Need help, URGENT maths question: (1 Viewer)

InteGrand · May 29, 2016

leehuan said:
With the p-test btw, just in terms of the actual boundaries

When we were taught it, he demonstrated it with $\int_1^\infty$

Does it have to be 1, or can it be any value of a provided the function remains continuous? $\int_a^\infty$

Yeah, can be any real number a > 0. This is a pretty classic fact about improper integrals in general (not just p-test), and is analogous with a result from infinite series (essentially you can start the series anywhere and it doesn't affect the convergence status; only the "long-run behaviour" determines this).

So if you wanted, you could restate the p-test as:

$\noindent Let $a>0$ be a fixed real number. Let $p\in \mathbb{R}$. Then $\int _a ^\infty \frac{1}{x^{p}}\text{ d}x$ converges iff $p > 1$ (diverges to $\infty$ otherwise).$

$\noindent Usually we don't state it like this because we just use that fact about being able to start anywhere as a separate fact if needed because it's so well-known. I guess we state it from $1$ because it's nicer like that.$

Paradoxica · May 29, 2016

leehuan said:
With the p-test btw, just in terms of the actual boundaries

When we were taught it, he demonstrated it with $\int_1^\infty$

Does it have to be 1, or can it be any value of a provided the function remains continuous? $\int_a^\infty$

Technically, integrating across/onto a singularity can also create non-convergence.

But yes, as long as the interval [a,∞) is continuous, it is the same thing.

1008 · May 29, 2016

Thanks everyone for your replies!

I have this other question. For the first part, I get V'(x) = u(x)(a-x). If this is right, how would you do the second part?

InteGrand · May 29, 2016

1008 said:
Thanks everyone for your replies!

I have this other question. For the first part, I get V'(x) = u(x)(a-x). If this is right, how would you do the second part?

$\noindent a) Note that $\frac{\mathrm{d}}{\mathrm{d} x}\left(\int _0 ^x U(t) \text{ d}t\right)= U(x)$ from the FTC. Now,$

$\begin{align*}V^\prime (x) &= -U(x) + \left(a-x\right)u(x) + U(x) \\ &= \left(a-x\right)u(x). \end{align*}$

(So your first answer is right.)

$\noindent b) Note that $V(0)=aU(0) + 0 = aU(0)$. Also note $V(a) = \int _0 ^a U(x) \text{ d}x$. Since $V^\prime (x) = \left(a-x\right)u(x) $, integrating both sides from $0$ to $a$ yields$

$\begin{align*}V(a) - V(0) &= \int_{0} ^{a} \left(a-x\right)u(x)\text{ d}x \\ \Rightarrow \int _0 ^a U(x) \text{ d}x - aU(0) &= \int _{0} ^{a} \left(a-x\right)u(x)\text{ d}x \\ \Rightarrow \int _0 ^a U(x) \text{ d}x &= aU(0)+\int _{0} ^{a} \left(a-x\right)u(x)\text{ d}x. \end{align*}$

1008 · May 30, 2016

InteGrand said:
$\noindent a) Note that $\frac{\mathrm{d}}{\mathrm{d} x}\left(\int _0 ^x U(t) \text{ d}t\right)= U(x)$ from the FTC. Now,$

$\begin{align*}V^\prime (x) &= -U(x) + \left(a-x\right)u(x) + U(x) \\ &= \left(a-x\right)u(x). \end{align*}$

(So your first answer is right.)

$\noindent b) Note that $V(0)=aU(0) + 0 = aU(0)$. Also note $V(a) = \int _0 ^a U(x) \text{ d}x$. Since $V^\prime (x) = \left(a-x\right)u(x) $, integrating both sides from $0$ to $a$ yields$

$\begin{align*}V(a) - V(0) &= \int_{0} ^{a} \left(a-x\right)u(x)\text{ d}x \\ \Rightarrow \int _0 ^a U(x) \text{ d}x - aU(0) &= \int _{0} ^{a} \left(a-x\right)u(x)\text{ d}x \\ \Rightarrow \int _0 ^a U(x) \text{ d}x &= aU(0)+\int _{0} ^{a} \left(a-x\right)u(x)\text{ d}x. \end{align*}$

Thanks

$\\\text{Suppose }f\text{ has a continuous 2nd order derivative }f''.\text{ Use integration by parts to show that }\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\int_{0}^{1}f''(x)\frac{1}{2}x(1-x)dx\\\\\text{Use the Mean Value theorem for integrals to deduce that there is a real number }c \text{, with }\\ 0<c<1,\text{ such that}\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\frac{1}{12}f''(c)$
For the above question, I've done the first part (using IBP on the RHS to obtain the LHS). But how would you do the MVT part?

Plus how would you do this?:
$\\\text{Use the transformation }t=x^{-1}\text{ to show that }\Gamma (s) = \int_{0}^{\infty }dt \text{ is convergent for }s>0. \text{ [This is the }\textbf{Gamma function }\text{and is one of the most important functions in mathematics. It is easy to show that }\\\Gamma (s+1) = s\Gamma (s) \text{ for }s>0\text{ so that }\Gamma (n+1) = n!]$

InteGrand · May 30, 2016

1008 said:
Thanks

$\\\text{Suppose }f\text{ has a continuous 2nd order derivative }f''.\text{ Use integration by parts to show that }\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\int_{0}^{1}f''(x)\frac{1}{2}x(1-x)dx\\\\\text{Use the Mean Value theorem for integrals to deduce that there is a real number }c \text{, with }\\ 0<c<1,\text{ such that}\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\frac{1}{12}f''(c)$
For the above question, I've done the first part (using IBP on the RHS to obtain the LHS). But how would you do the MVT part?

Plus how would you do this?:
$\\\text{Use the transformation }t=x^{-1}\text{ to show that }\Gamma (s) = \int_{0}^{\infty }dt \text{ is convergent for }s>0. \text{ [This is the }\textbf{Gamma function }\text{and is one of the most important functions in mathematics. It is easy to show that }\\\Gamma (s+1) = s\Gamma (s) \text{ for }s>0\text{ so that }\Gamma (n+1) = n!]$

For the Gamma function one:

$\noindent Let $I = \int _{0}^{\infty} t^{s-1} e^{-t}\text{ d}t$, where $s>0$ is fixed. Sub. $t = u^{-1}$, and $\mathrm{d}t = -u^{-2}\text{ d}u$. Updating the limits, this gives$

$\begin{align*}I &= \int _0 ^\infty u^{1-s} e^{-\frac{1}{u}}\times \frac{1}{u^2}\text{ d}u \\ &= \int _0 ^\infty u^{-\left(s+1\right)} e^{-u^{-1}} \text{ d}u.\end{align*}$

$The lower limit is of no worry here essentially because exponentials dominate polynomials, so the integrand just goes to 0 as $u\to 0^{+}$. Now we can use a comparison test. Since the integrand here is always positive and less than $u^{-(s+1)}$ (because $0<e^a<1$ for all $a<0$), and the improper integral of this from $1$ to $\infty$ converges (as $s>0$), our integral in question converges (recalling our comment that the lower limit here is irrelevant, so if you want you can just start it at $1$).$

In fact it converges for real s if and only if s > 0, due to the 'iff' nature of the p-test.

InteGrand · May 30, 2016

1008 said:
Thanks

$\\\text{Suppose }f\text{ has a continuous 2nd order derivative }f''.\text{ Use integration by parts to show that }\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\int_{0}^{1}f''(x)\frac{1}{2}x(1-x)dx\\\\\text{Use the Mean Value theorem for integrals to deduce that there is a real number }c \text{, with }\\ 0<c<1,\text{ such that}\\\int_{0}^{1}f(x) dx-\frac{1}{2}(f(0)+f(1)) = -\frac{1}{12}f''(c)$
For the above question, I've done the first part (using IBP on the RHS to obtain the LHS). But how would you do the MVT part?

Plus how would you do this?:
$\\\text{Use the transformation }t=x^{-1}\text{ to show that }\Gamma (s) = \int_{0}^{\infty }dt \text{ is convergent for }s>0. \text{ [This is the }\textbf{Gamma function }\text{and is one of the most important functions in mathematics. It is easy to show that }\\\Gamma (s+1) = s\Gamma (s) \text{ for }s>0\text{ so that }\Gamma (n+1) = n!]$

$\noindent For the MVT for integrals one, recall that this theorem states that if $F$ is continuous on $[a,b]$ and $G$ is an integrable function that doesn't change sign on this interval, then there is a $c \in (a,b)$ with $\int _a ^ b F(x) G(x)\text{ d}x = F(c) \int _a ^b G(x)\text{ d}x$. So apply this theorem to the RHS of the previous part, noting the quadratic term there doesn't change sign on the interval and we are told $f''$ is continuous, so we can pull out the $f''$ writing the integral as $f''(c)\times \text{ remaining integral }$, for some $c \in (0,1)$.$

1008 · May 30, 2016

InteGrand said:
$\noindent For the MVT for integrals one, recall that this theorem states that if $F$ is continuous on $[a,b]$ and $G$ is an integrable function that doesn't change sign on this interval, then there is a $c \in (a,b)$ with $\int _a ^ b F(x) G(x)\text{ d}x = F(c) \int _a ^b G(x)\text{ d}x$. So apply this theorem to the RHS of the previous part, noting the quadratic term there doesn't change sign on the interval and we are told $f''$ is continuous, so we can pull out the $f''$ writing the integral as $f''(c)\times \text{ remaining integral }$, for some $c \in (0,1)$.$

Thanks. There is another question similar to the MVT one above. I'll try to do attempt that one as well as I didn't get it before.

How would you do this one?

$\\\text{(The Error Estimate for the Midpoint Rule) If }x_1-x_0=h \text{ and }m_1\text{ is the midpoint of }[x_0,x_1]\text{ use the result }\\f(x) = f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(c)(x-x_0)^2\\\text{to show that, if } |f''(x)|\leq K\text{ for all }x,\\|f(x)-f(m_1)-f'(m_1)(x-m_1)|\leq \frac{K}{2}(x-m_1)^2\\\text{Show that }\\\left|\int_{x_0}^{x_1}f(x)dx-f(m_1)h\right|=\left|\int_{x_0}^{x_1}(f(x)-f(m_1)-f'(m_1)(x-m_1))dx\right|\leq\frac{K}{24}h^3$

Meanwhile for this question, I get parts (a) and (b), but could someone please verify if I am correct. Also, I don't get part (c) and (d).

$\\\text{a) Estimate the error in the Simpson's Rule approximation }\\e^{1/k}-e^{-1/k}=\int_{a}^{b}e^tdt \approx \frac{1}{3k}\left(e^{-1/k}+4+e^{1/k}\right) \\\\\text{I get part (a). This is a simple application of the formula }\\\left|\int_{a}^{b}f(x)dx-S_{n}\leq \frac{L(b-a)^5}{2880n^4}\right|\\\text{where n=1 and }\left|f^{(4)}(x)\right|\leq L\forall x \in [a,b]$

$\text{Answer to (a) }\frac{e^{1/k}}{90k^5}$

$\\\text{b) Estimate the error in the approximation }\\(3k-1)e^{2/k}-4e^{1/k}-(3k+1) \approx 0\\\text{I get this part as well. Rearranging (a) with the error in mind,}\\Error = \frac{e^{2/k}}{30k^4}$

$\\\text{c) Estimate the error in the approximation}\\e^{1/k}\approx \frac{2+\sqrt{3+9k^2}}{3k-1}$

$\\\text{d) Estimate the error in the approximation}\\e\approx \left[\frac{2+\sqrt{3+9k^2}}{3k-1}\right]^k.\\\text{Compare this estimate when }k=10\text{ with the estimate }e\approx \left(1+\frac{1}{k}\right)^k\text{ when k=10}$

leehuan · May 30, 2016

For that first question

Substitute x0=m1 without hesitation

Then just put absolute value brackets around it. Replace |f"(c)| with K and swap out the equal sign for an inequality.

To prove the harder part, show that the integral of f(m1) is just equal to h.f(m1) which is trivially true
And then prove that the integral of f'(m1)(x-m1) equals to 0

i.e. prove MHS = LHS

To prove MHS \le RHS use the fact that |\int f| \le \int |f|

1008 · May 30, 2016

leehuan said:
For that first question

Substitute x0=m1 without hesitation

Then just put absolute value brackets around it. Replace |f"(c)| with K and swap out the equal sign for an inequality.

To prove the harder part, show that the integral of f(m1) is just equal to h.f(m1) which is trivially true
And then prove that the integral of f'(m1)(x-m1) equals to 0

i.e. prove MHS = LHS

To prove MHS \le RHS use the fact that |\int f| \le \int |f|

Thanks. But I don't get the logic behind this...

And why does the integral of f'(m1)(x-m1) =0?

InteGrand · May 30, 2016

1008 said:
Thanks. But I don't get the logic behind this...

And why does the integral of f'(m1)(x-m1) =0?

Remember that m₁ is the midpoint of the interval. If we integrate a linear function over an interval and the function takes value 0 at the midpoint, the area above and below the curve on either side is equal, so the integral is 0.

(Alternatively, just integrate it normally and remember m1 is the midpoint of the interval, so (x₁ - m₁)² = (x₀ - m1)².)

And also, the f"(c) is bounded above in magnitude by K, i.e. |f"(c)| ≤ K. This allows us to get the inequality.

1008 · May 30, 2016

InteGrand said:
Remember that m₁ is the midpoint of the interval. If we integrate a linear function over an interval and the function takes value 0 at the midpoint, the area above and below the curve on either side is equal, so the integral is 0.

(Alternatively, just integrate it normally and remember m1 is the midpoint of the interval, so (x₁ - m₁)² = (x₀ - m1)².)

And also, the f"(c) is bounded above in magnitude by K, i.e. |f"(c)| ≤ K. This allows us to get the inequality.

Thanks, and what about the RHS of the inequality, K/24 h^3?

InteGrand · May 30, 2016

1008 said:
Thanks. There is another question similar to the MVT one above. I'll try to do attempt that one as well as I didn't get it before.

How would you do this one?

$\\\text{(The Error Estimate for the Midpoint Rule) If }x_1-x_0=h \text{ and }m_1\text{ is the midpoint of }[x_0,x_1]\text{ use the result }\\f(x) = f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(c)(x-x_0)^2\\\text{to show that, if } |f''(x)|\leq K\text{ for all }x,\\|f(x)-f(m_1)-f'(m_1)(x-m_1)|\leq \frac{K}{2}(x-m_1)^2\\\text{Show that }\\\left|\int_{x_0}^{x_1}f(x)dx-f(m_1)h\right|=\left|\int_{x_0}^{x_1}(f(x)-f(m_1)-f'(m_1)(x-m_1))dx\right|\leq\frac{K}{24}h^3$

Meanwhile for this question, I get parts (a) and (b), but could someone please verify if I am correct. Also, I don't get part (c) and (d).

1008 said:
Thanks, and what about the RHS of the inequality, K/24 h^3?

Remember h is the interval width. Have you tried using the triangle inequality as leehuan suggested? If so, what have you managed to get up to using that?

1008 · May 31, 2016

1008 said:
Thanks. There is another question similar to the MVT one above. I'll try to do attempt that one as well as I didn't get it before.

How would you do this one?

$\\\text{(The Error Estimate for the Midpoint Rule) If }x_1-x_0=h \text{ and }m_1\text{ is the midpoint of }[x_0,x_1]\text{ use the result }\\f(x) = f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(c)(x-x_0)^2\\\text{to show that, if } |f''(x)|\leq K\text{ for all }x,\\|f(x)-f(m_1)-f'(m_1)(x-m_1)|\leq \frac{K}{2}(x-m_1)^2\\\text{Show that }\\\left|\int_{x_0}^{x_1}f(x)dx-f(m_1)h\right|=\left|\int_{x_0}^{x_1}(f(x)-f(m_1)-f'(m_1)(x-m_1))dx\right|\leq\frac{K}{24}h^3$

Got it

$\\\text{We also know }\\\int_{x_0}^{x_1}|u(x)|dx\leq\left|\int_{x_0}^{x_1}u(x)dx\right|\\\therefore\text{The above stated inequality is true}$

For this question, I use the fact that as the curve of u(x) is always less than or equal to the curve of the RHS inequality, the area under the curve of u(x) would also be lesser than that of the RHS of the inequality (from x_0 to x_1 of course). Then I used leehuan's suggestion that |\int f| \le \int |f|, thus solving the inequality. Is this right? Oh yeah, BTW, forgot to change it, please ignore the absolute value signs around by substitution for u(x)

Also, any thoughts on the other question I put up?

InteGrand · May 31, 2016

1008 · May 31, 2016

InteGrand said:
$\noindent Yes, if $f(t) \leq g(t)$ for all $t\in \left[a,b]$, then $\int _a ^b f(t)\text{ d}t \leq \int _a ^b g(t)\text{ d}t$.$

Thanks

Any thoughts on this btw?

1008 said:
Meanwhile for this question, I get parts (a) and (b), but could someone please verify if I am correct. Also, I don't get part (c) and (d).

$\\\text{a) Estimate the error in the Simpson's Rule approximation }\\e^{1/k}-e^{-1/k}=\int_{a}^{b}e^tdt \approx \frac{1}{3k}\left(e^{-1/k}+4+e^{1/k}\right) \\\\\text{I get part (a). This is a simple application of the formula }\\\left|\int_{a}^{b}f(x)dx-S_{n}\leq \frac{L(b-a)^5}{2880n^4}\right|\\\text{where n=1 and }\left|f^{(4)}(x)\right|\leq L\forall x \in [a,b]$

$\text{Answer to (a) }\frac{e^{1/k}}{90k^5}$

$\\\text{b) Estimate the error in the approximation }\\(3k-1)e^{2/k}-4e^{1/k}-(3k+1) \approx 0\\\text{I get this part as well. Rearranging (a) with the error in mind,}\\Error = \frac{e^{2/k}}{30k^4}$

$\\\text{c) Estimate the error in the approximation}\\e^{1/k}\approx \frac{2+\sqrt{3+9k^2}}{3k-1}$

$\\\text{d) Estimate the error in the approximation}\\e\approx \left[\frac{2+\sqrt{3+9k^2}}{3k-1}\right]^k.\\\text{Compare this estimate when }k=10\text{ with the estimate }e\approx \left(1+\frac{1}{k}\right)^k\text{ when k=10}$

1008 · May 31, 2016

In continuation to my previous post, for this question:

I got the first two parts. However, when I did (c), this is what I get:

This has got all the terms that have been asked in (c), so I know I've expanded correctly, but, for some reason, there are two extra terms in the answer to (c) (refer question). Could someone please help?

1008 · May 31, 2016

When going from my second to the third line, I just substituted in the stuff from part (a) wherever I could, like dr/dx = cos (theta) etc. (excuse my d/dx's in this reply, they're meant to be partial)

1008 · May 31, 2016

IG any thoughts on these q's?

1008 said:
Meanwhile for this question, I get parts (a) and (b), but could someone please verify if I am correct. Also, I don't get part (c) and (d).

$\\\text{a) Estimate the error in the Simpson's Rule approximation }\\e^{1/k}-e^{-1/k}=\int_{a}^{b}e^tdt \approx \frac{1}{3k}\left(e^{-1/k}+4+e^{1/k}\right) \\\\\text{I get part (a). This is a simple application of the formula }\\\left|\int_{a}^{b}f(x)dx-S_{n}\leq \frac{L(b-a)^5}{2880n^4}\right|\\\text{where n=1 and }\left|f^{(4)}(x)\right|\leq L\forall x \in [a,b]$

$\text{Answer to (a) }\frac{e^{1/k}}{90k^5}$

$\\\text{b) Estimate the error in the approximation }\\(3k-1)e^{2/k}-4e^{1/k}-(3k+1) \approx 0\\\text{I get this part as well. Rearranging (a) with the error in mind,}\\Error = \frac{e^{2/k}}{30k^4}$

$\\\text{c) Estimate the error in the approximation}\\e^{1/k}\approx \frac{2+\sqrt{3+9k^2}}{3k-1}$

$\\\text{d) Estimate the error in the approximation}\\e\approx \left[\frac{2+\sqrt{3+9k^2}}{3k-1}\right]^k.\\\text{Compare this estimate when }k=10\text{ with the estimate }e\approx \left(1+\frac{1}{k}\right)^k\text{ when k=10}$

1008 said:
In continuation to my previous post, for this question:

I got the first two parts. However, when I did (c), this is what I get:

This has got all the terms that have been asked in (c), so I know I've expanded correctly, but, for some reason, there are two extra terms in the answer to (c) (refer question). Could someone please help?

InteGrand · May 31, 2016

1008 said:
When going from my second to the third line, I just substituted in the stuff from part (a) wherever I could, like dr/dx = cos (theta) etc. (excuse my d/dx's in this reply, they're meant to be partial)

I think you forgot to use product rules or quotient rules etc. where needed. Like for the (∂/∂r)[∂ƒ/∂θ* (sin(θ)/r)] term in that second line, you should use product rule because the second function here (sin(θ)/r) is dependent on r (the sin(θ) part is a constant wrt r, but the r isn't of course, so we do need to use product rule).

Need help, URGENT maths question: (1 Viewer)

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