Converting Sigma notation....


by Pranav-Arora
Tags: converting, notation, sigma
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#19
Jun29-11, 01:37 AM
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Quote Quote by Pranav-Arora View Post
I like Serena said that define a function of x and integrate it but i still don't get how he got x2k?
I defined an arbitrary function s(x) that looks a bit like your problem with the special property that if you substitute x=1, it is identical to your problem.

It's a trick to solve your problem.

The choice of the power 2k was inspired so that the factor (2k+1) would disappear during integration.
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#20
Jun29-11, 01:44 AM
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Quote Quote by I like Serena View Post
I defined an arbitrary function s(x) that looks a bit like your problem with the special property that if you substitute x=1, it is identical to your problem.

It's a trick to solve your problem.

The choice of the power 2k was inspired so that the factor (2k+1) would disappear during integration.
Ok i got it!!
But how i would integrate the expression. It involves sigma notation and i have never done integration of any expression which involves sigma notation.
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#21
Jun29-11, 01:57 AM
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Quote Quote by Pranav-Arora View Post
Ok i got it!!
But how i would integrate the expression. It involves sigma notation and i have never done integration of any expression which involves sigma notation.
So write out the terms of the summation, do the integration, and combine the resulting terms back into sigma notation.
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#22
Jun29-11, 02:08 AM
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Quote Quote by I like Serena View Post
So write out the terms of the summation, do the integration, and combine the resulting terms back into sigma notation.
I did as you said. After integrating, i got
[tex]x+nx^3+\frac{n(n-1)x^5}{2!}+\frac{n(n-1)(n-2)x^7}{3!}........x^{2n+1}[/tex]

Now what should i do next?
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#23
Jun29-11, 02:15 AM
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To be frank, I have never answered such questions myself either, but I took I like Serena's advice and it worked out wonderfully.

We'll start again just to make things clear,

To solve [tex]\sum_{k=0}^{n}(2k+1)^{n}C_k[/tex]

we define a function

[tex]f(x)=\sum_{k=0}^{n}(2k+1){^n}C_kx^{2k}[/tex]

we make it x2k because the integral of that is [itex]\frac{x^{2k+1}}{2k+1}[/itex] and notice how that denominator will cancel with the (2k+1) factor in the original question. So we have,

[tex]\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx^{2k+1}[/tex]

And here is the tricky part, we need to convert the right side into a binomial expression using the formula

[tex]\sum_{k=0}^{n}{^n}C_ka^kb^{n-k}=(a+b)^n[/tex]

It is clear that the b is again missing, which it is just hidden as b=1, but we need to convert the x2k+1 in such a way that it is equivalent to ak.

Use your rules for indices to convert it in such a way.
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#24
Jun29-11, 02:20 AM
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What might have been easier for you is you could've split the summation into

[tex]2\sum k{^n}C_k+\sum {^n}C_k[/tex]

and then defined [tex]s(x)=2\sum k{^n}C_kx^{k+1}+\sum {^n}C_kx^{k+1}[/tex]
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#25
Jun29-11, 02:24 AM
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Quote Quote by Pranav-Arora View Post
I did as you said. After integrating, i got
[tex]x+nx^3+\frac{n(n-1)x^5}{2!}+\frac{n(n-1)(n-2)x^7}{3!}........x^{2n+1}[/tex]

Now what should i do next?
Hmm, what I intended was this:

[tex]\sum_{k=0}^n {^n}C_k (2k+1) x^{2k}
= {^n}C_0 + {^n}C_1 \cdot 3 x^2 + {^n}C_2 \cdot 5 x^4 + ... + {^n}C_k (2k + 1) x^{2k} + ... [/tex]

Integration would give:
[tex]{^n}C_0 \cdot x + {^n}C_1 \cdot x^3 + {^n}C_2 \cdot x^5 + ... + {^n}C_k x^{2k+1} + ... [/tex]

Converting back to sigma notation:
[tex]\sum_{k=0}^n {^n}C_k x^{2k+1} [/tex]
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#26
Jun29-11, 02:27 AM
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Quote Quote by Mentallic View Post
To be frank, I have never answered such questions myself either, but I took I like Serena's advice and it worked out wonderfully.

We'll start again just to make things clear,

To solve [tex]\sum_{k=0}^{n}(2k+1)^{n}C_k[/tex]

we define a function

[tex]f(x)=\sum_{k=0}^{n}(2k+1){^n}C_kx^{2k}[/tex]

we make it x2k because the integral of that is [itex]\frac{x^{2k+1}}{2k+1}[/itex] and notice how that denominator will cancel with the (2k+1) factor in the original question. So we have,

[tex]\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx^{2k+1}[/tex]

And here is the tricky part, we need to convert the right side into a binomial expression using the formula

[tex]\sum_{k=0}^{n}{^n}C_ka^kb^{n-k}=(a+b)^n[/tex]

It is clear that the b is again missing, which it is just hidden as b=1, but we need to convert the x2k+1 in such a way that it is equivalent to ak.

Use your rules for indices to convert it in such a way.
Would it be like this (x2k+1+1)n.
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#27
Jun29-11, 02:30 AM
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Quote Quote by I like Serena View Post
Hmm, what I intended was this:

[tex]\sum_{k=0}^n {^n}C_k (2k+1) x^{2k}
= {^n}C_0 + {^n}C_1 \cdot 3 x^2 + {^n}C_2 \cdot 5 x^4 + ... + {^n}C_k (2k + 1) x^{2k} + ... [/tex]

Integration would give:
[tex]{^n}C_0 \cdot x + {^n}C_1 \cdot x^3 + {^n}C_2 \cdot x^5 + ... + {^n}C_k x^{2k+1} + ... [/tex]

Converting back to sigma notation:
[tex]\sum_{k=0}^n {^n}C_k x^{2k+1} [/tex]
I did the same way.
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#28
Jun29-11, 03:07 AM
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Quote Quote by Pranav-Arora View Post
Would it be like this (x2k+1+1)n.
Noo...

[tex]\sum{^n}C_kx^k1^{n-k}=(x+1)^n[/tex]

and not [tex](x^k+1)^n[/tex]

Use the fact that

[tex]a^{b+1}=a\cdot a^b[/tex] and [tex]a^{2b}=\left(a^2\right)^b[/tex]
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#29
Jun29-11, 03:39 AM
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Quote Quote by Mentallic View Post
Noo...

[tex]\sum{^n}C_kx^k1^{n-k}=(x+1)^n[/tex]

and not [tex](x^k+1)^n[/tex]

Use the fact that

[tex]a^{b+1}=a\cdot a^b[/tex] and [tex]a^{2b}=\left(a^2\right)^b[/tex]
I tried it but got stuck again. I did it like this:-

[tex]x^{2k+1}=x^{2k}.x=(x^2)^k.x[/tex]

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#30
Jun29-11, 04:54 AM
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Quote Quote by Pranav-Arora View Post
I tried it but got stuck again. I did it like this:-

[tex]x^{2k+1}=x^{2k}.x=(x^2)^k.x[/tex]

Why did you get stuck? That's exactly what it should be!

So now we have

[tex]\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx\cdot \left(x^2\right)^k[/tex]

And since x is independent of k, it can move out the front of the summation, so we have

[tex]x\sum_{k=0}^{n}{^n}C_k\left(x^2\right)^k1^{n-k}[/tex]

And now apply the formula to convert it into a binomial. And since you need to find f(1), take the derivative of both sides to get the expression for f(x).
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#31
Jun29-11, 05:07 AM
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Quote Quote by Mentallic View Post
Why did you get stuck? That's exactly what it should be!

So now we have

[tex]\int{f(x)dx}=\sum_{k=0}^{n}{^n}C_kx\cdot \left(x^2\right)^k[/tex]

And since x is independent of k, it can move out the front of the summation, so we have

[tex]x\sum_{k=0}^{n}{^n}C_k\left(x^2\right)^k1^{n-k}[/tex]

And now apply the formula to convert it into a binomial. And since you need to find f(1), take the derivative of both sides to get the expression for f(x).
If i convert it into binomial, i get
[tex]x.(x^2+1)^n[/tex]

I substitute the value 1 and i get
[tex]2^n[/tex]

But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?
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#32
Jun29-11, 05:50 AM
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Quote Quote by Pranav-Arora View Post
If i convert it into binomial, i get
[tex]x.(x^2+1)^n[/tex]
Good!

Quote Quote by Pranav-Arora View Post
But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?
Yep!
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#33
Jun29-11, 07:52 AM
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Quote Quote by Pranav-Arora View Post
But then how i would find out the derivative??
Do i have to first take the derivative and substitute the value 1?
Yep, that's what I meant by

Quote Quote by Mentallic View Post
And since you need to find f(1), take the derivative of both sides to get the expression for f(x).


You're nearly there!
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#34
Jun29-11, 09:39 AM
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Thanks!!
I think that this time i am right.
I took the derivative and found it to be
[tex]2nx(x^2+1)^{n-1}[/tex]

Now i substituted the value 1 and i got:-
[tex]2n.2^{n-1}[/tex]

Right...?
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#35
Jun29-11, 10:59 AM
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No, not quite.
What you have is not the derivative of [itex]x.(x^2+1)^n[/itex].

You need to apply the so called product rule.
That is: (u v)' = u' v + u v'
And you have to apply the so called chain rule.
That is: (u(v))' = u'(v) v'

Are you familiar with those rules?
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#36
Jun29-11, 12:39 PM
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I am not familiar with the product rule but when i calculated the derivative on Wolfram, it was the same as i got?


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