When is the Taylor series bounded by a partial sum?

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Cliff notes: Is there a nice, general property of a Taylor series that shows when it is everywhere bounded by its partial sum. E.g., is there a nice/simple way to show that
$$
cos(x) = sum_i=0^infty frac(-4pi^2 x^2)^i(2i)! geq sum_i=0^2k+1frac(-4pi^2 x^2)^i(2i)! qquad (*)
$$
for all integers $k geq 0$ and $x in mathbbR$, or say, that
$$
cos(x)cos(x^2) = 1-x^2/2 - 11x^4/24 + cdots geq 1-x^2/2 - 11x^4/24 qquad (**) ; .
$$
It is often extremely useful not just to approximate a Taylor series by its partial sum, but to bound it. E.g., we very frequently use the inequality
$$e^-x = 1 - x + x^2/2 - x^3/6 + x^4/24 - cdots geq 1-x
; .$$
(I'm using $e^-x$ instead of just $e^x$ to stress the fact that the summands can be negative, since the inequality is completely trivial otherwise.)
We typically prove this from the convexity of $e^-x$, which immediately implies that
$$
x^2/2 - x^3/6 + x^4/24 - cdots
$$
is non-negative. But, for even slightly more complicated cases, like the inequality
$$
cos(x) = 1- x^2/2 + x^4/24 - cdots geq 1-x^2/2 qquad (***)
; ,
$$
this simple technique fails.
In fact, the best proof of $(***)$ that I know observes that the inequality is trivial except in a small interval around zero, in which the fourth derivative of $cos(x)$ (which is just $cos(x)$) is positive. My main gripe with this proof is that it gets quite ugly when we try to generalize it. Proving $(*)$ with this strategy is already a pain, and generalizing to even slightly less pretty functions like that in $(**)$ seems like a nightmare.
Is there a nice way to handle such things?
power-series taylor-expansion
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up vote
2
down vote
favorite
Cliff notes: Is there a nice, general property of a Taylor series that shows when it is everywhere bounded by its partial sum. E.g., is there a nice/simple way to show that
$$
cos(x) = sum_i=0^infty frac(-4pi^2 x^2)^i(2i)! geq sum_i=0^2k+1frac(-4pi^2 x^2)^i(2i)! qquad (*)
$$
for all integers $k geq 0$ and $x in mathbbR$, or say, that
$$
cos(x)cos(x^2) = 1-x^2/2 - 11x^4/24 + cdots geq 1-x^2/2 - 11x^4/24 qquad (**) ; .
$$
It is often extremely useful not just to approximate a Taylor series by its partial sum, but to bound it. E.g., we very frequently use the inequality
$$e^-x = 1 - x + x^2/2 - x^3/6 + x^4/24 - cdots geq 1-x
; .$$
(I'm using $e^-x$ instead of just $e^x$ to stress the fact that the summands can be negative, since the inequality is completely trivial otherwise.)
We typically prove this from the convexity of $e^-x$, which immediately implies that
$$
x^2/2 - x^3/6 + x^4/24 - cdots
$$
is non-negative. But, for even slightly more complicated cases, like the inequality
$$
cos(x) = 1- x^2/2 + x^4/24 - cdots geq 1-x^2/2 qquad (***)
; ,
$$
this simple technique fails.
In fact, the best proof of $(***)$ that I know observes that the inequality is trivial except in a small interval around zero, in which the fourth derivative of $cos(x)$ (which is just $cos(x)$) is positive. My main gripe with this proof is that it gets quite ugly when we try to generalize it. Proving $(*)$ with this strategy is already a pain, and generalizing to even slightly less pretty functions like that in $(**)$ seems like a nightmare.
Is there a nice way to handle such things?
power-series taylor-expansion
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Cliff notes: Is there a nice, general property of a Taylor series that shows when it is everywhere bounded by its partial sum. E.g., is there a nice/simple way to show that
$$
cos(x) = sum_i=0^infty frac(-4pi^2 x^2)^i(2i)! geq sum_i=0^2k+1frac(-4pi^2 x^2)^i(2i)! qquad (*)
$$
for all integers $k geq 0$ and $x in mathbbR$, or say, that
$$
cos(x)cos(x^2) = 1-x^2/2 - 11x^4/24 + cdots geq 1-x^2/2 - 11x^4/24 qquad (**) ; .
$$
It is often extremely useful not just to approximate a Taylor series by its partial sum, but to bound it. E.g., we very frequently use the inequality
$$e^-x = 1 - x + x^2/2 - x^3/6 + x^4/24 - cdots geq 1-x
; .$$
(I'm using $e^-x$ instead of just $e^x$ to stress the fact that the summands can be negative, since the inequality is completely trivial otherwise.)
We typically prove this from the convexity of $e^-x$, which immediately implies that
$$
x^2/2 - x^3/6 + x^4/24 - cdots
$$
is non-negative. But, for even slightly more complicated cases, like the inequality
$$
cos(x) = 1- x^2/2 + x^4/24 - cdots geq 1-x^2/2 qquad (***)
; ,
$$
this simple technique fails.
In fact, the best proof of $(***)$ that I know observes that the inequality is trivial except in a small interval around zero, in which the fourth derivative of $cos(x)$ (which is just $cos(x)$) is positive. My main gripe with this proof is that it gets quite ugly when we try to generalize it. Proving $(*)$ with this strategy is already a pain, and generalizing to even slightly less pretty functions like that in $(**)$ seems like a nightmare.
Is there a nice way to handle such things?
power-series taylor-expansion
Cliff notes: Is there a nice, general property of a Taylor series that shows when it is everywhere bounded by its partial sum. E.g., is there a nice/simple way to show that
$$
cos(x) = sum_i=0^infty frac(-4pi^2 x^2)^i(2i)! geq sum_i=0^2k+1frac(-4pi^2 x^2)^i(2i)! qquad (*)
$$
for all integers $k geq 0$ and $x in mathbbR$, or say, that
$$
cos(x)cos(x^2) = 1-x^2/2 - 11x^4/24 + cdots geq 1-x^2/2 - 11x^4/24 qquad (**) ; .
$$
It is often extremely useful not just to approximate a Taylor series by its partial sum, but to bound it. E.g., we very frequently use the inequality
$$e^-x = 1 - x + x^2/2 - x^3/6 + x^4/24 - cdots geq 1-x
; .$$
(I'm using $e^-x$ instead of just $e^x$ to stress the fact that the summands can be negative, since the inequality is completely trivial otherwise.)
We typically prove this from the convexity of $e^-x$, which immediately implies that
$$
x^2/2 - x^3/6 + x^4/24 - cdots
$$
is non-negative. But, for even slightly more complicated cases, like the inequality
$$
cos(x) = 1- x^2/2 + x^4/24 - cdots geq 1-x^2/2 qquad (***)
; ,
$$
this simple technique fails.
In fact, the best proof of $(***)$ that I know observes that the inequality is trivial except in a small interval around zero, in which the fourth derivative of $cos(x)$ (which is just $cos(x)$) is positive. My main gripe with this proof is that it gets quite ugly when we try to generalize it. Proving $(*)$ with this strategy is already a pain, and generalizing to even slightly less pretty functions like that in $(**)$ seems like a nightmare.
Is there a nice way to handle such things?
power-series taylor-expansion
power-series taylor-expansion
asked Feb 28 at 19:46
Noah Stephens-Davidowitz
1112
1112
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2 Answers
2
active
oldest
votes
up vote
3
down vote
Is there a nice way to handle such things?
Not for all sums. But for sums coming from objects like sine and cosine expansions, you can use the bound coming from alternating series. More formally, if you are interested in
$$ sum_n geq 1 (-1)^n a(n)$$
and the $a(n)$ satisfy
- $a(n) geq 0$
- $a(n) geq a(n+1)$
- $a(n) to 0$
then you have that
$$ a(1) - a(2) + cdots - a(N) leq sum_n geq 1 (-1)^n a(n) leq a(1) - a(2) + cdots - a(N) + a(N+1)$$
for any $N$. This is commonly proved while proving the alternating series test for convergence.
In practice, it is so much easier to understand the convergence of alternating series that sometimes is is better to figure out how to understand how to decompose an object into alternating series.
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
 |Â
show 2 more comments
up vote
0
down vote
I guess the original question that I asked was far too general and vague, but with Siyao Guo and Alex Lombardi we at least found a cute proof that, e.g.,
$$1 - x + fracx^22 - fracx^36 + cdots - fracx^2k-1(2k-1)! leq e^-x leq 1-x + fracx^22 + cdots + fracx^2k(2k)!
; $$
for integer $k geq 1$ and $x geq 0$. We can similarly prove bounds for $cos(x)$ and $sin(x)$. I doubt this is original, but since I couldn't find it by Googling, I figured it was worth posting.
Equivalently, we want to prove that the tail of the series satisfies
$$f_k(x) := sum_i=0^infty (-1)^i fracx^k+i(k+i)! geq 0
; $$
for $x geq 0$.
The proof is by induction on $k$. The base case is $k = 0$, which is trivial since $f_k(x) = e^-x$. For $k geq 1$, we simply note that the derivative satisfies $fracrm drm d x f_k(x) = f_k-1(x)$. By induction, the derivative is positive for $x geq 0$, and the result follows from the fact that $f_k(0) = 0$.
The proof for $cos(x)$ and $sin(x)$ is essentially the same and works by just replacing $f_k$ by
$$g_k(x) := sum_i=0^infty (-1)^i fracx^k + 2i(k+2i)!
; ,$$
so that $g_0(x) = cos(x)$, $g_1(x) = sin(x)$, and $fracrm drm d x g_k(x) = g_k-1(x)$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Is there a nice way to handle such things?
Not for all sums. But for sums coming from objects like sine and cosine expansions, you can use the bound coming from alternating series. More formally, if you are interested in
$$ sum_n geq 1 (-1)^n a(n)$$
and the $a(n)$ satisfy
- $a(n) geq 0$
- $a(n) geq a(n+1)$
- $a(n) to 0$
then you have that
$$ a(1) - a(2) + cdots - a(N) leq sum_n geq 1 (-1)^n a(n) leq a(1) - a(2) + cdots - a(N) + a(N+1)$$
for any $N$. This is commonly proved while proving the alternating series test for convergence.
In practice, it is so much easier to understand the convergence of alternating series that sometimes is is better to figure out how to understand how to decompose an object into alternating series.
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
 |Â
show 2 more comments
up vote
3
down vote
Is there a nice way to handle such things?
Not for all sums. But for sums coming from objects like sine and cosine expansions, you can use the bound coming from alternating series. More formally, if you are interested in
$$ sum_n geq 1 (-1)^n a(n)$$
and the $a(n)$ satisfy
- $a(n) geq 0$
- $a(n) geq a(n+1)$
- $a(n) to 0$
then you have that
$$ a(1) - a(2) + cdots - a(N) leq sum_n geq 1 (-1)^n a(n) leq a(1) - a(2) + cdots - a(N) + a(N+1)$$
for any $N$. This is commonly proved while proving the alternating series test for convergence.
In practice, it is so much easier to understand the convergence of alternating series that sometimes is is better to figure out how to understand how to decompose an object into alternating series.
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
 |Â
show 2 more comments
up vote
3
down vote
up vote
3
down vote
Is there a nice way to handle such things?
Not for all sums. But for sums coming from objects like sine and cosine expansions, you can use the bound coming from alternating series. More formally, if you are interested in
$$ sum_n geq 1 (-1)^n a(n)$$
and the $a(n)$ satisfy
- $a(n) geq 0$
- $a(n) geq a(n+1)$
- $a(n) to 0$
then you have that
$$ a(1) - a(2) + cdots - a(N) leq sum_n geq 1 (-1)^n a(n) leq a(1) - a(2) + cdots - a(N) + a(N+1)$$
for any $N$. This is commonly proved while proving the alternating series test for convergence.
In practice, it is so much easier to understand the convergence of alternating series that sometimes is is better to figure out how to understand how to decompose an object into alternating series.
Is there a nice way to handle such things?
Not for all sums. But for sums coming from objects like sine and cosine expansions, you can use the bound coming from alternating series. More formally, if you are interested in
$$ sum_n geq 1 (-1)^n a(n)$$
and the $a(n)$ satisfy
- $a(n) geq 0$
- $a(n) geq a(n+1)$
- $a(n) to 0$
then you have that
$$ a(1) - a(2) + cdots - a(N) leq sum_n geq 1 (-1)^n a(n) leq a(1) - a(2) + cdots - a(N) + a(N+1)$$
for any $N$. This is commonly proved while proving the alternating series test for convergence.
In practice, it is so much easier to understand the convergence of alternating series that sometimes is is better to figure out how to understand how to decompose an object into alternating series.
answered Feb 28 at 20:18
davidlowrydudaâ¦
72.7k6113244
72.7k6113244
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
 |Â
show 2 more comments
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
Note that the monotonicity is essential here, merely alternating in signs and eventually converging to zero is not enough, even though our intuition might think it should be ("the next term has the opposite sign and surely the next partial sum is closer to the infinite sum than this one is")
â Ian
Feb 28 at 20:22
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I see. That's a very nice answer, but I'm not sure it's powerful enough to give a simple proof of even $cos(x) geq 1-x^2$, or am I missing something? It seems that in order to apply it directly, we would need $x^/24, x^6/720, ldots$ to be a non-increasing sequence, which is false for large $x$. --- --- We can use your technique to show that the identity holds for small $x$ and handle large $x$ separately. I think this is probably better than the convexity approach, but not dramatically so.
â Noah Stephens-Davidowitz
Feb 28 at 20:27
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
I guess this technique also has some trouble handling series with rather complicated sign patterns such as the $cos(x) cos(x^2)$ example, whose series has a rather complex sign pattern. But, again, maybe I'm missing something, since I've used convexity to prove such statements quite a bit but haven't played much with this approach.
â Noah Stephens-Davidowitz
Feb 28 at 20:33
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@NoahStephens-Davidowitz $cos(x) geq 1-x^2/2$ for large $x$ can't be deduced solely by inspecting the first few terms of the Maclaurin series. After all, $1-x^2/2+x^4/24-x^6/120$ does not have this property (nor do any of the other truncations of the series with an even number of terms). Instead the breaking point of the inequality is pushed further and further away from zero as you take larger even numbers of terms.
â Ian
Feb 28 at 20:38
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
@Ian, I'm not sure what it means "to deduce something from the Maclaurin series." I'm happy to use properties of the cosine, such as the fact that it is bounded or that its second derivative is its negative. (I guess something like this is necessary in some sense.) I think we've discussed two proofs of this inequality that are pretty simple. I'd just like a proof that generalizes to inequalities like $(*)$ or $(**)$ without too much hassle, since they come up quite a bit. (Maybe simple proofs don't exist, but that would be disappointing for such natural questions.)
â Noah Stephens-Davidowitz
Feb 28 at 20:46
 |Â
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up vote
0
down vote
I guess the original question that I asked was far too general and vague, but with Siyao Guo and Alex Lombardi we at least found a cute proof that, e.g.,
$$1 - x + fracx^22 - fracx^36 + cdots - fracx^2k-1(2k-1)! leq e^-x leq 1-x + fracx^22 + cdots + fracx^2k(2k)!
; $$
for integer $k geq 1$ and $x geq 0$. We can similarly prove bounds for $cos(x)$ and $sin(x)$. I doubt this is original, but since I couldn't find it by Googling, I figured it was worth posting.
Equivalently, we want to prove that the tail of the series satisfies
$$f_k(x) := sum_i=0^infty (-1)^i fracx^k+i(k+i)! geq 0
; $$
for $x geq 0$.
The proof is by induction on $k$. The base case is $k = 0$, which is trivial since $f_k(x) = e^-x$. For $k geq 1$, we simply note that the derivative satisfies $fracrm drm d x f_k(x) = f_k-1(x)$. By induction, the derivative is positive for $x geq 0$, and the result follows from the fact that $f_k(0) = 0$.
The proof for $cos(x)$ and $sin(x)$ is essentially the same and works by just replacing $f_k$ by
$$g_k(x) := sum_i=0^infty (-1)^i fracx^k + 2i(k+2i)!
; ,$$
so that $g_0(x) = cos(x)$, $g_1(x) = sin(x)$, and $fracrm drm d x g_k(x) = g_k-1(x)$.
add a comment |Â
up vote
0
down vote
I guess the original question that I asked was far too general and vague, but with Siyao Guo and Alex Lombardi we at least found a cute proof that, e.g.,
$$1 - x + fracx^22 - fracx^36 + cdots - fracx^2k-1(2k-1)! leq e^-x leq 1-x + fracx^22 + cdots + fracx^2k(2k)!
; $$
for integer $k geq 1$ and $x geq 0$. We can similarly prove bounds for $cos(x)$ and $sin(x)$. I doubt this is original, but since I couldn't find it by Googling, I figured it was worth posting.
Equivalently, we want to prove that the tail of the series satisfies
$$f_k(x) := sum_i=0^infty (-1)^i fracx^k+i(k+i)! geq 0
; $$
for $x geq 0$.
The proof is by induction on $k$. The base case is $k = 0$, which is trivial since $f_k(x) = e^-x$. For $k geq 1$, we simply note that the derivative satisfies $fracrm drm d x f_k(x) = f_k-1(x)$. By induction, the derivative is positive for $x geq 0$, and the result follows from the fact that $f_k(0) = 0$.
The proof for $cos(x)$ and $sin(x)$ is essentially the same and works by just replacing $f_k$ by
$$g_k(x) := sum_i=0^infty (-1)^i fracx^k + 2i(k+2i)!
; ,$$
so that $g_0(x) = cos(x)$, $g_1(x) = sin(x)$, and $fracrm drm d x g_k(x) = g_k-1(x)$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I guess the original question that I asked was far too general and vague, but with Siyao Guo and Alex Lombardi we at least found a cute proof that, e.g.,
$$1 - x + fracx^22 - fracx^36 + cdots - fracx^2k-1(2k-1)! leq e^-x leq 1-x + fracx^22 + cdots + fracx^2k(2k)!
; $$
for integer $k geq 1$ and $x geq 0$. We can similarly prove bounds for $cos(x)$ and $sin(x)$. I doubt this is original, but since I couldn't find it by Googling, I figured it was worth posting.
Equivalently, we want to prove that the tail of the series satisfies
$$f_k(x) := sum_i=0^infty (-1)^i fracx^k+i(k+i)! geq 0
; $$
for $x geq 0$.
The proof is by induction on $k$. The base case is $k = 0$, which is trivial since $f_k(x) = e^-x$. For $k geq 1$, we simply note that the derivative satisfies $fracrm drm d x f_k(x) = f_k-1(x)$. By induction, the derivative is positive for $x geq 0$, and the result follows from the fact that $f_k(0) = 0$.
The proof for $cos(x)$ and $sin(x)$ is essentially the same and works by just replacing $f_k$ by
$$g_k(x) := sum_i=0^infty (-1)^i fracx^k + 2i(k+2i)!
; ,$$
so that $g_0(x) = cos(x)$, $g_1(x) = sin(x)$, and $fracrm drm d x g_k(x) = g_k-1(x)$.
I guess the original question that I asked was far too general and vague, but with Siyao Guo and Alex Lombardi we at least found a cute proof that, e.g.,
$$1 - x + fracx^22 - fracx^36 + cdots - fracx^2k-1(2k-1)! leq e^-x leq 1-x + fracx^22 + cdots + fracx^2k(2k)!
; $$
for integer $k geq 1$ and $x geq 0$. We can similarly prove bounds for $cos(x)$ and $sin(x)$. I doubt this is original, but since I couldn't find it by Googling, I figured it was worth posting.
Equivalently, we want to prove that the tail of the series satisfies
$$f_k(x) := sum_i=0^infty (-1)^i fracx^k+i(k+i)! geq 0
; $$
for $x geq 0$.
The proof is by induction on $k$. The base case is $k = 0$, which is trivial since $f_k(x) = e^-x$. For $k geq 1$, we simply note that the derivative satisfies $fracrm drm d x f_k(x) = f_k-1(x)$. By induction, the derivative is positive for $x geq 0$, and the result follows from the fact that $f_k(0) = 0$.
The proof for $cos(x)$ and $sin(x)$ is essentially the same and works by just replacing $f_k$ by
$$g_k(x) := sum_i=0^infty (-1)^i fracx^k + 2i(k+2i)!
; ,$$
so that $g_0(x) = cos(x)$, $g_1(x) = sin(x)$, and $fracrm drm d x g_k(x) = g_k-1(x)$.
answered Aug 30 at 22:57
Noah Stephens-Davidowitz
1112
1112
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