpr

What im given: (2-e)/2 + (2-e)^2/8 + (2-e)^3/48
Im not sure about the taylor series that I wrote down.

16. Nov 17, 2013

### Dick

The taylor series for e^x starts at k=0. What's the first term? Isn't it 1? Does putting x=(2-e)/2 change that?

17. Nov 17, 2013

### xtrubambinoxpr

e^x at k=0 is 1
For 2-e / 2 it's something different

18. Nov 17, 2013

### Dick

No, it not different. Your given series is different from the series from e^x where x=(2-e)/2. It differs in only one term because of the k=1. I'm running out of ways to explain this.

19. Nov 17, 2013

### xtrubambinoxpr

It's the same but with different starting places? Is the only way I think of it from what you're telling me

20. Nov 17, 2013

### Dick

Yes! It's the same with different starting places. That means the difference between the two series is the difference between the starting places. (1+2+4)=7. (2+4)=6. The difference is the first term 1. They aren't the same. They differ by 1. Same with your series.