Solving Summation Question with Floor & Ceiling Functions

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In summary, the conversation discusses an easier way to solve a problem involving floor and ceiling functions using integration, but it is not applicable in this case as it is a summation problem. The conversation also mentions using graphing to find the area under the curve, and shows two different methods of solving the problem using summation. The final answer is 39.75.
  • #1
mohabitar
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Not exactly sure how they went from the first step to the 2nd step? Is there an easier way to solve this?

(keep in mind we're dealing with floor and ceiling functions)
 
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  • #2
There is an easier way to do the problem. You can use what's called integration but that is a calculus topic.

One way you can solve it is by graphing.

You can draw the graph of the function (which is linear) and find the area under the curve. When you calculate the area under the curve (which is a triangle) you only take into account the area from 0 to 30. Also, if any piece of the graph from 0 to 30 is below the x-axis then you have to subtract that portion from the area that is above the x axis.

If you have a graphing calculator I can show you an even simpler way.
 
  • #3
I know calculus, I just posted it in here because I thought its a precalc question. How would I use integrals to solve this?
 
  • #4
How much calculus do you know?
 
  • #5
Lol I'm in Calc II now, will that be enough?
 
  • #6
Oh wait...I have to apologize for being retarded. You can't solve this with integration because it's a summation (it's been awhile). Unfortunately, you have to add all those #'s together because a summation is not the exact answer but an estimation. An integral, on the other hand, is the exact answer.
 
  • #7
So the question now is, I see that they did it manually in step 1, where its just the sum of all those values, but step 2 I see some multiplication going on-where did that come from?
 
  • #8
I get: S[ (i/20 – 1/2) + (i/10 – 1/2 ) ] = S[ i/20 –1/2 + i/10 – 1/2 ] = S[ (3/20)i – 1 ] = 3/20*S -S[1] = 3*(30^2 + 30)/40 – 30 = 39.75

or

S[ (i/20 – 1/2) + (i/10 – 1/2 ) ] = S[ i/20 ] – S[1/2] + S[i/10] – S[1/2 ] = 1/20S – 1/2S[1] + 1/10S – 1/2S = 1/20*465 – 1/2*30 – 1/10*465 – 1/2*30 = 39.75
 

FAQ: Solving Summation Question with Floor & Ceiling Functions

What is a summation question with floor and ceiling functions?

A summation question with floor and ceiling functions is a mathematical problem that involves finding the sum of a series of numbers, where the terms are modified by floor (rounding down) and ceiling (rounding up) functions.

Why are floor and ceiling functions used in summation questions?

Floor and ceiling functions are used in summation questions to account for rounding and to ensure the sum is an integer. This is particularly useful in real-life applications where the terms being summed may represent discrete quantities, such as the number of items in a collection.

How do I approach solving a summation question with floor and ceiling functions?

To solve a summation question with floor and ceiling functions, you can first evaluate the sum without the functions. Then, make adjustments to the answer based on the floor and ceiling functions. Alternatively, you can break down the summation into smaller, simpler parts and apply the functions to each part.

What are some common mistakes to avoid when solving summation questions with floor and ceiling functions?

One common mistake is forgetting to apply the floor or ceiling function to each term in the summation. Another mistake is incorrectly applying the floor or ceiling function, such as rounding up when it should be rounding down. It is also important to fully understand the properties of floor and ceiling functions, such as their relationship to rounding, to avoid errors.

Can the use of floor and ceiling functions in summation questions result in a different answer?

Yes, the use of floor and ceiling functions can result in a different answer compared to not using them. This is because the functions modify the terms being summed, and thus can change the overall value of the sum. It is important to carefully consider when and how to use these functions in order to obtain the correct answer.

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