Mastering Series Problems: Tips and Tricks for Solving Tricky Questions

[Physics lover]

Homework Statement

Let a=$\frac x {1+x^2}$ + $\frac {x^3} {3(1+x^2)^3}$ +...
and b=$x-\frac {2x^3} 5 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$+...
then find relation between a and b.
a)a=2b b)a=b c)2a=b d)2a=3b

Relevant Equations

None

a

 

[Physics lover]

Homework Statement:: Let a=x/(1+x^2)+x^3/{3(1+x^2)^3}+...
and b=x-2x^3/5+x^5/5+x^7/7-2x^9/9+...
then find relation between a and b.
Relevant Equations:: None

a

I forget to post my attempt so here it is-:
I tried hours to find a pattern but I could not find it.I also wrote the rth term of 'a' and then tried to sum it but of no use.Also I am not able to find rth term of 'b'.Please help me

 

[Delta2]

I think the problem is not well defined. I am able to see the $n-th$ term of a which should be $\frac{x^{2n+1}}{(2n+1)(1+x^2)^{2n+1}}$ but I am not able to see the $n-th$ term of b .

 

[FactChecker]

I think the problem is not well defined. I am able to see the $n-th$ term of a which should be $\frac{x^{2n+1}}{(2n+1)(1+x^2)^{2n+1}}$ but I am not able to see the $n-th$ term of b .

And even your n-th term of a might be the best guess, but it is still a guess. The OP needs to be more definitive.

 

[Physics lover]

I think the problem is not well defined. I am able to see the $n-th$ term of a which should be $\frac{x^{2n+1}}{(2n+1)(1+x^2)^{2n+1}}$ but I am not able to see the $n-th$ term of b .

Sorry i was in a hurry that time so i forgot to right the third term in a.By the way You are right the third term is $\frac {x^5} {5(1+x^2)^5},$

 

[Physics lover]

I think the problem is not well defined. I am able to see the $n-th$ term of a which should be $\frac{x^{2n+1}}{(2n+1)(1+x^2)^{2n+1}}$ but I am not able to see the $n-th$ term of b .

I too am not able to see the rth term of b that's the main problem occurring.

 

[FactChecker]

I too am not able to see the rth term of b that's the main problem occurring.

Then I think you should write down more terms of b so someone can help you to determine a formula.

 

[lurflurf]

Homework Statement:: Let a=$\frac x {1+x^2}$ + $\frac {x^3} {3(1+x^2)^3}$ +...
and b=$x-\frac {2x^3} 5 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$+...
then find relation between a and b.
Relevant Equations:: None

a

looks like a typo should be

b=$x-\frac {2x^3} 3 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$+...
then the pattern is obvious
$$a=\sum_{k=1}^\infty \frac{1}{2k-1}\left(\frac{x}{1+x^2}\right)^{2k-1}$$
$$b=\sum_{k=1}^\infty \left(\frac{1}{2k-1}x^{2k-1}-\frac{1}{2k-1}x^{6k-3}\right)$$

now to compare them you need to write both sums in the same form so try to do that

ie either expand a in x or b in x/(1+x^2)

 

[Physics lover]

Then I think you should write down more terms of b so someone can help you to determine a formula.

I cannot produce them on my own.I just posted the same question I cane across.This was the only data given in the question.

 

[Physics lover]

looks like a typo should be

b=$x-\frac {2x^3} 3 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$+...
then the pattern is obvious
$$a=\sum_{k=1}^\infty \frac{1}{2k-1}\left(\frac{x}{1+x^2}\right)^{2k-1}$$
$$b=\sum_{k=1}^\infty \left(\frac{1}{2k-1}x^{2k-1}-\frac{1}{2k-1}x^{6k-3}\right)$$

now to compare them you need to write both sums in the same form so try to do that

ie either expand a in x or b in x/(1+x^2)

I think I wrote the same thing in b.

 

[FactChecker]

looks like a typo should be

b=$x-\frac {2x^3} 3 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$+...
then the pattern is obvious

You mean that a possible pattern is obvious. I would like to see a little more evidence.

 

[Physics lover]

Edit:I have provided the options in the problem.

 

[lurflurf]

What class are you in? Did you learn geometric series yet?
try using

$$\frac{1}{1+x^2}=\sum_{k=0}^\infty (-x)^{2k}$$​

try expanding​

$$\left(\frac{x}{1+x^2}\right)+\frac{1}{3}\left(\frac{x}{1+x^2}\right)^3+\frac{1}{5}\left(\frac{x}{1+x^2}\right)^5$$​

as​

$$x(1-x^2+x^4)+\frac{1}{3}x^3(1-x^2)^3+\frac{1}{5}x^5(1-x^2)^5$$​

discard powers of x bigger than 5​

 

[Physics lover]

What class are you in? Did you learn geometric series yet?
try using

$$\frac{1}{1+x^2}=\sum_{k=0}^\infty (-x)^{2k}$$​

try expanding​

$$\left(\frac{x}{1+x^2}\right)+\frac{1}{3}\left(\frac{x}{1+x^2}\right)^3+\frac{1}{5}\left(\frac{x}{1+x^2}\right)^5$$​

as​

$$x(1-x^2+x^4)+\frac{1}{3}x^3(1-x^2)^3+\frac{1}{5}x^5(1-x^2)^5$$​

discard powers of x bigger than 5​

yes i have learned that
I am in 12th standard.

 

[SammyS]

I think I wrote the same thing in b.

No, @lurflurf is correct

For Series b you had $\displaystyle x-\frac {2x^3} 5 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$.

The typo is in the second term.

 

[Physics lover]

No, @lurflurf is correct

For Series b you had $\displaystyle x-\frac {2x^3} 5 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$.

The typo is in the second term.

ok.
By the way the good news is that I have solved the problem.Though I made a different approach and was not sure that it would work or not.But at last it worked!

 

[SammyS]

ok.
By the way the good news is that I have solved the problem.Though I made a different approach and was not sure that it would work or not.But at last it worked!

Thanks for letting us know that you solved the problem. Homework Helpers here at PF are volunteers. One of our few rewards is knowing that you have been helped, so yes, it's good that you indicted that you have a solution.

What approach did you use?
Answering that will bring the thread to a nice conclusion and help others to learn from this exercise.

 

[Physics lover]

Thanks for letting us know that you solved the problem. Homework Helpers here at PF are volunteers. One of our few rewards is knowing that you have been helped, so yes, it's good that you indicted that you have a solution.

What approach did you use?
Answering that will bring the thread to a nice conclusion and help others to learn from this exercise.

ok so what I did was
I just differentiated a and b and found a relation between a' and b'.Since there is no constant term in the expression of both a and b,so I can say that the same relation applies to a and b also.And the problem was solved.
By the way if anybody find finds another method without using differentiation,please do post here.

 

[Physics lover]

No, @lurflurf is correct

For Series b you had $\displaystyle x-\frac {2x^3} 5 + \frac {x^5} 5 + \frac{x^7} 7- \frac {2x^9} 9$.

The typo is in the second term.

Yes you were correct.There was a misprint in my book.I took the second term as $-\frac {2{x^3}} 3$ and the problem was solved.