Why can't I convert this sum into an integral??

[Radium84]

Hi! This is my first time posting. I haven't taken calculus for almost 20 years, but I'm an Aerospace Engineer (I just have a job that doesn't really use calculus all that much). I was presented with a non-work, whimsical math problem that I was trying (and failing) to take to the next level. The problem goes like this:

1 + 4 = 5
2 + 5 = 12
3 + 6 = 21
8 + 11 = ?

It was pretty easy to see, once I wrote down the numbers in between 3 and 8, that the answer was 96. That's fine, so I have an answer to this one problem. Then I thought it would be cool to be able to come up with the answer for ANY number, so I quickly wrote down the characteristic equation as (2x + 3). Finding the answer for any number, then, just becomes a simple sum:

A:

If N = 8, I can sum all of these up and get 96. But this would obviously be tedious for a large value of N (say, 100). This is where my memory gets murky. I think I can represent this as an integral:

B:

This definite integral would then be solved to be the following:

C:

And that's where I'm stumped, because for N = 8 this equation produces 84, not 96. I must be doing something wrong, but I'm not sure what! As it turns out, looking at this empirically I know the answer has to be:

D:

This equation (D) and my sum above (A) always produce the same result. But I'm not sure how they relate to one another, unless I'm seriously screwing up the integral (B).

My wife and I are really scratching our heads on this one. Anyone know what I'm doing wrong?

 

[Deleted member 4993]

Hi! This is my first time posting. I haven't taken calculus for almost 20 years, but I'm an Aerospace Engineer (I just have a job that doesn't really use calculus all that much). I was presented with a non-work, whimsical math problem that I was trying (and failing) to take to the next level. The problem goes like this:

1 + 4 = 5
2 + 5 = 12
3 + 6 = 21
8 + 11 = ?

It was pretty easy to see, once I wrote down the numbers in between 3 and 8, that the answer was 96. That's fine, so I have an answer to this one problem. Then I thought it would be cool to be able to come up with the answer for ANY number, so I quickly wrote down the characteristic equation as (2x + 3). Finding the answer for any number, then, just becomes a simple sum:

A: View attachment 28154

If N = 8, I can sum all of these up and get 96. But this would obviously be tedious for a large value of N (say, 100). This is where my memory gets murky. I think I can represent this as an integral:

B: View attachment 28155

This definite integral would then be solved to be the following:

C:View attachment 28156

And that's where I'm stumped, because for N = 8 this equation produces 84, not 96. I must be doing something wrong, but I'm not sure what! As it turns out, looking at this empirically I know the answer has to be:

D:View attachment 28157

This equation (D) and my sum above (A) always produce the same result. But I'm not sure how they relate to one another, unless I'm seriously screwing up the integral (B).

My wife and I are really scratching our heads on this one. Anyone know what I'm doing wrong?

Use sum of arithmetic sequence:

\(\displaystyle \sum_{i = 1}^{n}i \ = \ \frac{n * (n+1)}{2} \)

\(\displaystyle \sum_{i = 1}^{n}(2i + 3) \ = \ n * (n+1) + 3*n\)

..... continue

I would NOT try to use "integration" here. Integration is efficient for continuous functions. Here you are dealing with discrete sequence.

 

[lex]

As @Subhotosh Khan rightly says, this is not a job for integration.
If however you want to see what the integration was doing:

The shaded area is what you would be integrating:
[MATH]\int_0^N 2x+3 \,\,dx=N^2+3N[/MATH]but the real sum you want is the sum of the areas of the rectangles, and your integral is short by [MATH]N[/MATH] little triangles, each of area [MATH] \tfrac{1}{2}\times(2)\times(1)=1[/MATH]So the true sum you want is [MATH]\hspace1ex N^2+3N \hspace2ex +\hspace2ex N\times 1[/MATH][MATH]=N^2 +4N[/MATH]
You could do the same for
[MATH] \int_0^N ax+b \,\,dx \hspace4ex +\hspace4ex \tfrac{1}{2}\times a \times (1)\times N\\ \hspace20ex =\tfrac{a}{2}N^2+bN+\tfrac{a}{2}N\\ \hspace20ex =\tfrac{a}{2}N^2+\left(\tfrac{a}{2}+b\right)N [/MATH]

 

[lookagain]

Radium84, there is an integral for the sum:

\(\displaystyle \displaystyle\int_{0.5}^{N+0.5} \bigg(2x + 3\bigg)dx \)

After taking the antiderivative and substituting:

\(\displaystyle [(N + 0.5)^2 + 3(N + 0.5)] - [0.5^2 + 3(0.5)] \)

When N = 8, this expression equals 96.

 

[lex]

@lookagain
Very neat

 

[Radium84]

As @Subhotosh Khan rightly says, this is not a job for integration.
If however you want to see what the integration was doing:
View attachment 28159
The shaded area is what you would be integrating:
[MATH]\int_0^N 2x+3 \,\,dx=N^2+3N[/MATH]but the real sum you want is the sum of the areas of the rectangles, and your integral is short by [MATH]N[/MATH] little triangles, each of area [MATH] \tfrac{1}{2}\times(2)\times(1)=1[/MATH]So the true sum you want is [MATH]\hspace1ex N^2+3N \hspace2ex +\hspace2ex N\times 1[/MATH][MATH]=N^2 +4N[/MATH]
You could do the same for
[MATH] \int_0^N ax+b \,\,dx \hspace4ex +\hspace4ex \tfrac{1}{2}\times a \times (1)\times N\\ \hspace20ex =\tfrac{a}{2}N^2+bN+\tfrac{a}{2}N\\ \hspace20ex =\tfrac{a}{2}N^2+\left(\tfrac{a}{2}+b\right)N [/MATH]

Thanks for taking the time to really explain this! I had some similar rectangles on the scratch paper I was using with my wife, but couldn't really put two and two together.

 

[Radium84]

Radium84, there is an integral for the sum:

\(\displaystyle \displaystyle\int_{0.5}^{N+0.5} \bigg(2x + 3\bigg)dx \)

After taking the antiderivative and substituting:

\(\displaystyle [(N + 0.5)^2 + 3(N + 0.5)] - [0.5^2 + 3(0.5)] \)

When N = 8, this expression equals 96.

Okay, yep - this makes sense to me I think. This basically constructs the rectangles so that each one intersects with the line at the midpoint of the rectangle, rather than the rightmost or leftmost point of the rectangle (right?)

 

[lex]

@Radium84
You seem to have a good understanding of this. Yes, just move my rectangles over a half.
(For the integral, the shaded area should then start at [MATH]\tfrac{1}{2}[/MATH] and end at [MATH]N+\tfrac{1}{2}[/MATH], rather than 0 and N).

 

[Cubist]

Red: [MATH]y=2\lceil x \rceil+3\,[/MATH], blue: [MATH]y=2(x+0.5)+3[/MATH]
The area under the blue line is [MATH]\int_0^82(x+0.5)+3\,dx=\int_{0.5}^{8.5}2u+3\,du\,[/MATH] after the substitution \(u=x+0.5\)

 

[Steven G]

Why are we using integrals at all?

Define + as follows: a+b = a + ab
Now
1 + 4 = 1 + 1*4 = 5
2 + 5 = 2 + 2*5 = 12
3 + 6 = 3 + 3*6 = 21
8 + 11 = 8 + 8*11 = 96
...
100 + 103 = 100 + 100*103 = 10400

 

[Steven G]

1 + 4 = 1 + 1*4 = 1*5 = 5
2 + 5 = 2 + 2*5 = 2*6 = 12
3 + 6 = 3 + 3*6 = 3*7 = 21
8 + 11 = 8 + 8*11 = 8*12 = 96

So a + b = a*(b+1)