Dachekar Mombrun
2022-07-05

You can still ask an expert for help

asked 2022-01-20

Prove that $\frac{x}{x+1}<\mathrm{ln}(1+x)<x$ for $x>-1,x\ne 0$

asked 2021-11-05

Use the properties of logarithms to expand the logarithmic expression.

$x\mathrm{ln}\sqrt{x-4}$

asked 2022-08-19

Avoiding subtraction for finite difference with log and exp

I want to approximate the derivative of f(x)

Finite difference

${f}^{\prime}(x)\approx \frac{f(x+h)-f(x)}{h}$

I was taught that the error from the subtraction is blown up for small h. This I can verify with MATLAB.

Smartass method

So I though maybe the following would fix the problem:

${f}^{\prime}(x)\approx \frac{\mathrm{log}\frac{\mathrm{exp}(f(x+h))}{\mathrm{exp}(f(x))}}{h}$

However, I get the same relative errors. (Which also start increasing for h smaller than approx. ${\u03f5}_{mach}/2$

Why is this?

I want to approximate the derivative of f(x)

Finite difference

${f}^{\prime}(x)\approx \frac{f(x+h)-f(x)}{h}$

I was taught that the error from the subtraction is blown up for small h. This I can verify with MATLAB.

Smartass method

So I though maybe the following would fix the problem:

${f}^{\prime}(x)\approx \frac{\mathrm{log}\frac{\mathrm{exp}(f(x+h))}{\mathrm{exp}(f(x))}}{h}$

However, I get the same relative errors. (Which also start increasing for h smaller than approx. ${\u03f5}_{mach}/2$

Why is this?

asked 2022-08-26

Find the inverse of the function

Find the inverse of the function $f(x)=-2\cdot {4}^{2(x-3)}-1$

Find the inverse of the function $f(x)=-2\cdot {4}^{2(x-3)}-1$

asked 2021-03-08

Find
$\mathrm{log}5\left(0.0016\right)$

asked 2022-09-04

Rewrite as a single log with a coefficient of one.

$(4\mathrm{ln}(x))-(3\mathrm{ln}(x))+(1/4\mathrm{ln}(x))$

$(4\mathrm{ln}(x))-(3\mathrm{ln}(x))+(1/4\mathrm{ln}(x))$

asked 2022-06-04

$\frac{a(a+b)}{4{a}^{2}+ab+{b}^{2}}+\frac{b(b+c)}{4{b}^{2}+bc+{c}^{2}}+\frac{c(c+a)}{4{c}^{2}+ca+{a}^{2}}\le 1$

I've got stuck at this problem:

Firstly, I've thought this :

${a}^{2}-2ab+{b}^{2}\ge 0$

$4{a}^{2}+ab+{b}^{2}\ge 3(ab+{a}^{2})$

$\frac{a(a+b)}{4{a}^{2}+ab+{b}^{2}}\le \frac{1}{3}$

Similarly we obtain that

$\frac{b(b+c)}{4{b}^{2}+bc+{c}^{2}}\le \frac{1}{3}$

Summing all, we obtain the inequality.

Is this way correct? (I have doubts about my solution because this problem was found in a math magazine which usually has difficult problems(at least for me)). Or is there a another way?

Thanks!

I've got stuck at this problem:

Firstly, I've thought this :

${a}^{2}-2ab+{b}^{2}\ge 0$

$4{a}^{2}+ab+{b}^{2}\ge 3(ab+{a}^{2})$

$\frac{a(a+b)}{4{a}^{2}+ab+{b}^{2}}\le \frac{1}{3}$

Similarly we obtain that

$\frac{b(b+c)}{4{b}^{2}+bc+{c}^{2}}\le \frac{1}{3}$

Summing all, we obtain the inequality.

Is this way correct? (I have doubts about my solution because this problem was found in a math magazine which usually has difficult problems(at least for me)). Or is there a another way?

Thanks!