Closed form of a power series solution to a differential equation
$begingroup$
Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?
real-analysis sequences-and-series power-series
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
$endgroup$
add a comment |
$begingroup$
Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?
real-analysis sequences-and-series power-series
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
$endgroup$
add a comment |
$begingroup$
Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?
real-analysis sequences-and-series power-series
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
$endgroup$
Given the differential equation $$f''(x)-2xf'(x)-2f(x)=2$$ I've managed to find the power series solution $$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}.$$ Now how can I find the closed form of this function? It seems a bit similar to $$sum_{n=0}^{infty}frac{x^n}{n!}$$ which is the exponential function. But with a difference of shifted powers. Substitution didn't seem to get me anywhere. Does it require integrating or differentiating the expression? Also, is there a general strategy for tackling this kind of problems? What should be the first thing to look at?
real-analysis sequences-and-series power-series
real-analysis sequences-and-series power-series
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
asked Jan 31 at 18:38
Sei SakataSei Sakata
10810
10810
add a comment |
add a comment |
2 Answers
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You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$
So we get
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!} = x sum_{m=1}^infty frac{(x^2)^m}{m!} = x (exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
answered Jan 31 at 18:48
flawrflawr
11.5k32446
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add a comment |
$begingroup$
Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
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2 Answers
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2 Answers
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$begingroup$
You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$
So we get
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!} = x sum_{m=1}^infty frac{(x^2)^m}{m!} = x (exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
answered Jan 31 at 18:48
flawrflawr
11.5k32446
$endgroup$
add a comment |
$begingroup$
You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$
So we get
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!} = x sum_{m=1}^infty frac{(x^2)^m}{m!} = x (exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
answered Jan 31 at 18:48
flawrflawr
11.5k32446
$endgroup$
add a comment |
$begingroup$
You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$
So we get
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!} = x sum_{m=1}^infty frac{(x^2)^m}{m!} = x (exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
answered Jan 31 at 18:48
flawrflawr
11.5k32446
$endgroup$
You can try to "factor" out terms until they start to look similar enough, in your case you have a suspicion what it could look like so we can try to give your powerseries this form:
$$frac{x^{2(n+1)+1}}{(n+1)!} overset{m=n+1}{=} frac{x^{2m+1}}{m!} = x frac{(x^2)^m}{m!}$$
So we get
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!} = x sum_{m=1}^infty frac{(x^2)^m}{m!} = x (exp(x^2) - 1)$$
Note that this does require practice and you need to be able to recognize the power series. At the same time you have to keep in mind that most power series do not admit a closed form, at least the ones you will encounter outside of textbooks and problem sets.
EDIT: Many similar techniques arise in the study of generating functions. There is a great free textbook by Herbert S. Wilf on this topic: https://www.math.upenn.edu/~wilf/DownldGF.html
answered Jan 31 at 18:48
flawrflawr
11.5k32446
edited Jan 31 at 18:59
answered Jan 31 at 18:48
flawrflawr
11.5k32446
answered Jan 31 at 18:48
flawrflawr
11.5k32446
answered Jan 31 at 18:48
flawrflawr
11.5k32446
11.5k32446
add a comment |
add a comment |
$begingroup$
Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
$endgroup$
add a comment |
$begingroup$
Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
$endgroup$
add a comment |
$begingroup$
Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
$endgroup$
Yes, you're right — it is similar enough, so we can manipulate it into the series for the exponential functions, roughly speaking. Here's one way to look at it:
$$sum_{n=0}^infty frac{x^{2(n+1)+1}}{(n+1)!}=xcdotsum_{n=0}^infty frac{(x^2)^{n+1}}{(n+1)!}=xcdotsum_{n=1}^infty frac{(x^2)^n}{n!}=xcdotleft(sum_{n=0}^infty frac{(x^2)^n}{n!}-1right)=cdots$$
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
answered Jan 31 at 18:50
zipirovichzipirovich
11.3k11631
11.3k11631
add a comment |
add a comment |
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