Taylor Series Help: Determine Real Number

[xtrubambinoxpr]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[Dick]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

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?

 

[xtrubambinoxpr]

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?

Well because i need to have a x in the equation for it to maintain a taylor polynomial wouldn't 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

 

[Dick]

Well because i need to have a x in the equation for it to maintain a taylor polynomial wouldn't 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

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?

 

[xtrubambinoxpr]

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?

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


is this correct??

 

[Dick]

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


is this correct??

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.

 

[xtrubambinoxpr]

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.

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?

 

[Dick]

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?

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!+...

 

[xtrubambinoxpr]

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!+...

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?

 

[Dick]

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?

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.

 

[xtrubambinoxpr]

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.

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.

 

[Dick]

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.

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.

 

[xtrubambinoxpr]

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.

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 :/

 

[Dick]

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 :/

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?

 

[xtrubambinoxpr]

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?

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

 

[Dick]

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

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?

 

[xtrubambinoxpr]

The taylor series for e^x starts at k=0. What's the first term? Does putting x=(2-e)/2 change that?

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

 

[Dick]

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

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.

 

[xtrubambinoxpr]

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.

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

 

[Dick]

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

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.

 

[xtrubambinoxpr]

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.

I just realized k can't be = 0 ok my term because it make denominator 0 -___- which is why it started at k=1
But I get the started stopping thing in relation to that concrete example, but would that mean since it differs by one I have to add one to e^(2-e)/2 since there is a diff of one? Or am I just overthinking it now?
Thanks so much for your patience btw! Just want to pass this class.

 

[Dick]

I just realized k can't be = 0 ok my term because it make denominator 0 -___- which is why it started at k=1
But I get the started stopping thing in relation to that concrete example, but would that mean since it differs by one I have to add one to e^(2-e)/2 since there is a diff of one? Or am I just overthinking it now?
Thanks so much for your patience btw! Just want to pass this class.

You are close, but 0!=1. So the first term of the taylor expansion of e^x with x=((e-2)/2) is not undefined. It 1. e^((e-2)/2) is larger than the series you are given. By 1. I don't think you add 1.

 

[xtrubambinoxpr]

You are close, but 0!=1. So the first term of the taylor expansion of e^x with x=((e-2)/2) is not undefined. It 1. e^((e-2)/2) is larger than the series you are given. By 1. I don't think you add 1.

Ok I get that now, but just figure it would be my series plus 1 *or* my e^x x=(2-e)/2 expression minus 1 to make it equal to my series. Just my thoughts.

On a completely different note the series then converges to my expression right? Because it is a number when I plug it into a calc.

 

[Dick]

Ok I get that now, but just figure it would be my series plus 1 *or* my e^x x=(2-e)/2 expression minus 1 to make it equal to my series. Just my thoughts.

On a completely different note the series then converges to my expression right? Because it is a number when I plug it into a calc.

Right. The sum of the series converges to (2-e)/2-1. 1+x+x^2/2!+x^3/3!+... converges for any value of x and it's value is e^x.