When to stop to be sure of the pattern of signs in a Taylor expansion?
It is common practice to derive just "enough" terms of a series expansion to illustrate the pattern. For example, upon seeing some terms one can be certain how the negative sign appears alternatingly or how the coefficient relates to the power.
$$log(1 + x) = x - frac{x^2}2 + frac{ x^3 }3 - frac{ x^4 }4 + cdots \
tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots $$
How does one know when is "enough" when deriving the series?
Some series appears to have the signs go like $++--++--cdots$ (perhaps with zeros interleaved). How does one know the pattern truly is "alternating by two"?
Suppose the series starts with the first four terms of plus signs $++++$, how does one know if it's all plus sign or if it's $++++----++++----cdots$?
Note 1: my question is clearly different from this post with a similar title.
Note 2: I'm not sure how my question can be addressed by the Descartes rule of signs.
calculus power-series taylor-expansion
add a comment |
It is common practice to derive just "enough" terms of a series expansion to illustrate the pattern. For example, upon seeing some terms one can be certain how the negative sign appears alternatingly or how the coefficient relates to the power.
$$log(1 + x) = x - frac{x^2}2 + frac{ x^3 }3 - frac{ x^4 }4 + cdots \
tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots $$
How does one know when is "enough" when deriving the series?
Some series appears to have the signs go like $++--++--cdots$ (perhaps with zeros interleaved). How does one know the pattern truly is "alternating by two"?
Suppose the series starts with the first four terms of plus signs $++++$, how does one know if it's all plus sign or if it's $++++----++++----cdots$?
Note 1: my question is clearly different from this post with a similar title.
Note 2: I'm not sure how my question can be addressed by the Descartes rule of signs.
calculus power-series taylor-expansion
3
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
1
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago
add a comment |
It is common practice to derive just "enough" terms of a series expansion to illustrate the pattern. For example, upon seeing some terms one can be certain how the negative sign appears alternatingly or how the coefficient relates to the power.
$$log(1 + x) = x - frac{x^2}2 + frac{ x^3 }3 - frac{ x^4 }4 + cdots \
tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots $$
How does one know when is "enough" when deriving the series?
Some series appears to have the signs go like $++--++--cdots$ (perhaps with zeros interleaved). How does one know the pattern truly is "alternating by two"?
Suppose the series starts with the first four terms of plus signs $++++$, how does one know if it's all plus sign or if it's $++++----++++----cdots$?
Note 1: my question is clearly different from this post with a similar title.
Note 2: I'm not sure how my question can be addressed by the Descartes rule of signs.
calculus power-series taylor-expansion
It is common practice to derive just "enough" terms of a series expansion to illustrate the pattern. For example, upon seeing some terms one can be certain how the negative sign appears alternatingly or how the coefficient relates to the power.
$$log(1 + x) = x - frac{x^2}2 + frac{ x^3 }3 - frac{ x^4 }4 + cdots \
tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots $$
How does one know when is "enough" when deriving the series?
Some series appears to have the signs go like $++--++--cdots$ (perhaps with zeros interleaved). How does one know the pattern truly is "alternating by two"?
Suppose the series starts with the first four terms of plus signs $++++$, how does one know if it's all plus sign or if it's $++++----++++----cdots$?
Note 1: my question is clearly different from this post with a similar title.
Note 2: I'm not sure how my question can be addressed by the Descartes rule of signs.
calculus power-series taylor-expansion
calculus power-series taylor-expansion
edited 3 hours ago
Charlie Mosby
asked 3 hours ago
Charlie MosbyCharlie Mosby
5116
5116
3
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
1
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago
add a comment |
3
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
1
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago
3
3
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
1
1
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago
add a comment |
2 Answers
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To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.
That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement
$$tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots = sum_{k=0}^infty frac{(-1)^{k}}{2k+1} x^{2k+1}$$
means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.
I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of the statement - that after making up however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding
$$x ;;; , ;;; - frac{x^3}3 ;;; , ;;; frac{ x^5 }5 ;;; , ;;; - frac{ x^7 }7 ;;; , ;;; ...$$
that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.
add a comment |
Well, do recall where the Taylor expansion comes from.
$$f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2+cdots$$
Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $sin x$ cycles every 4 derivatives.
For one of your examples, $log(1+x)$, note that the first derivative is $frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.
For the other example, $tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.
You can apply this reasoning to most common Taylor series expansions.
add a comment |
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2 Answers
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2 Answers
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To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.
That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement
$$tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots = sum_{k=0}^infty frac{(-1)^{k}}{2k+1} x^{2k+1}$$
means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.
I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of the statement - that after making up however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding
$$x ;;; , ;;; - frac{x^3}3 ;;; , ;;; frac{ x^5 }5 ;;; , ;;; - frac{ x^7 }7 ;;; , ;;; ...$$
that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.
add a comment |
To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.
That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement
$$tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots = sum_{k=0}^infty frac{(-1)^{k}}{2k+1} x^{2k+1}$$
means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.
I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of the statement - that after making up however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding
$$x ;;; , ;;; - frac{x^3}3 ;;; , ;;; frac{ x^5 }5 ;;; , ;;; - frac{ x^7 }7 ;;; , ;;; ...$$
that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.
add a comment |
To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.
That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement
$$tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots = sum_{k=0}^infty frac{(-1)^{k}}{2k+1} x^{2k+1}$$
means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.
I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of the statement - that after making up however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding
$$x ;;; , ;;; - frac{x^3}3 ;;; , ;;; frac{ x^5 }5 ;;; , ;;; - frac{ x^7 }7 ;;; , ;;; ...$$
that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.
To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.
That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement
$$tan^{-1} x = x - frac{x^3}3 + frac{ x^5 }5 - frac{ x^7 }7 + cdots = sum_{k=0}^infty frac{(-1)^{k}}{2k+1} x^{2k+1}$$
means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.
I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of the statement - that after making up however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding
$$x ;;; , ;;; - frac{x^3}3 ;;; , ;;; frac{ x^5 }5 ;;; , ;;; - frac{ x^7 }7 ;;; , ;;; ...$$
that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.
answered 2 hours ago
Eevee TrainerEevee Trainer
5,2291834
5,2291834
add a comment |
add a comment |
Well, do recall where the Taylor expansion comes from.
$$f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2+cdots$$
Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $sin x$ cycles every 4 derivatives.
For one of your examples, $log(1+x)$, note that the first derivative is $frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.
For the other example, $tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.
You can apply this reasoning to most common Taylor series expansions.
add a comment |
Well, do recall where the Taylor expansion comes from.
$$f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2+cdots$$
Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $sin x$ cycles every 4 derivatives.
For one of your examples, $log(1+x)$, note that the first derivative is $frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.
For the other example, $tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.
You can apply this reasoning to most common Taylor series expansions.
add a comment |
Well, do recall where the Taylor expansion comes from.
$$f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2+cdots$$
Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $sin x$ cycles every 4 derivatives.
For one of your examples, $log(1+x)$, note that the first derivative is $frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.
For the other example, $tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.
You can apply this reasoning to most common Taylor series expansions.
Well, do recall where the Taylor expansion comes from.
$$f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2+cdots$$
Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $sin x$ cycles every 4 derivatives.
For one of your examples, $log(1+x)$, note that the first derivative is $frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.
For the other example, $tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.
You can apply this reasoning to most common Taylor series expansions.
answered 2 hours ago
QuintecQuintec
22625
22625
add a comment |
add a comment |
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3
Instead of "guessing" a pattern by just looking at the series, do the actual computation and see where the pattern is coming from.
– Michael Barz
3 hours ago
1
Special case math.stackexchange.com/q/2924175.
– user514787
2 hours ago
@user514787 Whoa that is a very interesting example. Thank you.
– Charlie Mosby
2 hours ago