/sci/ - Science & Math

[Anonymous No.5916029]

I just...
I can't...
I've been able to blow past through most of the maths this sem, but I just can't wrap my head around this one.

How the hell am I supposed to do this MacLauren and Taylor thing? I understand the Fourier series, but not this.

Sorry I don't have any other example question.

[Anonymous No.5916035]

Here's the working that I did.
I understand that you're supposed to simplify it from here, but after that, what are you supposed to do to end up with an integrat-able form?
Also,
<span class="math">testing[/spoiler]

[Anonymous No.5916038]

Here's the working that I did.
I understand that you're supposed to simplify it from here, but after that, what are you supposed to do to end up with an integrat-able form?
Also,
testing

[Anonymous No.5916055]

>>5916029
<span class="math">e^u=\sum_{k\geq 0} \frac{u^k}{k!}[/spoiler] and here <span class="math">u=-x^2[/spoiler] so <span class="math">e^{-x^2}=1-x^2+\frac{x^4}{2}-\frac{x^6}{3!}-\ldots[/spoiler], now prooving the first equality should be easy ...

to integrate just replace <span class="math">\frac{1-e^{-x^2}}{x^2}[/spoiler] by the infinite sum and commute the sum and the integral ( you have to justify you can do that ).

[Anonymous No.5916068]

>>5916055
Hey, thanks for helping out.

Your math tags are a little broken, though.

By the way, I TOTALLY didn't know you can susbtitution integrate that.

Also, how exactly do I replace it with the infinite sum? Sorry for being such a mathbabby. Not used to dealing with infinites.

[Anonymous No.5916088]

Pic related

[Anonymous No.5916092]

>>5916088
>because integration is linear
This is not enough.

[Anonymous No.5916093]

>>5916088
Damn dude. Thanks a lot.
I'll go through this right away. Not going to understand immediately, so I'll let it sink in first.

[Anonymous No.5916099]

>>5916068
>Your math tags are a little broken, though.
what do you mean ?

there exists theorems on how you can intervet the infinite sum and the integral but if you've never seen it it's gonna be a little more tricky...

but in this case you have

<span class="math">\int_0^1 \frac{1-e^{-x^2}}{x^2} dx= \int_0^1 1-x^2+\frac{x^4}{2}-\frac{ x^6}{3!}-\ldots dx = \int_0^1 1 dx -\int_0^1 x^2 dx+\int_0^1 x^4/2dx-\ldots[/spoiler]

but i'm not sure yet about how you should proov those equalities...

[Anonymous No.5916101]

>>5916092
How/why not? In the reformulated sum (which converges, I guess you might have to clarify that) there is no singularity in the path of integration, so you can just go ahead and take the integral, no?

[Anonymous No.5916106]

>>5916101
>>5916088 here, ignore that "0 when n=0" bit, idk what I was thinking there.

[Anonymous No.5916107]

>>5916099
Wait...
Then what the hell kind of final answer am I supposed to get?

The tags seem kinda broken, but that's probably because of my shitty 1336x768 res.

[Anonymous No.5916112]

>>5916107
Your final answer will be an infinite sum as far as I can see. It's probably possible to interpret it somehow, but I don't think that's the point of the exercise.

[Anonymous No.5916120]

>>5916112
I'm in my first year, dude. Also, I'm just a chem engineer babby.
My final answer is probably going to have a value. The only thing I've been "proving" this sem is why structures often fail with a crack that's 45 degree to the surface.

[Anonymous No.5916133]

>>5916107
probably something like 1-1/3+1/10-1/42+... ( an infinite sum )

>>5916101
linearity only work with a finite number of terms

[Anonymous No.5916135]

>>5916088
>>5916106
Sorry man, but I'm still not understanding this.
Particularly this part.
Did you just divide the whole sum with an [mathx^2[/math]?

[Anonymous No.5916139]

>>5916120
I'm in my first year as well, EE to be precise. Just because the result is an infinite series doesn't mean it's wrong, and you won't have to prove shit for that. It'll definitely be exact though, which is what the question asked for.

[Anonymous No.5916144]

>>5916135
Yes, you're allowed to do that because division is linear. (1+2+3+4....)/5 = (1/5+2/5+3/5...) and x is not the variable of summation (which would be n in this case) so you're allowed to multiply and divide by it as you please.

[Anonymous No.5916153]

>>5916133
So what do you use to justify changing the sum and integral operators? That the sum converges? We never covered this in university, we always just used it...

[Anonymous No.5916169]

>>5916144
Okay...
One more thing I noticed.
Did you just take out the negative from the x?

Is that...legal?

[Anonymous No.5916174]

>>5916153
in this case, it's because you're dealing with a converging power series: the antiderivative exists on the same domain, as well as the derivative. (and by extension, derivatives and antiderivatives to any order)

[Anonymous No.5916175]

>>5916169
just think about it a little more before asking brah

try on an example

[Anonymous No.5916177]

>>5916175
Sorry about that. Just woke up, and had to ask this soon as possible. Got a fluids presentation in a few hours, so I can't really ask later.

[Anonymous No.5916181]

>>5916153
<div class="math">\int_0^1 \sum_{n=1}^\infty f_n(x) \mathrm{d}x = \int_0^1 \lim_{N \to \infty} \sum_{n=1}^N f_n(x) \mathrm{d}x = \lim_{N \to \infty} \int_0^1\sum_{n=1}^N f_n(x) \mathrm{d}x = \lim_{N \to \infty} \sum_{n=1}^N \int_0^1 f_n(x) \mathrm{d}x = \sum_{n=1}^\infty \int_0^1 f_n(x) \mathrm{d}x</div>

To swap the integral with the sum you just need to be able to pull a limit out of the integral. Lebesgue's Dominated Convergence Theorem tells you under which conditions you can do it.

http://en.wikipedia.org/wiki/Dominated_convergence_theorem

[Anonymous No.5916182]

>>5916177
well yeah it's legal, you just multiply -x^2 n times with itself.
And as for an integer n, (a*b)^n = a^n * b^n, you can apply this with a=-1 and b=x^2

[Anonymous No.5916184]

>>5916181
Thanks, very interesting! My engineering analysis classes don't cover such details, unfortunately.

[Anonymous No.5916190]

>>5916175
>>5916182

Wait...
<span class="math">-(x^2) =/= (-x^2)[/spoiler]

This is why I thought it was illegal, anyway.

For me, at least, working with double powers is quite new, so I might not be grasping something essential here...

[Anonymous No.5916193]

>>5916190
What exactly are you confused about? -x^2=(-1)*x^2, so (-x^2)^n=(-1)^n*(x^2)^n=(-1)^n*x^(2n)

[Anonymous No.5916204]

>>5916193
Uh, it would help if you used the math tags. It's a little dense there, for a babby like myself.

<span class="math">-x^2=(-1)*x^2, so (-x^2)^n=(-1)^n*(x^2)^n=(-1)^n*x^(2n)[/spoiler]

But isn't the first part supposed to be:
<span class="math">-x^2=(-1)^2(x^2)[/spoiler]

[Anonymous No.5916207]

>>5916204
Wait, now I'm just confusing myself.
Fuck that, fuck that.
I'm not going to delete that, but I'll keep looking at your equation.

[Anonymous No.5916210]

>>5916207
>>5916204
Okay, so my qualm is here:
<span class="math">(-x^2)^n=(-1)^n*(x^(2n))[/spoiler]

It seemed like you pulled the (-1)^n out of nowhere.

[Anonymous No.5916211]

>>5916204
so if I write -x^2, it's always a positive number?

Boy you're in trouble

-x^2 is just as you'd read it: minus the square of x.

[Anonymous No.5916213]

>>5916211
and I say minus the square of x, not (minus x) squared

[Anonymous No.5916217]

>>5916211
But...
It doesn't make sense.
Without going into complex numbers, does that mean the square of anything negative can STILL be a negative?

[Anonymous No.5916218]

>>5916217
the square doesn't encompass the minus sign.
Otherwise you'd need parenthesis.

am I being trolled?

[Anonymous No.5916223]

>>5916217
>>5916213
>>5916211
>>5916218
Goddamnit, no. Not trolling. Just a little fucked in the head.

It's like this:
<span class="math"> -x^2=(-1)*x^2[/spoiler]
Using x = 2,
<span class="math"> -2^2=(-1)*2^2[/spoiler]
<span class="math"> -4 = -4[/spoiler]

This is the case?

Then fuck, what am I confusing myself about.

Okay, okay. I get it now. Just a little case of brainfart.

After you get the final summy-sigma thing, you integrate the values within n=0,1 only, or...?

[Anonymous No.5916227]

>>5916204
Exponents have a higher precedence than negative signs; -x^2 will always be negative, (-x)^2 will always be positive.
So -x^2 = (-1)(x^2)

[Anonymous No.5916234]

>>5916227
>-x^2 will always be negative, (-x)^2 will always be positive
Unless x is negative, in which case the roles are reversed.

[Anonymous No.5916240]

OP here.
I have to leave for uni soon, so I gotta go now.

Thanks a lot to all my niggas.

[Anonymous No.5916255]

>>5916223
No, you take the SUM of all the definite integrals from 0 to 1 of the term within the sigma thingy. You probably won't be able to evaluate it and get a proper number as a result, but you'll have an infinite sum describing the result accurately.

[Anonymous No.5916337]

>>5916029
Here's a step by step, with the easy terms first.

1) Evaluate the Maclaurin series with e^x. Give the series in sigma notation. This will help with step 2.
2) Evaluate the Maclaurin series with e^x^2 using the Sigma notation from step 1. Keep this in the form of the actual written series. (don't use sigma notation)
3) Subtract the series from step 2 from the value 1. This will give you the actual numerator of the original problem. Keep the answer in the form of a series. (Don't use sigma notation)
4) Take the written series from step 3 and divide each term by the denominator, x^2. You have now proved the series.
5) Take the series written in step 4 and convert to sigma notation. Solve.

And you're done!

[Anonymous No.5916339]

>>5916337
Ah dang, small correction here.

For step 2, it's evaluate the Maclaurin series with e^-x^2. I forgot the negative sign in the exponent.

[Anonymous No.5916355]

>>5916088
That is an awesome pen man.
What kind is it?