To the Mathematical Geniuses out there. (1 Viewer)

[Omiums Troll]

I was hoping for some help.

I just need to know if what I've done is mathematically legal.

I have "A" which is function of x given by :

A(x) = [Integral from 0 to infinity] cos(t^3 + x*t)) dt

t is an unknown.


Lets say i wish to find the value of x, at which A(x) = 0

Can i simply do this:

A(x) =0

0 = cos(t^3 + x*t)

hence x = something ugly
?

 

[madsam]

no, first you have to integrate it all as an indefinate integral, then sub in t = o, then solve using the limits to find A(0)

 

[Omiums Troll]

no, first you have to integrate it all as an indefinate integral, then sub in t = o to find A(0)

I don't want A(0)

I would like A(x) = 0

i.e. my function looks like this

A(x )= [Integral from 0 to infinity] cos(t^3 + x*t)) dt

I want

A(x) = 0

Thus i have

0 = [Integral from 0 to infinity] cos(t^3 + x*t)) dt

Can't i simply say "Differentiate both sides with respect to 't' "

and viola integral thingy is gone ?

Iruka or other maths prodigies, any thoughts ?

 

[Omiums Troll]

I will cry if this doesn't work.

 

[_santa]

I don't want A(0)

I would like A(x) = 0

i.e. my function looks like this

A(x )= [Integral from 0 to infinity] cos(t^3 + x*t)) dt

I want

A(x) = 0

Thus i have

0 = [Integral from 0 to infinity] cos(t^3 + x*t)) dt

Can't i simply say "Differentiate both sides with respect to 't' "

and viola integral thingy is gone ?

Iruka or other maths prodigies, any thoughts ?

I dont think you can, because you can only "Differentiate both sides with respect to 't'" when that int is indefinite, here it is (0 to infinity). So yeah you have to calculate the int.

 

[Omiums Troll]

I dont think you can, because you can only "Differentiate both sides with respect to 't'" when that int is indefinite, here it is (0 to infinity). So yeah you have to calculate the int.

http://www.usyd.edu.au/stuserv/documents/maths_learning_centre/definiteintegral.pdf

Page 9.

It is possible to differentiate a definite integral, however the infinity is worrying me.

 

[undalay]

Is this question possible ?

edit: like frm a textbook or something?

 

[undalay]

Okay got it (possibly?):

A(x) = [Integral from 0 to infinity] cos(t^3 + x*t)) dt

t^3 + x*t = (pi/2 + kpi) k e Z

x = ((pi/2+kpi) - t^3)/t

now t^3 + x*t is always zero.

So basically you're going to be finding the area under a curve that is always zero.
Which must be zero.

??

 

[Omiums Troll]

Is this question possible ?

edit: like frm a textbook or something?

Ok i will provide some background.

I'm doing a Quantum mechanics question, I need to solve the schroedinger equation for the potential,

I need to assume that the wavefunction can be solved by an Airy function.

To find the quantised heights i need to find the first 4 roots of the Airy function which is :

Airy function - Wikipedia, the free encyclopedia

 

[undalay]

Is my above solution correct?

I think it's the only way, since with any non zero function you're trying to integrate:

(For the above equation)
It's oscillating (parts will be positive, parts will be negative), and so a limit to infinity will inevitably lead to an oscillating value (and by that I mean no limit)

 

[Omiums Troll]

Okay got it (possibly?):

A(x) = [Integral from 0 to infinity] cos(t^3 + x*t)) dt

t^3 + x*t = (pi/2 + kpi) k e Z

x = ((pi/2+kpi) - t^3)/t

now t^3 + x*t is always zero.

So basically you're going to be finding the area under a curve that is always zero.
Which must be zero.

??

t^3 + x*t is not always zero.

There are infinitely many roots of which, i need to find the first "4"

Your solution is EXACTLY what i have yes.

But that's not really what I'm after, I still don't know if its legal to do what we've done.

Then again, this seems like the only way to do this problem so it must be legal.

 

[undalay]

t^3 + x*t is not always zero.

There are infinitely many roots of which, i need to find the first "4"

Your solution is EXACTLY what i have yes.

But that's not really what I'm after, I still don't know if its legal to do what we've done.

Then again, this seems like the only way to do this problem so it must be legal.

Oh yeah it is : )

You may need to email your tutor/lecturer they are always helpful.

 

[lolokay]

having x in terms of t doesn't really make sense does it lol? i would think x is some constant (i.e. can't vary with t)

http://mathworld.wolfram.com/AiryFunctionZeros.html

if it was solvable exactly it would say there wouldn't it?

 

[Iruka]

I assume that this is for a physics assigment? If so, a sufficiently accurate numerical approximation to the problem should suffice.

The Airy function is defined as the solution to the DE

y'' = xy.

That equation you have is an alternate representation of the Airy function. In this case, not necessarily the most helpful one.

You can use the method of Frobenius to find a series solution to this equation. (If you have done the second year maths subject about DEs you should know what I am talking about.) According to one of my maths books, for small x the series solution converges fairly quickly. This means that you can find an approximate solution by truncating the infinite series after a *reasonable* number of terms. You should probably use Maple or Matlab for this. Maple may already know about the Airy function - you can check the help menu.

Alternately, you can just cite the values given on the wolfram website that lolokay linked to.

 

[Omium]

Thanks guys. I figured it out.

One last question dumb math question.

I don't really understand the change of derivative maths rules.

i.e. we will be doing this mathematical problem such as: (THIS IS NOT THE EXACT QUESTION)

x = integral (r^2)/(sqrt(a^2+r^2) d/dr.

Then they suddenly say "Its simpler if we take d/d(r^2)"

Then they write something like

1/sqrt(a^2 + r^2 ) d/d(r^2) and it becomes much simpler to solve.

A large ammount of times with questions they "change the differentiation variable" from lets say "r" to 1/r or 1/r^2

Could someone just tell me what mathematics topic this is (probably something like "change of differentaion variable or similar")

Or explain to me the finer details of what's going on.

I've tried searching on the net for practise problem, but no luck.

 

[Uncle]

no, first you have to integrate it all as an indefinate integral, then sub in t = o, then solve using the limits to find A(0)

lol hsc 2009

Thanks guys. I figured it out.

One last question dumb math question.

I don't really understand the change of derivative maths rules.

i.e. we will be doing this mathematical problem such as: (THIS IS NOT THE EXACT QUESTION)

x = integral (r^2)/(sqrt(a^2+r^2) d/dr.

Then they suddenly say "Its simpler if we take d/d(r^2)"

Then they write something like

1/sqrt(a^2 + r^2 ) d/d(r^2) and it becomes much simpler to solve.

A large ammount of times with questions they "change the differentiation variable" from lets say "r" to 1/r or 1/r^2

Could someone just tell me what mathematics topic this is (probably something like "change of differentaion variable or similar")

Or explain to me the finer details of what's going on.

I've tried searching on the net for practise problem, but no luck.

that integral is a bit hard to read itself but,
ummmm...all i can think of is leibniz's rule:

give it a try:

from m.pahor's math2019 lecture

 

[LordPc]

I need help too, although this question is from 1st year maths, so its not purely restricted to "Mathematical Geniuses".

if f is differentiable at a, find

lim h->0 of {f(a+ph) - f(a-ph)} / h

 

[ianc]

I need help too, although this question is from 1st year maths, so its not purely restricted to "Mathematical Geniuses".

if f is differentiable at a, find

lim h->0 of {f(a+ph) - f(a-ph)} / h

sorry i haven't done maths for a while (so it might be wrong), but here goes....

also, im assuming p is just a real constant






hope this helps

 

[LordPc]

ahh, thx.

although why does { f(a-ph) - f(a) } / ph
become -f'(a)

is this some sort of result that I should know

 

[ianc]

i spose you could memorise the result, but i'll try to explain it:

well basically the derivative = gradient = rise/run

to work out the rise between a and a+h, you use f(a+h)-f(a).

to work out the rise between (a-h) and (a), the definition of rise is f(a)-f(a-h)

but in our equation previously, we had f(a-h)-f(a)
= - [f(a)-f(a-h)]
= - rise
= -f'(a)

it's a dodgy explanation, but does that help?