# a) If displaystyle f{{left({t}right)}}={t}^{m}{quadtext{and}quad} g{{left({t}right)}}={t}^{n}, where m and n are positive integers. show that displaystyle{f}ast{g}={t}^{{{m}+{n}+{1}}}{int_{{0}}^{{1}}}{u}^{m}{left({1}-{u}right)}^{n}{d}{u} b) Use the convolution theorem to show that displaystyle{int_{{0}}^{{1}}}{u}^{m}{left({1}-{u}right)}^{n}{d}{u}=frac{{{m}!{n}!}}{{{left({m}+{n}+{1}right)}!}} c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.

Question
Integrals
a) If $$\displaystyle f{{\left({t}\right)}}={t}^{m}{\quad\text{and}\quad} g{{\left({t}\right)}}={t}^{n}$$, where m and n are positive integers. show that $$\displaystyle{f}\ast{g}={t}^{{{m}+{n}+{1}}}{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}$$
b) Use the convolution theorem to show that
$$\displaystyle{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}=\frac{{{m}!{n}!}}{{{\left({m}+{n}+{1}\right)}!}}$$
c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.

2021-01-31
(a)
By definition,
$$\displaystyle{f}\cdot{g}={\int_{{0}}^{{t}}} f{{\left(\tau\right)}} g{{\left({t}-\tau\right)}}{d}\tau={\int_{{0}}^{{t}}}\tau^{m}{\left({t}-\tau\right)}^{n}{d}\tau$$
Since we must get $$\displaystyle{u}^{m}{\left({1}-{u}\right)}^{n}$$, we can try with the substitution $$\displaystyle\tau={u}{t}$$:
$$\displaystyle{\int_{{0}}^{{t}}}\tau^{m}{\left({t}-\tau\right)}^{n}{d}\tau={\left\lbrace{\left(\tau={u}{t}\ {0}\mapsto{0}\right)},{\left({d}\tau={t}{d}{u}\ {t}\mapsto{1}\right)}\right\rbrace}$$
$$\displaystyle={\int_{{0}}^{{1}}}{\left({u}{t}\right)}^{m}{\left({t}-{u}{t}\right)}^{n}{t}{d}{u}$$
$$\displaystyle={\int_{{0}}^{{1}}}{u}^{m}{t}^{m}{\left({t}{\left({1}-{u}\right)}\right)}^{n}{t}{d}{u}$$
$$\displaystyle={t}^{{{m}+{n}+{1}}}{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}$$
(b) and (c)
Notice that
$$\displaystyle{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}={B}{\left({m},{n}\right)},$$
where B is $$\beta$$ function. Recall that,
$$\displaystyle{B}{\left({m},{n}\right)}=\frac{{\Gamma{\left({m}+{1}\right)}\Gamma{\left({n}+{1}\right)}}}{{\Gamma{\left({m}+{n}+{2}\right)}}},$$
where $$\displaystyle\Gamma{i}{s}\gamma$$ function. When m and n are positive integers,
$$\displaystyle{B}{\left({m},{n}\right)}=\frac{{{m}!{n}!}}{{{\left({m}+{n}+{1}\right)}!}}$$
The finally result for (a), consider the subtitution $$\displaystyle\tau={u}{t}$$.
For (b) and (c), notice that this integral is a $$\beta$$ function

### Relevant Questions

Give the correct answer and solve the given equation
Let $$\displaystyle{p}{\left({x}\right)}={2}+{x}{\quad\text{and}\quad}{q}{\left({x}\right)}={x}$$. Using the inner product $$\langle\ p,\ q\rangle=\int_{-1}^{1}pqdx$$ find all polynomials $$\displaystyle{r}{\left({x}\right)}={a}+{b}{x}\in{P}{1}{\left({R}\right)}{P}$$
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$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
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$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
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State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
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$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
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The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
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What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
You need to prove that question
$$\displaystyle{\int_{{0}}^{{1}}} \sin{{\left(\pi{m}{x}\right)}} \sin{{\left(\pi{n}{x}\right)}}{\left.{d}{x}\right.}={\left\lbrace{\left(\begin{matrix}{0}{m}\ne{n}\\{1}\text{/}{2}{m}={n}\end{matrix}\right)}\right.}$$
Two drugs, Abraxane and Taxol, are both cancer treatments, yet have differing rates at which they leave a patient’s system. Using terminology from pharmacology, Abraxane leaves the system by so-called “first-order elimination”, which means that the concentration decreases at a constant percentage rate for each unit of time that passes. Taxol leaves the system by “zero-order elimination”, which means that the concentration decreases by a constant amount for each unit of time that passes.
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(b) As soon as the infusion of Abraxane is completed, the drug concentration in a patient’s blood is 1000 nanograms per milliliter $$\displaystyle{\left(\frac{{{n}{g}}}{{{m}{l}}}\right)}$$. 24 hours later there is $$\displaystyle{50}\frac{{{n}{g}}}{{{m}{l}}}$$ left in the patient’s system. Use the data to construct an appropriate formula modeling the blood concentration of Abraxane as a function of time after the infusion is completed.
(c) Find the long-term behavior of the function from part (b). Is this behavior meaningful in the context of the model?
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Part B.) Find the vertical displacement D(t) between the raised hands of the two players for the time period after Boris has jumped ($$\displaystyle{t}{>}{t}_{{R}}$$) but before Arabella has landed. (Express youranswer in terms of t,$$\displaystyle{t}_{{R}}$$, g,and H)
Part C.) What advice would you give Arabella To minimize the chance of her shot being blocked?
Scientists are working with a sample of cobalt-56 in their laboratory. They begin with a sample that has 60 mg of cobalt-56, and they measure that after 31 days, the mass of cobalt-56 sample is 45.43 mg. Recall that the differential equation which models exponential decay is $$\frac{dm}{dt}=-km$$ and the solution of that differential equation if $$m(t)=m_0e^{-kt}$$, where $$m_0$$ is the initial mass and k is the relative decay rate.
Scientists are working with a sample of cobalt-56 in their laboratory. They begin with a sample that has 60 mg of cobalt-56, and they measure that after 31 days, the mass of cobalt-56 sample is 45.43 mg. Recall that the differential equation which models exponential decay is $$\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}$$ and the solution of that differential equation if $$\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}$$, where $$\displaystyle{m}_{{0}}$$ is the initial mass and k is the relative decay rate.