The discussion revolves around determining the sum of the series \(\sum^{∞}_{k=1} \frac{(2-e)^k}{2^k k!}\), which is related to the Taylor series expansion of the exponential function.
Participants are actively engaging with the problem, clarifying the relationship between the given series and the Taylor series for \(e^x\). Some guidance has been offered regarding the simplification of terms and the importance of recognizing differences in series starting points. There is an ongoing exploration of how to reconcile the two series.
Participants note that the original series starts at \(k=1\), which affects the comparison with the Taylor series that starts at \(k=0\). There are also concerns about the implications of substituting values and the potential for division by zero.
xtrubambinoxpr said: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 said: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 said: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 said: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 said:\frac{(2-e)^k}{2^k}*\frac{1}{(k!)} = (\frac{2-e}{2})^k * \frac{1}{k!}
is this correct??
Dick said: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 said: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 said: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 said: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 said: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 said: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 said: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 said: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 said: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 said: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.
e^x at k=0 is 1Dick said:The taylor series for e^x starts at k=0. What's the first term? Does putting x=(2-e)/2 change that?
xtrubambinoxpr said:e^x at k=0 is 1
For 2-e / 2 it's something different
Dick said: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 said:It's the same but with different starting places? Is the only way I think of it from what you're telling me
Dick said: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 said: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 said: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 said: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.