leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

leehuan · Jan 31, 2016

turntaker said:
�� calculus shit tHat’s sum cAlcuLUς $#!t�� thats∰some good∰∭integration right∳∰there �� right⨌there⨍find the limit��find the��limit��that’s not showing your work right tᴴᵉᴿe �� show w0rk(chorus: ᵍᴵᵛᴱn ��⁽ˣ⁾��)mMMMMMM��∰∯∰ НO0ОଠOOOOOОଠଠOoooᵒᵒᵒᵒᵒᵒᵒᵒᵒ��∰��CAlcUlUs $#!t

Go home you're drunk.
____________________________________
$\\ $If I memorise $\sinh { x } =\frac { { e }^{ x }-{ e }^{ -x } }{ 2 } \, , \, \cosh { x } =\frac { { e }^{ x }+{ e }^{ -x } }{ 2 } , \\ $the Pythagorean identities and the derivatives, do you reckon that's all I really need until I start doing inverse hyperbolic functions?$

leehuan · Jan 31, 2016

Drsoccerball said:
The first part is easy but the second how can we get it in that form we can't multiply integrals.

My guess

$\\ $Let $u={e}^{-x}$

Otherwise, let me actually try doing the integral

Edit: This is horrifically wrong lol. But it was an uneducated guess.

Drsoccerball · Jan 31, 2016

leehuan said:
Go home you're drunk.
____________________________________
$\\ $If I memorise $\sinh { x } =\frac { { e }^{ x }-{ e }^{ -x } }{ 2 } \, , \, \cosh { x } =\frac { { e }^{ x }+{ e }^{ -x } }{ 2 } , \\ $the Pythagorean identities and the derivatives, do you reckon that's all I really need until I start doing inverse hyperbolic functions?$

$Bro just treat $ \sinh{x} $ as $ \sin{x} $ and $ \cosh{x} $ as $ \cos{x} $ it works.$

leehuan · Jan 31, 2016

Drsoccerball said:
$Bro just treat $ \sinh{x} $ as $ \sin{x} $ and $ \cosh{x} $ as $ \cos{x} $ it works.$

Not really.

sinh^2(x)-cosh^2(x)=1

and d/dx cosh(x) = sinh(x) [NO NEGATIVE]

InteGrand · Jan 31, 2016

Drsoccerball said:
$Bro just treat $ \sinh{x} $ as $ \sin{x} $ and $ \cosh{x} $ as $ \cos{x} $ it works.$

Not exactly, some signs change sometimes. This due to the following facts:

$\noindent $\cosh z = \cos (iz)$ for all $z\in \mathbb{C}$ and $\sinh z= -i\sin (iz)$ for all $z\in \mathbb{C}$. Equivalently, $\cos z= \cosh (iz)$ and $\sin z = -i \sinh (iz)$ for all $z\in \mathbb{C}$.$

$\noindent The integral of $\mathrm{sech}\, x$ becomes dissimilar to the one for $\sec$. One form of it is $\tan^{-1} (\sinh x)$. Also, the derivative of $\mathrm{sech}\, x$ is $-\mathrm{sech}\,x \tan x$ (negative sign in front; the other reciprocal hyperbolic functions, namely $\mathrm{csch}, \coth$, as well as $\tanh$, have analogous integrals and derivatives as their circular trig. counterparts, i.e. no sign changes or anything completely dissimilar.)$

leehuan · Jan 31, 2016

$x={ e }^{ u }\rightarrow dx={ e }^{ u }du\\ \int _{ x=0 }^{ x=1 }{ \frac { dx }{ 1+{ x }^{ 4 } } } \\ =\int _{ u=1 }^{ u=e }{ \frac { { e }^{ u }du }{ 1+{ e }^{ 4u } } } \\ =\int _{ 1 }^{ e }{ \frac { { e }^{ -u }du }{ { e }^{ -2u }+{ e }^{ 2u } } }$

Hmm...

Paradoxica · Jan 31, 2016

Drsoccerball said:
$$ Using $ : \int{\frac{\sin h(x)}{\cos h(2x)}}dx $ and $ \int{\frac{\cos h(x)}{\cos h(2x)}}dx$

$$ Find: $ \int_0^1{\frac{1}{1+x^4}}dx$

This approach is extremely unwieldy. I do not recommend going about the integral in this manner.

Paradoxica · Jan 31, 2016

Paradoxica said:
$$\noindent $ \\ 2I = \int \frac{1+t^2+1-t^2}{1+t^4}$d$t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t\\2I = \int \frac{\text{d}(t-\frac{1}{t})}{(t-\frac{1}{t})^2+2} + \int \frac{\text{d}(t+\frac{1}{t})}{(t+\frac{1}{t})^2-2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+\frac{1}{2\sqrt{2}}\log\left(\frac{t^2 + \sqrt{2}t+1}{t^2-\sqrt{2}t+1}\right)$

Knew this would come in handy someday.

Drsoccerball · Jan 31, 2016

Paradoxica said:
Knew this would come in handy someday.

How does this prove the integral using info given ?

EDIT: ahhh okay after the substitution !

Paradoxica · Feb 1, 2016

Okay this is weird. Somehow I managed to prove the integral equals $\frac{\pi}{2\sqrt{2}}$ . This is not right. Must find flaw in logic.

leehuan · Feb 1, 2016

$\\ $To write $ \sinh^{-1} {x}$ in it's logarithmic form would my assumption to use inverse functions and quadratics to derive the result be correct?$

Also is there no such LaTeX for arcsinh(x)

Paradoxica · Feb 1, 2016

Yes, the logarithmic forms of inverse trigonometric functions are identical and valid for real numbers.

$\sinh ^{-1}{x}$

drsabz101 · Feb 1, 2016

What is a cardinal number. ?

InteGrand · Feb 1, 2016

drsabz101 said:
What is a cardinal number. ?

From Google, it means:

''noun
a number denoting quantity (one, two, three, etc.), as opposed to an ordinal number (first, second, third, etc.).''

(Further reading: http://www.mathsisfun.com/definitions/cardinal-number.html)

drsabz101 · Feb 1, 2016

How do I express a fraction eg. 5/8 as a recurring decimal without using a calculator

Drsoccerball · Feb 1, 2016

drsabz101 said:
How do I express a fraction eg. 5/8 as a recurring decimal without using a calculator

$\frac{5}{8} $ Isn't a recurring decimal though ? $$

drsabz101 · Feb 1, 2016

true sorry different example:

17/6

InteGrand · Feb 1, 2016

drsabz101 said:
true sorry different example:

17/6

$\noindent Use primary school long division. Doing this, we see the answer is $2.8333...$.$

si2136 · Feb 1, 2016

Let Alpha and Beta be the roots of lx^2 + mx + n = 0.

Evaluate "(lα + m) (lβ + m)" in terms of l,m and n.

Thanks.

InteGrand · Feb 1, 2016

si2136 said:
Let Alpha and Beta be the roots of lx^2 + mx + n = 0.

Evaluate "(lα + m) (lβ + m)" in terms of l,m and n.

Thanks.

$\noindent We will assume $l\neq 0$ (if $l=0$, the product is clearly $m^2$). We have$

$\begin{align*}(l\alpha + m)(l\beta + m) &= l^2 \alpha \beta + ml (\alpha + \beta) + m^2 \quad \text{(expanding)} \\ &= l^2 \cdot \frac{n}{l} + ml \cdot \left( -\frac{m}{l}\right) + m^2 \quad (\text{sum and product of roots})\\ &= nl - m^2 + m^2 \\ &= nl.\end{align*}$

leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

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