Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 7 years ago.
Improve this question
I always have trouble with factorials. Can someone walk me through simplifying this expression?
(x+1)! - 1 + (x+1)(x+1)!
I'm trying to get it to equal to (x+2)! - 1.
It can be solved as :-
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)! + (x+1)(x+1)! - 1
= (x+1)!.{1+(x+1)} - 1
= (x+1)!.{x+2} - 1
= (x+2)! - 1. // since n!.(n+1) = (n+1)!
Hence proved.
Note that (x+1)! = (x+1)*x!
(x+1)! - 1 + (x+1)(x+1)!
= (x+1)!((x+1)+1) - 1
= (x+1)!(x+2) - 1
= (x+2)! - 1
Related
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 6 years ago.
Improve this question
I'm doing Tim Roughgarden's Algorithms course and he has a slide with an integer multiplication algorithm.
Whats the rule that makes 10(n/2)a * 10(n/2)c become 10(n)ac ?
What do you do when multiplying fractional exponents like that?
It's based on the First Index Law, where:
am * an = am + n
in your case, the powers add to give n/2 + n/2 = 2n/2 = n
Base is the same so you just add the power of 10 i.e., (n/2) + (n/2) = n. Then it's basic multiplication 10(n)ac= 10(n)ac.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 7 years ago.
Improve this question
i saw this picture over social network
put into equation that will be:
Given:
A=rabit
B=Dog
C=Cat
A+C=10
A+B=20
C+B=24
solve A+B+C = ?
some programming logic? what is your answer?
I do not see any programming logic in this, its just Math.
B + C = 24
-A-C = -20
Adding these two,
B-A = 4
Now taking the 1st equation,
A+B = 10
-A+B = 4
Adding both again
2B= 14
B = 7
Then, A = 3 ( from A+B = 10) and C = 17kgs (from B + C = 24)
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 8 years ago.
Improve this question
I cannot find a function which I can use.
It has to have diminishing returns and f(0)=1/2 and when x-->infinity f(x)-->1
Do any of you have a suggestion?
Thank you in advance!
2/pi atan(x + 1)
Simple. 2/pi for being a 1 on infinity and then solving equation: 2/pi atan(x) == 1/2 to get the offset: 1.
Wolfram:
f[x_] := 2/\[Pi] ArcTan[x + 1];
f[0] (* 1/2 *)
Limit[2/\[Pi] ArcTan[x - Tan[1/2] - 1], x -> \[Infinity]] (* 1 *)
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 9 years ago.
Improve this question
limit x-->3 2x^2 + 7x-15/x-3
What i Simplified
Step 1 : 2x^2 + 10x -3x -15 / x-3
Step 2 : 2x( x + 5)-3( x + 5)/x-3
Step 3 : (2x - 3)(x + 5)/x-3
but unable to move further.. im thinking either the question ix wrong or there ix some trick which im unable to understand
thanx in advance
as x -> 3, the numerator goes to 3 * 8 = 24 and the denominator goes to 0, so the limit goes to +infinity if you approach 3 from the right, and -infinity if you approach 3 from the left
since you didn't specify which direction, the limit does not exist.
try graphing it: https://www.desmos.com/calculator
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 9 years ago.
Improve this question
http://doyourmath.com/web-algebrator/#c=expand_algexpand&v5=2%5E(k%2B1)
Anyone can explain why does expand 2^(k + 1) equal to (2^k) + 1?
That's not actually possible. 2^(k+1) is always going to be an even number. 2^k + 1 is always going to be an odd number.
I think you mean
2^(k+1) = 2^k * 2^1 = 2^k * 2.
One way of looking at it is the associative property of multiplication:
(2 X 3) X 4 = 2 X (3 X 4)
No matter how you group the numbers, the outcome will always be equal. In this case we're dealing with exponents, which is a shorthand notation for multiplying a number by itself.
It is not!!!
2^(k+1) = 2^k * 2 which is greater than 2^k + 1
Instead (k+1)^2 expands to (k^2)+2k+1
http://doyourmath.com/web-algebrator/#c=expand_algexpand&v5=2%5E(k%2B1) has ERRORS!