Change $0.75$ to a fraction and reduce to lowest terms

Deborah Wyatt
2022-07-23
Answered

Change $0.75$ to a fraction and reduce to lowest terms

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asked 2022-04-02

Question involving approximation, taylor series and proving

Question: Consider the approximation

$\mathrm{ln}\left(2\right)\approx 2(\frac{1}{3}+\frac{1}{3\times {3}^{3}}+\frac{1}{5\times {3}^{5}})$

Prove that the error in this approximation is less than

$\frac{1}{7\times {2}^{2}\times {3}^{5}}$

Attempt: It looks like the expression comes from the taylor series expansion so:$\mathrm{ln}(1+x)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\dots \text{}\text{for}\text{}\text{}-1x1$

$\mathrm{ln}(1-x)=-x-\frac{{x}^{2}}{2}-\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\dots$

$\therefore \mathrm{ln}\left(\frac{1+x}{1-x}\right)=2(x+\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}+\frac{{x}^{7}}{7})$

$\text{Now let}\text{}x=\frac{1}{3}$

$\therefore \mathrm{ln}\left(2\right)=2(\frac{1}{3}+\frac{1}{3\times {3}^{3}}+\frac{1}{5\times {3}^{5}}+\frac{1}{7\times {3}^{7}})$

So we have to prove that:

$2(\frac{1}{7\times {3}^{7}}+\frac{1}{9\times {3}^{9}}+\frac{1}{11\times {3}^{11}}\cdots )<\frac{1}{7\times {2}^{2}\times {3}^{5}}$

Question: Consider the approximation

Prove that the error in this approximation is less than

Attempt: It looks like the expression comes from the taylor series expansion so:

So we have to prove that:

asked 2021-09-30

If the probability of a certain event not happening is $\frac{49}{82}$ , what is the probability of the event happening?

Write all answers as simplified fractions.

Write all answers as simplified fractions.

asked 2022-05-22

Inequality - GM, AM, HM and SM means

I've got stuck at this problem :

Prove that for any $a>0$ and any $b>0$ the following inequality is true:

$3(\frac{{a}^{3}}{{b}^{3}}+\frac{{b}^{3}}{{a}^{3}})\ge \frac{a}{b}+\frac{b}{a}+4$

The first thing that I've thought was the AM-GM inequality (the extended version - heard that is also known as The power mean inequality):

$HM\le GM\le AM\le SM$

where $HM$, $GM$, $AM$, and $SM$ refer to the harmonic, geometric, arithmetic, and square mean, respectively. CBS(Cauchy - Buniakowsky - Schwartz) also come to my mind, but I think it isn't helpful in this case.

I would be greatful for some hints.

Thanks!

I've got stuck at this problem :

Prove that for any $a>0$ and any $b>0$ the following inequality is true:

$3(\frac{{a}^{3}}{{b}^{3}}+\frac{{b}^{3}}{{a}^{3}})\ge \frac{a}{b}+\frac{b}{a}+4$

The first thing that I've thought was the AM-GM inequality (the extended version - heard that is also known as The power mean inequality):

$HM\le GM\le AM\le SM$

where $HM$, $GM$, $AM$, and $SM$ refer to the harmonic, geometric, arithmetic, and square mean, respectively. CBS(Cauchy - Buniakowsky - Schwartz) also come to my mind, but I think it isn't helpful in this case.

I would be greatful for some hints.

Thanks!

asked 2022-04-06

Compute 4*15*14-9*7 using the correct order of operations.

asked 2022-09-26

Fractions with sum 1

Using all numbers 0 to 9 only once, form two fractions whose sum is 1.

I have tried every possible combination but with no luck. I believe the fractions must be xx/xx + xxx/xxx but I am not sure. Any ideas are most welcome. I even tried getting all different ways to select 3 out of 10 digits, then omitting all the primes and trying to make fractions with simple values 1/3, 1/4, 2/5 etc using only different digits but again with no luck! By the way, two years ago I was given a similar one, with three fractions and without the digit 0, for which I found a solution 7/68+9/12+5/34 but now I am stuck!!

Using all numbers 0 to 9 only once, form two fractions whose sum is 1.

I have tried every possible combination but with no luck. I believe the fractions must be xx/xx + xxx/xxx but I am not sure. Any ideas are most welcome. I even tried getting all different ways to select 3 out of 10 digits, then omitting all the primes and trying to make fractions with simple values 1/3, 1/4, 2/5 etc using only different digits but again with no luck! By the way, two years ago I was given a similar one, with three fractions and without the digit 0, for which I found a solution 7/68+9/12+5/34 but now I am stuck!!

asked 2022-05-20

Express the following product as a single fraction:

$(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots (1+\frac{1}{{3}^{(2n)}})$

I'm having difficulty with this problem:

What i did was:

I rewrote the $1$ as $\frac{3}{3}$

here is what i rewrote the whole product as:

$(\frac{3}{3}+\frac{1}{3})(\frac{3}{3}+\frac{1}{{3}^{2}})(\frac{3}{3}+\frac{1}{{3}^{4}})\cdots (\frac{3}{3}+\frac{1}{{3}^{{2}^{n}}})$

but how would i proceed after this?

$(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots (1+\frac{1}{{3}^{(2n)}})$

I'm having difficulty with this problem:

What i did was:

I rewrote the $1$ as $\frac{3}{3}$

here is what i rewrote the whole product as:

$(\frac{3}{3}+\frac{1}{3})(\frac{3}{3}+\frac{1}{{3}^{2}})(\frac{3}{3}+\frac{1}{{3}^{4}})\cdots (\frac{3}{3}+\frac{1}{{3}^{{2}^{n}}})$

but how would i proceed after this?

asked 2022-09-17

Solving without squaring

If $\mathrm{sec}\theta -\mathrm{csc}\theta ={\displaystyle \frac{4}{3}}$, then prove that $\theta ={\displaystyle \frac{1}{2}}\mathrm{arcsin}{\displaystyle \frac{3}{4}}$

I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.

The closest thing I reached is $\mathrm{sin}\theta ={\displaystyle \frac{3(1-2{\mathrm{sin}}^{2}(\theta /2))}{8{\mathrm{sin}}^{2}(\theta /2)-1}}$

I also obtained $\mathrm{sin}2\theta ={\displaystyle \frac{-3}{2}}{\displaystyle \frac{(1-\mathrm{tan}\theta /2{)}^{2}}{1+{\mathrm{tan}}^{2}(\theta /2)}}$

If $\mathrm{sec}\theta -\mathrm{csc}\theta ={\displaystyle \frac{4}{3}}$, then prove that $\theta ={\displaystyle \frac{1}{2}}\mathrm{arcsin}{\displaystyle \frac{3}{4}}$

I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.

The closest thing I reached is $\mathrm{sin}\theta ={\displaystyle \frac{3(1-2{\mathrm{sin}}^{2}(\theta /2))}{8{\mathrm{sin}}^{2}(\theta /2)-1}}$

I also obtained $\mathrm{sin}2\theta ={\displaystyle \frac{-3}{2}}{\displaystyle \frac{(1-\mathrm{tan}\theta /2{)}^{2}}{1+{\mathrm{tan}}^{2}(\theta /2)}}$