ReIndexing a Series(non-infinite)

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Homework Statement



reindex.jpg


Homework Equations


The Attempt at a Solution


#7 only
I'm stumped, i have no idea as to what to do.

I can't even find help on the internet/book. Its almost like this topic doesn't even exist.

I know both upper and lower limits have been decreased by 2
 
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jack612blue said:

Homework Statement



Homework Equations


The Attempt at a Solution


#7 only
I'm stumped, i have no idea as to what to do.

I can't even find help on the internet/book. Its almost like this topic doesn't even exist.

I know both upper and lower limits have been decreased by 2

Ok, then also change k to k+2. Doesn't that make sense? It would offset the other change.
 
Last edited:
Dick said:
Ok, then also change k to k+2. Doesn't that make sense? It would offset the other change.

Hello,

Thank you for responding.

I understand that part

6
sigma
1

I'm just confused as to what to put in the inside.

I found this on wikipedia. is this relevant?

e3114b84602343dddffd631e9abd9047.png
 
Define a new variable, say, j. How should j and k be related so that j goes from 1 to 6 when k goes from 3 to 8?
 
jack612blue said:
I found this on Wikipedia. Is this relevant?

e3114b84602343dddffd631e9abd9047.png
Yes, it's relevant .
 
It's really just an algebraic substitution. The series, as given, starts with k= 3 and we want to change that to a series starting at 1. Rather than use "k" to mean two different things, I am going to call this second index "i". That is, we want i= 1 to correspond to k= 3. That is the same as saying k- i= 3- 1= 2 so that k= i+ 2 or i= k- 2.

When k= 8, i= 8- 2= 6 so the series goes from i= 1 to 6 as desired. And in the formula for the terms of the series, since, as above, k=i+ 2, replace each k with i+ 2. d What do you get when you replace "k" in "2k- 1" with "i+2"?

You will get a series in i. Since the "index" is a dummy, and has no meaning in the total sum, you can, to completely match what is required, simply replace "i" with "k" again.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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