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Got it thanks

Taylor Series --> Maclaurin Expansion

Thanks , this is so interesting. I will explore.

I appreciate your inputs.

Thanks

Nakul

Hmm ok.

In which case would it be applicable?

Hi,

For problem no 6, I am taking a different method.

The general term for this is = n/ (n+1)^2.

Now to find the sum of the first 6 terms = integration this general term from 0 to 6.

What am i doing wrong?

Integration of

n/ (n+1)^2 = ln (n+1) + 1 / (n+1) + C

yes the lower one

**NakulG**- Replies: 5

Can anyone help solve this limit.

Thanks

Lim x-->0, [1 - (x^2)/2) - Cos ( x / (1 - x^2 ) ) ] / x^4

sinθ + cos θ = x and sinθcosθ=y

substituting we get

x^3 - 6.4x^2+x+8.4=0

Solve for x, so x = -1, 1.4 & 6

Hope this helps...

Sin2θ = -1, 0.96, 26

Thanks it worked :-)

ganesh wrote:

Hi;

622. Evaluate:

.

Hi, Can someone show me the working for the above question?

Thanks

Nakul

thanks

Where are these used? In Physics?

yes - silly mistake in my calculation :-)

X= (4a-1)b/5 and Y=(4b-a)/5b

Thanks Bob, I will look into it.

Also in this what is meant by -1, any thoughts.

**NakulG**- Replies: 7

:-)

http://qz.com/627467/china-just-announced-one-of-the-largest-single-layoffs-in-history/

In the question that I had mentioned, what is the significance of the xy term in the equation? Can you suggest.