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Why ignoring higher order terms in Taylor's series won't cause a problem?
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Please don't mark this as a duplicate question as I have read similar questions on SE but couldn't find my answer.
I have been studying numerical methods and an important part of the introduction to numerical methods is Taylor's series especially when studying numerical methods applied to D.Es(also PDEs). I have come to understand that Taylor's series expansion of a function is an expansion around a pivot point and if this pivot point is taken to be some '$a$' in the domain of the function and we have to find the value at some '$x$' close to 'a' then the expansion is given as:
$$f(x)=f(a)+(x-a)f'(a)+(x-a)^2f''(a)/2!+(x-a)^3f'''(a)/3!.....$$
where $|x-a|$ can be said to be the step size.
If the pivot point is given as any '$x$' in the domain of function $f(x)$ then the approximation of the function near to the given pivot point is given by:
$$f(x) or f(x+h)=f(x)+hf'(x)+h^2f''(x)/2!+h^3f'''(x)/3!.....$$
where '$h$' is the step size.
What I haven't been able to understand is how can we truncate off after a particular number of terms in the expansion...especially when the step size i.e. $|x-a|$ or $h$ is greater than $1$. I know that in the expansion of certain function we get alternating positive and negative terms...which potentially balances out...but let's consider that no negative is present in the expansion. Also, I have read that the value of the factorial increases with the order of the term and $1/factorial$ grows smaller and smaller but its growth cannot be compared to how the exponential terms grow (as here in this question the exponential is taken for the $step-size>1$).
I am studying the numerical methods to get a better grip of CFD codes. The explanations in CFD books and elsewhere the truncation-error is given in terms of the big-O notion and is always given as some exponent of the step-size value i.e if we truncate after $2nd$ order terms then the order of error is given as $O(h^3)$ (where the exponent 3 is called the order of error). Now if $h>1$ shouldn't the order of error be infinity as $h^3$ won't be the greatest term among the truncated terms any more(when h>1), but the subsequent order terms i.e. $h^4$ or $h^5$...and so on would be greater. Is this because when we use the concept of taylor's series in CFD we are assuming that the next point would be with in the radius of convergece of the polynomial after we truncate the series?
taylor-expansion
asked Aug 2 at 11:17
GRANZER
1849
add a comment |Â
up vote
1
down vote
favorite
Please don't mark this as a duplicate question as I have read similar questions on SE but couldn't find my answer.
I have been studying numerical methods and an important part of the introduction to numerical methods is Taylor's series especially when studying numerical methods applied to D.Es(also PDEs). I have come to understand that Taylor's series expansion of a function is an expansion around a pivot point and if this pivot point is taken to be some '$a$' in the domain of the function and we have to find the value at some '$x$' close to 'a' then the expansion is given as:
$$f(x)=f(a)+(x-a)f'(a)+(x-a)^2f''(a)/2!+(x-a)^3f'''(a)/3!.....$$
where $|x-a|$ can be said to be the step size.
If the pivot point is given as any '$x$' in the domain of function $f(x)$ then the approximation of the function near to the given pivot point is given by:
$$f(x) or f(x+h)=f(x)+hf'(x)+h^2f''(x)/2!+h^3f'''(x)/3!.....$$
where '$h$' is the step size.
What I haven't been able to understand is how can we truncate off after a particular number of terms in the expansion...especially when the step size i.e. $|x-a|$ or $h$ is greater than $1$. I know that in the expansion of certain function we get alternating positive and negative terms...which potentially balances out...but let's consider that no negative is present in the expansion. Also, I have read that the value of the factorial increases with the order of the term and $1/factorial$ grows smaller and smaller but its growth cannot be compared to how the exponential terms grow (as here in this question the exponential is taken for the $step-size>1$).
I am studying the numerical methods to get a better grip of CFD codes. The explanations in CFD books and elsewhere the truncation-error is given in terms of the big-O notion and is always given as some exponent of the step-size value i.e if we truncate after $2nd$ order terms then the order of error is given as $O(h^3)$ (where the exponent 3 is called the order of error). Now if $h>1$ shouldn't the order of error be infinity as $h^3$ won't be the greatest term among the truncated terms any more(when h>1), but the subsequent order terms i.e. $h^4$ or $h^5$...and so on would be greater. Is this because when we use the concept of taylor's series in CFD we are assuming that the next point would be with in the radius of convergece of the polynomial after we truncate the series?
taylor-expansion
asked Aug 2 at 11:17
GRANZER
1849
1
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
2
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Please don't mark this as a duplicate question as I have read similar questions on SE but couldn't find my answer.
I have been studying numerical methods and an important part of the introduction to numerical methods is Taylor's series especially when studying numerical methods applied to D.Es(also PDEs). I have come to understand that Taylor's series expansion of a function is an expansion around a pivot point and if this pivot point is taken to be some '$a$' in the domain of the function and we have to find the value at some '$x$' close to 'a' then the expansion is given as:
$$f(x)=f(a)+(x-a)f'(a)+(x-a)^2f''(a)/2!+(x-a)^3f'''(a)/3!.....$$
where $|x-a|$ can be said to be the step size.
If the pivot point is given as any '$x$' in the domain of function $f(x)$ then the approximation of the function near to the given pivot point is given by:
$$f(x) or f(x+h)=f(x)+hf'(x)+h^2f''(x)/2!+h^3f'''(x)/3!.....$$
where '$h$' is the step size.
What I haven't been able to understand is how can we truncate off after a particular number of terms in the expansion...especially when the step size i.e. $|x-a|$ or $h$ is greater than $1$. I know that in the expansion of certain function we get alternating positive and negative terms...which potentially balances out...but let's consider that no negative is present in the expansion. Also, I have read that the value of the factorial increases with the order of the term and $1/factorial$ grows smaller and smaller but its growth cannot be compared to how the exponential terms grow (as here in this question the exponential is taken for the $step-size>1$).
I am studying the numerical methods to get a better grip of CFD codes. The explanations in CFD books and elsewhere the truncation-error is given in terms of the big-O notion and is always given as some exponent of the step-size value i.e if we truncate after $2nd$ order terms then the order of error is given as $O(h^3)$ (where the exponent 3 is called the order of error). Now if $h>1$ shouldn't the order of error be infinity as $h^3$ won't be the greatest term among the truncated terms any more(when h>1), but the subsequent order terms i.e. $h^4$ or $h^5$...and so on would be greater. Is this because when we use the concept of taylor's series in CFD we are assuming that the next point would be with in the radius of convergece of the polynomial after we truncate the series?
taylor-expansion
asked Aug 2 at 11:17
GRANZER
1849
Please don't mark this as a duplicate question as I have read similar questions on SE but couldn't find my answer.
I have been studying numerical methods and an important part of the introduction to numerical methods is Taylor's series especially when studying numerical methods applied to D.Es(also PDEs). I have come to understand that Taylor's series expansion of a function is an expansion around a pivot point and if this pivot point is taken to be some '$a$' in the domain of the function and we have to find the value at some '$x$' close to 'a' then the expansion is given as:
$$f(x)=f(a)+(x-a)f'(a)+(x-a)^2f''(a)/2!+(x-a)^3f'''(a)/3!.....$$
where $|x-a|$ can be said to be the step size.
If the pivot point is given as any '$x$' in the domain of function $f(x)$ then the approximation of the function near to the given pivot point is given by:
$$f(x) or f(x+h)=f(x)+hf'(x)+h^2f''(x)/2!+h^3f'''(x)/3!.....$$
where '$h$' is the step size.
What I haven't been able to understand is how can we truncate off after a particular number of terms in the expansion...especially when the step size i.e. $|x-a|$ or $h$ is greater than $1$. I know that in the expansion of certain function we get alternating positive and negative terms...which potentially balances out...but let's consider that no negative is present in the expansion. Also, I have read that the value of the factorial increases with the order of the term and $1/factorial$ grows smaller and smaller but its growth cannot be compared to how the exponential terms grow (as here in this question the exponential is taken for the $step-size>1$).
I am studying the numerical methods to get a better grip of CFD codes. The explanations in CFD books and elsewhere the truncation-error is given in terms of the big-O notion and is always given as some exponent of the step-size value i.e if we truncate after $2nd$ order terms then the order of error is given as $O(h^3)$ (where the exponent 3 is called the order of error). Now if $h>1$ shouldn't the order of error be infinity as $h^3$ won't be the greatest term among the truncated terms any more(when h>1), but the subsequent order terms i.e. $h^4$ or $h^5$...and so on would be greater. Is this because when we use the concept of taylor's series in CFD we are assuming that the next point would be with in the radius of convergece of the polynomial after we truncate the series?
taylor-expansion
asked Aug 2 at 11:17
GRANZER
1849
edited Aug 2 at 11:24
asked Aug 2 at 11:17
GRANZER
1849
asked Aug 2 at 11:17
GRANZER
1849
asked Aug 2 at 11:17
GRANZER
1849
1849
1
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
2
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45
add a comment |Â
1
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
2
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45
1
1
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
2
2
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45
add a comment |Â
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1
Well $n!$ can be compared to an exponential $e^n$. The $n!$ growth like $(n/e)^nsqrt2pi n$. This is, much faster. The factors $f^(n)(x)$ are also relevant, though. The convergence of the Taylor series of a function to the values of the function need to be studied for each case. The remainder of the series should tend to zero, and should tend to zero fast enough to justify deleting it for a specific value of $n$ and a specific range of $h$.
– spiralstotheleft
Aug 2 at 11:24
2
When you study Taylor series, a good amount of time is spent on the "interval of convergence." If $h$ is less than the radius of convergence, then the series converges with error $O(h^n).$ If not, then all your worries come true.
– B. Goddard
Aug 2 at 11:25
It's a classical exercise to prove that $frach^nn!xrightarrow[ntoinfty]0$ for any complex $h$. So the only possible cause of trouble would come from the factor $f^(n)(a)$ going larger and larger.
– Nicolas FRANCOIS
Aug 2 at 11:28
The truncated taylor-expansion will only be a good approximation near the point $a$. Even if the Taylor-series converges everywhere to the function , for very large $x$, you will need many terms to get a good approximation. Some series however converge very fast, including $e^x,sin(x),cos(x)$ , but even then, values far enough from $a$ will cause trouble if we truncate.
– Peter
Aug 2 at 11:45