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Apr
3 |
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revised |
how to delete two-dimentional double array added answer |
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Apr
3 |
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asked | how to delete two-dimentional double array |
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Apr
2 |
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accepted | C++ casting static two-dimensional double array to double** |
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Apr
2 |
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comment |
C++ casting static two-dimensional double array to double** That won't work in my case. Array may have size, for example, 100x100. And I won't write {m[0],m[1],m[2], ..., m[99]}; in my code :)
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Apr
2 |
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revised |
C++ casting static two-dimensional double array to double** modifing question |
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Apr
2 |
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asked | C++ casting static two-dimensional double array to double** |
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Jan
10 |
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comment |
IntelliSense: no operator "+" matches these operands Can't you just replace + with << ?
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Jan
6 |
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accepted | inserting Silverlight app into Symfony 1.4 layout |
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Jan
5 |
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comment |
inserting Silverlight app into Symfony 1.4 layout @1ed - thanks, that was the issue! What a dumb question I asked :). Could you post that as a full answer? I'll accept that and topic will be closed. |
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Jan
5 |
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asked | inserting Silverlight app into Symfony 1.4 layout |
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Dec
4 |
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comment |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ $ 2^1 + 2^2 + ... + 2^n = \sum\limits_{i=1}^{n}(2^i) = \sum\limits_{i=0}^{n}(2^i) - 1 = (2^{n+1} -1) - 1 = 2^{n+1}-2$. And after putting this into final formula, I have $S-2S = 2^{n+1}-2 - n2^n$, so finally $ S = n2^n - 2^{n+1} + 2 = 2^n(n-2) + 2$ |
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Dec
4 |
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awarded | Supporter |
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Dec
4 |
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accepted | Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ |
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Dec
4 |
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comment |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ I still think that there is a small mistake in your calculations (I have $(2^1 + 2^2 + ... + 2^{n-1} + 2^n) - n*2^n$), but the final answer is the same. Thanks a lot for your efford! |
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Dec
4 |
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comment |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ @anorton - okay, I'll remember about that in the future. It's part of algorithmic exercise, so $i$ stands for iterator, like in C++. |
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Dec
4 |
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comment |
Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ Did I miss something, or you forgot about $-2^n$ from $(n-1)*2^n$ multiplication? |
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Dec
4 |
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asked | Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$ |
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Dec
2 |
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accepted | Linux: permission to read file from /sys/kernel/debug wihout root privileges |
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Nov
24 |
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comment |
Linux: permission to read file from /sys/kernel/debug wihout root privileges I'd love to, but I don't have 15 rep :( |
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Nov
24 |
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awarded | Scholar |
