Deleted User's Maths Thread

[Deleted User]

How do I do this question thanks!

If lim_{x -> infinity} [sqrt(x^2+1) - sqrt(x^2-kx)] = 1, find k.

[psyxwar]

Ok so I tried a really iffy method, not sure if it's legit at all.

as x approaches infinity is just , so the square root of that is
, so , which is . Constants can be disregarded as x is approaching infinity, so , therefore .

What's wrong with this method exactly? I know there's a better way to do it.

Proper method courtesy of the boys at MHS: http://imgur.com/hTwBn0N (one correction though: I think k= -2 has to be rejected as )

[lzxnl]

How do I do this question thanks!

If lim_{x -> infinity} [sqrt(x^2+1) - sqrt(x^2-kx)] = 1, find k.

So let's take your limit expression sqrt(x^2+1) - sqrt(x^2-kx)
Rewrite as (sqrt(1+1/x^2) - sqrt(1-k/x))/(1/x)

We now have a scenario where as x => infinity, both top and bottom approach zero. Use L'Hopital's Rule.
Derivative of the top is 1/sqrt(x^2+1) -sqrt(x^2+1)/x^2-k/(2x*sqrt(x^2-kx)) using CAS as I really ceebs doing this one by hand.
Derivative of the bottom is just -1/x^2
Let's now clear the fraction.

Forgetting the negative sign for now, we get x^2/(sqrt(x^2+1) - sqrt(x^2+1)-k/2sqrt(1-k/x)

The last term simply becomes k/2 in the limit as x approaches infinity.
Let's try and simplify the first two terms. x^2/(sqrt(x^2+1) - sqrt(x^2+1) = 1/sqrt(x^2+1) * (x^2-(x^2+1)) = -1/(sqrtx^2+1)

In the limit as x approaches infinity, this drops off, so we're just left with -k/2. Now putting the negative sign back, we have the limit of your function = k/2 = 1. k=2

Admittedly I cheated using CAS to differentiate.

Edit: there's a much easier way of doing this.

sqrt(x^2+1) - sqrt(x^2-kx) = ((x^2+1) - (x^2-kx))/(sqrt(x^2+1)+sqrt(x^2-kx)) using DOTS
This simplifies to (1+kx)/(sqrt(x^2+1)+sqrt(x^2-kx)
Obviously the 1 is insignificant.
We have kx/(sqrt(x^2+1)+sqrt(x^2-kx))
Dividing by x, we get k/(sqrt(1+1/x^2)+sqrt(1-k/x))
Letting x approach infinity, we get k/(sqrt(1)+sqrt(1) = k/2
k/2 = 1 => k=2

So much simpler

[Deleted User]

Thanks guys!

[Deleted User]

How do I find the antiderivative of e^-2xsin(11x)? I know that I have to use the complex exponential but I can't quite work through the calculations till the end.

[Phy124]

How do I find the antiderivative of e^-2xsin(11x)? I know that I have to use the complex exponential but I can't quite work through the calculations till the end.

Spoiler

[Deleted User]

Thanks!

[b^3]

EDIT: The question seems to have disappeared.

HINT:

Spoiler

[Deleted User]

Haha yeah thanks. I deleted the comment cos I figured out how to do it.

[Deleted User]

How do I antidiff

sqrt(1 + 4x^2)

and

sqrt (4 - x^2)

Is there a formula/rule for these?

[lzxnl]

Assuming a>0


Integral becomes
=

Continue from here

[b^3]

For try making use of .

Spoiler

Then see if you can take it from there

[lzxnl]

For (1+4x^2)^1/2, try using the substitution x=1/2*sinh u
dx=1/2 cosh u du
(1+4x^2)^1/2=(1+sinh^2 u)^1/2 = cosh u
So you're integrating 1/2 cosh^2 u du
cosh^2 u=(1+cosh 2u)/2, the same as for trig functions
so your integral becomes u/2 + sinh(2u)/4 + c
x=1/2 sinh u, 2x = sinh u
u/2 = 1/2 arsinh (2x) = 1/2 ln(2x + sqrt(4x^2+1))
and sinh(2u)=2sinh u cosh u = 4x*sqrt(1+4x^2)^1/2
So your result should be 1/2 ln(2x+sqrt(4x^2+1))+x*sqrt(1+4x^2) + c or something like that

I did not mis-spell sin and cos, nor did I forget my differentiation rules. sinh u = (e^u-e^-u)/2, cosh u = (e^u+e^-u)/2
You can verify that the derivatives of sinh u is cosh u and vice versa, with no sign change.
All the regular trig identities with sin and cos work with sinh and cosh, except you have to flip the sign of any term which contains a multiple of 2, 6, 10, 14 etc sinh terms.
So cos 2u = 2cos^2 u - 1 = 1-2sin^2 u
but cosh 2u = 2cosh^2 u - 1 = 1+2sinh^2 u = cosh^2 u + sinh^2 u
and cos^2 x + sin^2 x = 1 but cosh^2 x - sinh^2 x = 1

These are hyperbolic functions; I personally haven't used them outside calculus, but they sure make some integrals a lot easier. You'd have to integrate sec^3 u by parts and then a little trick to be able to solve it, whereas hyperbolic functions make the original integral quite trivial.

[b^3]

All the regular trig identities with sin and cos work with sinh and cosh, except you have to flip the sign of any term which contains a multiple of 2, 6, 10, 14 etc sinh terms.
So cos 2u = 2cos^2 u - 1 = 1-2sin^2 u
but cosh 2u = 2cosh^2 u - 1 = 1+2sinh^2 u = cosh^2 u + sinh^2 u
and cos^2 x + sin^2 x = 1 but cosh^2 x - sinh^2 x = 1

Otherwise known as Osborn's rule Just to add a bit to this, it will affect as well, since you change the sign of any product (or implied product) of two sines. has the implied product of two sines. (as nlui1995 has said, it works out that for each even set of sinh you flip the sign).

EDIT: One source says 'Osborn's' while the other says 'Osborne's', hmmm

EDIT: Spelled sign as sine... fixed.

[lzxnl]

Otherwise known as Osborn's rule Just to add a bit to this, it will affect as well, since you change the sign of any product (or implied product) of two sines. has the implied product of two sines. (as nlui1995 has said, it works out that for each even set of sinh you flip the sign).

EDIT: One source says 'Osborn's' while the other says 'Osborne's', hmmm

EDIT: Spelled sign as sine... fixed.

I don't think it's every even set; for something which has sinh^4 x, no sign flip would occur; imagine it as sinh^2 x then squared; the sign flip is negated by the square.